Day By Day Notes for MATH 201

Fall 2009

 

Day 1

Activity:    Go over syllabus.  Take roll.  Overview example: Kristin Gilbert trial.

Goals:        Review course objectives: collect data, summarize information, and make inferences.

I have divided this course into three ÒunitsÓ.  Unit 1 (Days 2 through 14) is about summarizing data.  Unit 2 (Days 15 through 29) is about sampling and probability.  Unit 3 (Days 30 through 42) is about statistical inference.  Throughout the course, we will also focus on the abuses people often make of statistics and statistical methods.  Sometimes these abuses are unintentional.  My hope is that at the end of the course, you will have an appreciation for the ÒtricksÓ that people use to lie to you using numbers.

I use the Gilbert trial example because it demonstrates all three of the courseÕs main ideas in action.  The charts we see are examples of how to organize and summarize information that may be complicated by several variables or dimensions (Unit 1).  The argument about whether we would see results such as this one if there really were no relationship between the variables is an example of the probability we will study in Unit 2.  And the trial strategy itself is what statistical inference is all about (Unit 3).  Many of you will encounter inference when you read professional journals in your field and the researchers use statistics to support their conclusions of improved treatments or to estimate a particular proportion or average.

I believe to be successful in this course you must READ the text carefully, working many practice problems.  Our classroom activities will sometimes be unrelated to the homework you practice and/or turn in for the homework portion of your grade; instead they will be for understanding of the underlying principles.  For example, on Day 17 we will simulate Simple Random Sampling by taking 60 samples from each group in the class.  This is something you would never do in practice, but which I think will demonstrate several lessons for us.  In these notes, I will try to point out to you when weÕre doing something to gain understanding, and when weÕre doing something to gain skills.

Each semester, I am disappointed with the small number of students who come to me for help outside of class.  I suspect some of you are embarrassed to seek help, or you may feel I will think less of you for not Ògetting itÓ on your own.  Personally, I think that if you are struggling and cannot make sense of what we are doing, and donÕt seek help, you are cheating yourself out of your own education.  I am here to help you learn statistics.  Please ask questions when you have them; there is no such thing as a stupid question.  Often other students have the same questions but are also too shy to ask them in class.  If you are still too shy to ask questions in class, come to my office hours or make an appointment.  Incidentally, when I first took statistics, I didnÕt understand it all on my own either, and I too didnÕt go to the instructor for help.  I also didnÕt get as high a grade as I could have!

I believe you get out of something what you put into it.  Very rarely will someone fail a class by attending every day, doing all the assignments, and working many practice problems; typically people fail by not applying themselves enough - either through missing classes, or by not allocating enough time for the material.  Obviously I cannot tell you how much time to spend each week on this class; you must all find the right balance for you and your lifeÕs priorities.  One last piece of advice: donÕt procrastinate.  I believe statistics is learned best by daily exposure.  Cramming for exams may get you a passing grade, but you are only cheating yourself out of understanding and learning.

Reading:    To The Student, pages xxxi-xxxiv.  Bring to class on Day 2 a list of numbers (the source is not relevant, but knowing where the numbers came from and what they represent will help when you try to explain them).  Sample lists are on pages 26 to 30.

Day 2

Activity:    Discussion of variables and graphs.

Unit 1 is about summarization.  We will work on several types of summary:  graphical summaries of a list of numbers, graphical summaries of a list of characteristics, numerical summaries of a list of numbers, and the summaries for two variables.  Sometimes we will summarize situations with just a few numbers, called summary statistics.  Other times, we need more than that, such as a picture, or many summary statistics.  One of your tasks in this course is to gain enough experience to know what is enough summarization for a particular situation.

The first summaries we will work on are the Òone-variable graphical techniquesÓ.  If you are working with a list of characteristics (attribute data or categorical data), you will need to use a bar graph, or a frequency table, or a pie graph.  If you are working with a list of numbers, then you have many more options: frequency tables, histograms, quantile plots, stem plots, box plots (later), to name a few.  Your goal in Unit 1 is to become adept at producing these displays, either by hand or with the calculator, and to interpret the main features from either your own creations or someone elseÕs work.

In these notes, I will put the daily task in gray background.

Your task today is to communicate the important information from your list of numbers to the person next to you.  (Work in pairs or groups.)  Specifically, make and interpret a frequency table and a histogram.  In your description to your neighbor, keep in mind these terms: symmetry, skew, center, spread, mode, outlier.  On your calculator, make sure that you try different window settings for your histogram.  Remember, you are trying to describe your data to your neighbor without having to tell them every single value; you are trying to point out the main features that you have identified.  (For example, deciding whether some list is ÒskewedÓ or ÒsymmetricÓ will often be a matter of opinion; you need to look at many lists before you can make good judgment calls.)

In these notes, I will put sections of computer commands in boxes, like this one.  In these notes, I refer to the calculator as the TI-83.  The same commands apply to the TI-84.

Frequently we will use the TI-83 to make our work easier.  At first, you may think of the machines as burdens, confusing and intimidating.  I believe that if you stick with it and become experienced with the tools, you will discover that the calculators are indispensable for calculating statistics.  I will show you what I can about the TI-83Õs, but it is up to you to practice using them.  I cannot do that part for you, of course.

Generally, we have three chores to perform before our machine will show us the graphical display we want.  We must: 1) Enter the data into the calculator.  2) Choose the right options for the display we want.  And 3) Set up the proper window settings.  The commands to do these activities on the calculator are:

1) STAT EDIT  Use one of the lists to enter data, L1 for example; the other LÕs can be used too.  The LÕs are convenient work lists.  At times, you may find that you want more meaningful names.  One way to do this is to store the list in a new named list after entering numbers.  The syntax for this is L1 -> NEWL, assuming the data was entered in L1 and you want the new name to be NEWL.  (The -> key is in the lower left, directly above the ON key.)  Note: list names are limited to five letters.

2) 2nd STAT PLOT 1 On  Use this screen to designate the plot settings.  You have a choice of six displays:  scatter plot, line graph (we wonÕt use this), histogram, modified box plot, box plot, and normal plot.  You can have up to three plots on the screen at once.  For histograms, we will only use one at a time.  Later, when we see box plots, we will make multiple displays.

3) ZOOM 9  This command centers the window ÒonÓ your data.  It is always a good idea to see what the WINDOW settings are.  If you then change any of the WINDOW settings, you will then press GRAPH to see the changes.  (If you use ZOOM 9 again, the changes you just made donÕt get used!)

Goals:  (In these notes, I will summarize each dayÕs activity with a statement of goals for the day.)

                  Begin graphical summaries (describing data with pictures).  Be able to use the calculator to make a histogram.

Skills:  (In these notes, each day I will identify skills I believe you should have after working the dayÕs activity, reading the appropriate sections of the text, and practicing exercises in the text.

¥                Identify types of variables.  To choose the proper graphical displays, it is important to be able to differentiate between Categorical (or Qualitative) and Numerical (or Quantitative) variables.  Of course, it is also necessary to know exactly what a variable is.  Quite simply, a variable is an attribute about a collection of subjects.  As an example, for the students in this class, we could measure your height (a quantitative/numerical variable) or your major (a qualitative/categorical variable).

¥                Be familiar with types of graphs.  To graph categorical variables we use bar graphs or pie graphs.  To graph numerical variables, we use histograms, stem plots, or QUANTILE plots (using a TI-83 program we will explore on Day 3).  In practice, most of our variables will be numerical but it is still important to know the categorical displays.

¥                Summarize data into a frequency table and produce a histogram.  A frequency table is a list of class intervals (sometimes called bins or classes) and the number of data values in the class intervals.  You can make a frequency table yourself by just counting how many data values are in each interval, but the easiest way is to first use the TI-83 to make a histogram and then to TRACE over the boxes and record the classes and counts.  In fact, the histogram is just a ÒpictureÓ of the frequency table.

¥                Know how to modify the TI-83 default histogram.  You can control the size and number of the classes with Xscl and Xmin in the WINDOW menu.  The decision as to how many classes to create is arbitrary; there isnÕt a ÒrightÓ answer, or rather all choices of Xscl and Xmin are ÒrightÓ answers.  You should experiment every time you use the TI-83 to make a histogram by changing the interval width Xscl and starting point Xmin to see what happens to the display.

¥                Know how to create and interpret graphs for categorical variables.  The two main graphs for categorical variables are pie graphs and bar charts.  Pie graphs are difficult to make by hand, but are popular on computer programs like Excel.  Bar charts are also common on spreadsheets.  Data represented by pie graphs and bar charts usually are expressed as percents of the whole; thus they add to 100 %.  The ordering for categories is arbitrary; therefore concepts such as skew and center make no sense.

Reading:    Section 1.1. (Skip Time Plots and Time Series.)

Day 3

Activity:    Creating and interpreting stem plots, histograms, and quantile plots.

Use the Arizona Temps data set to practice creating the histograms, stem plots, and quantile plots for several lists.  Compare and interpret the graphs.  Identify shape, center, and spread.  TodayÕs new methods, the stem plot and the quantile plot, are alternate ways to summarize a list of numbers.

Answer these questions about the Arizona Temps:

1)   Do any of the lists have outliers?  (There is not a firm definition of exactly what an outlier is.  You will have to develop some personal judgment; the best way is to look at many lists and displays.)

2)     What information does the stem plot show that the histogram hides?

3)     What information does the quantile plot show that the stem plot and histogram hide?

4)     Which display is most ÒtrustworthyÓ?  That is, which one has the smallest likelihood of misleading you?

The stem plot is a hand technique, most useful for small data sets (under 40 entries).  It is basically a quick way to make a frequency chart, but always using class intervals based on the base 10 system.  This means the intervals will always be ten ÒunitsÓ wide, such as 10 to 19, or 1 to 9, or .01 to .09.  The ÒunitÓ chosen is called the stem, and the next digit after the stem is called the leaf.

QUANTILE is a program I wrote that plots the sorted data in a list and ÒstacksÓ the values up.  This is known as a quantile plot.  Basically we are graphing the data value versus the rank, or percentile, in the data set.  The syntax is PRGM EXEC QUANTILE ENTER.  The program will ask you for the list where youÕve stored the data.  A and B are temporary lists used by the program, so if you have data in these lists already, store them in another list before executing.

Goals:        Be able to make and interpret a quantile plot, using the TI-83 program QUANTILE.  Be able to make and interpret a stem plot by hand.

Skills:

¥                Use the TI-83 to create an appropriate histogram or quantile plot.  STAT PLOT and QUANTILE are our two main tools for viewing distributions of data on the TI-83.  Histograms are common displays, but have flaws; the choice of class intervals is troubling as it is not unique.  The quantile plot is more reliable, but less common.  For interpretation purposes, remember that in a histogram tall boxes represent places with lots of data, while in a quantile plot a steep slope represents those same high-density data places.

¥                Create a stem plot by hand.  The stem plot is a convenient manual display; it is most useful for small data sets, but not all data sets make good stem plots.  Choosing the ÒstemÓ and ÒleafÓ to make reasonable displays will require some practice.  Some ideas for a proper choice of stems: if you have many empty rows, you have too many stems.  Move one column to the left and try again.  If you have too few rows (all the data is on just one or two stems) you have too few stems.  Move to the right one digit and try again.  Some data sets will not give good pictures for any choice of stem, while some benefit from splitting or rounding (see the example on page 11).

¥                Describe shape, center, and spread.  From each of your graphs, you should be able to make general statements about the shape, center, and spread of the distribution of the variable being explored.  One of the main conclusions we want to make about lists of data when we are doing inference (Chapters 6 to 8) is whether the data is close to symmetric; many times Òclose enoughÓ is, well, close enough!  We will discuss this in more detail when we see the Central Limit Theorem in Chapter 5.

Reading:    Section 1.2.

Day 4

Activity:    Dance Fever example.  Using the TI-83 to calculate summary statistics and to make the box plots display.

Our second topic is numerical measures of data.  I hope that you are already familiar with some of these measures, such as the mean and median.  Others, like the standard deviation, will perhaps be mysterious.  Of course our goal is to unmask the mystery!

When a graphical display shows that a data set is well behaved, perhaps symmetrical with no outliers, we may want to summarize further.  We often want two measurements about numerical data: the center and the spread.  The reason for these two kinds of measurements will be clearer after we see the Central Limit Theorem in Unit 2.  Briefly, ÒcenterÓ is the typical data value, and ÒspreadÓ refers to how variable the data values are.  We will first practice using technology to produce the calculations for us.  Along the way, I hope we gain some understanding too.

Use the Arizona Temps data set to calculate the mean, the standard deviation, the 5-number summary, and the associated box plot for any of the variables.

Compare the box plots and numerical measures with the corresponding histograms and quantile plots you made on Day 2.  Note the similarities (where the data values are dense, and where they are sparse) but especially note the differences.  The box plots and numerical measures cannot describe shape very well.  On the other hand, histograms are messy to use to compare two lists.  The stem and leaf is tedious to modify.

Answer these questions about the Arizona Temps:

1)   Are high and low temperatures distributed the same way, other than the obvious fact that highs are higher than lows?  (When we talk of a distribution we mean a description of how the data values are centered and how they are spread out.  Sometimes this will be a simple statement; other times we need to see a graph.)

2)   How does a single case influence the calculatorÕs answers?  (What if there was an outlier in the list?  If you didnÕt have an outlier, change one of the values to a ridiculously large value and recalculate the graphs and measures.  Notice the effects of these extreme values.)

3)   What information does the box plot disguise?

To calculate our summary statistics, we will use 1-Var Stats (to use List 1) or 1-Var Stats L2 for List 2, for example.  There are two screens of output; we will be mostly concerned with the mean (pronounced Òx barÓ), the standard deviation Sx, and the five-number summary from screen two.

To make the box plot, we use STAT PLOT Type and pick the 4th or 5th icons.  The fifth icon is the true box plot, but the fourth one (the modified box plot) has a routine built in to flag possible outliers.  I recommend using the modified box plot as it shows at least as much information as the regular box plot, but includes the potential outliers, as defined by this procedure.

Goals:        Compare numerical measures of center.  Summarize data with numerical measures and box plots.  Compare these new measures with the histograms, stem plots, and quantile plots you made on Day 3.

Skills:

¥                Understand the effect of outliers on the mean and median.  The mean (or average) is unduly influenced by outlying (unusual) observations, whereas the median is not.  Therefore, knowing when your distribution is skewed is helpful.  The mean is attracted to the outliers, so for lists with a high outlier, the mean will be higher than the median, and higher than if the outlier is removed.  The degree of influence is a matter of judgment. The median is almost completely unaffected by outliers.  For technical reasons, though, the median is not as common in scientific applications as the mean.

¥                Use the TI-83 to calculate summary statistics.  Calculating summary statistics using the TI-83 is as simple as entering numbers into STAT EDIT and then choosing STAT CALC 1-Var Stats L#, where L# represents the list you are interested in.  If you are doing some work by hand, you may have to organize information in a correct way, such as listing the numbers from low to high to find the median.  Important note: I recommend that you get used to using the statistical features of your calculator to produce the mean.  While I realize many of you are in the habit of calculating the mean by adding up all the numbers and dividing by the sample size, you will not be using the full features of your machine.  Later on you will find yourself entering data twice: once to calculate the mean, then again when you discover you needed the standard deviation and the data should be in a list.

¥                Compare several lists of numbers using stem plots and box plots.  For two short lists, the best simple approach is the back-to-back stem plot.  For longer lists, or for more than two lists, use box plots, side-by-side, or stacked.  At a glance, you can then assess which lists have typically larger values or more spread out values, etc.  To graph up to three box plots on the TI-83, enter a different list in each of the 3 plots you can display using STAT PLOT.

¥                Understand box plots.  You should know that the box plots for some lists donÕt tell the interesting part of those lists.  For example, box plots do not describe the shape of a distribution very well; you can only see where the quartiles are.  Alternatively, you should know that the box plot gives a very good first impression.

Reading:    Section 1.2.

Day 5

Activity:    Interpreting standard deviation.  Quiz 1 today.

Today we will practice the ideas of mean and standard deviation by letting you play with some lists of your own.  There are many possible answers to some of the following questions; you should try to see if you understand why each of them may have many answers.  Also, the fourth list is an attempt to get you thinking about how spread out a data set is.  This is a requirement for understanding the standard deviation.  As usual, ask questions as you have them; we will be mostly working in groups today and I will listen to you work.

Create the following lists:

1)   A list of 10 numbers that has only one number below the mean.  (This seems to violate our interpretation of the average as being ÒrepresentativeÓ.  You should understand how the mean might mislead us!)

2)   A list of 10 numbers that has the standard deviation greater than the mean.  (Note that the standard deviation seems to be more affected by outlying values than the mean is.)

3)   A list of 10 numbers that has a standard deviation of zero.

4)   For your fourth list, start with any 21 numbers.  Find a number N such that 14 of the numbers in your list are within N units of the average.  For example, pick an arbitrary number N (say 4), calculate the average plus 4, the average minus 4, and count how many numbers in your list are between those two values.  If the count is less than 14, try a larger number for N (bigger than 4).  If the count is more than 14, try a smaller number for N (smaller than 4).  Continue guessing and counting until you get it close to 14 out of 21.  This number you settle on should be fairly close to the standard deviation.

Finally, compare the standard deviation to the Inter Quartile Range (IQR = Q3 - Q1).

Another way to understand standard deviation is to associate the graphical plots with the value of s.  Then change a few points, and reexamine the graphs and value of s.  Experimentation such as this cannot be learned by reading about it; you must do it yourself!  I hope that you use class time in this way.

(You may use any extra time today to discuss Presentation 1 in your groups.)

Goals:        Interpret standard deviation as a measure of spread.

Skills:

¥                Understand standard deviation.  At first, standard deviation will seem foreign to you, but I believe that it will make more sense the more you become familiar with it.  In its simplest terms, the standard deviation is non-negative number that measures how ÒwideÓ a dataset is.  One common interpretation is that the range of a dataset is about 4 or 5 standard deviations.  Another interpretation is that the standard deviation is roughly ¾ times the IQR; that is the standard deviation is a bit smaller than the IQR.  Eventually we will use the standard deviation in our calculations for statistical inference; until then, this measure is just another summary statistic, and getting used to this number is your goal.  The normal curve of Section 1.3 will further help us understand standard deviation.

Reading:    Section 1.3.

Day 6

Activity:    Introduce the TI-83Õs normal calculations.  Homework 1 due today.

So far, we have used graphs to summarize distributions and summary statistics to describe certain features such as center and spread.  Today we will look at a specific distribution that is so common it has a name, the normal distribution.  We will see that data following this curve are extremely easy to describe: we only need to know the mean and the standard deviation.  The difference between this new material and yesterdayÕs work is that if the data really does follow the normal curve, we can draw the distributionÕs curve by knowing just those two measures.  In yesterdayÕs work, although we knew the mean and the standard deviation, that still gave us no indication of the shape of the distribution.

Generally, finding areas under curves is a calculus problem; ideally you end up with a formula where you simply plug in the endpoints desired.  Unfortunately, the normal curve is an example of a function for which the calculus approach doesnÕt give a useful answer.  Therefore to find areas under the normal curve, we must use some sort of approximation.  The normal table in the book is one approximation.  I donÕt recommend using that table though, because our calculator is easier, and it happens to be more precise!  However, this means that as you read the problems in the textbook, you must skip over the descriptions given about using z-scores to find areas.  Also, remember this discrepancy when checking answers in the back of the book for the odd problems: often the bookÕs answers are different from the calculatorÕs answers because they used the (inaccurate) table instead of the (accurate) calculator.

My advice to you when calculating areas using the normal curve is to always draw a picture.  See your class notes for the actual diagram and the way to label it.  I will try to always put two scales on my normal curves: the z-scale and the scale of real units.  I do this so that I can make sense of answers I get.  For example, if the shaded area on my diagram is obviously less than half of the total area, and if my answer is 0.75, then that makes no sense, and I must have entered something into the calculator incorrectly (or drawn the picture wrong!).

Find the following areas:

1)    The area between -1 and +1 on the standard normal curve.

2)    The area between -2 and +2 on the standard normal curve.

3)    The value on the standard normal curve that gives the lower 80 % (the 80th percentile).

4)    The area on the normal curve with a mean of 100 and a standard deviation of 15 above 150.

5)    The values from a normal curve with a mean of 100 and a standard deviation of 15 that contain the central 95 % of the area.  (We will see this idea again when we do confidence intervals in Chapter 6.)

DISTR normalcdf( lower, upper ) calculates the area under a standard normal curve between lower and upper.  If you specify just the 2 values, then the mean is 0 and the standard deviation is 1 (this is called the standard normal curve).  If you want a different mean or standard deviation, add a third and fourth parameter, as in the next paragraph.

DISTR normalcdf( lower, upper, mean, standard deviation ) calculates the area under the normal curve between lower and upper that has the given mean and standard deviation.  The examples in the text where a z-score is calculated can be done much more efficiently with this flexible command.

Examples: DISTR normalcdf( -10, 20, 5, 10 ) finds the area between -10 and +20 on a normal curve with mean 5 and standard deviation 10 while DISTR normalcdf( -2, 2 ) finds the area on the standard normal curve between -2 and +2.

DISTR invNorm( works backwards, but only gives upper as an answer.  It is also referred to as a percentile.  The 90th percentile is that point at which 90 % of the observations are below.  The syntax is DISTR invNorm( .90 ) or DISTR invNorm( .90, 5, 10 ); the first example assumes the standard normal curve and reports the 90th percentile.  The second example gives the 90th percentile for a normal curve with a mean of 5 and a standard deviation of 10.

Important note: If the desired area is above a certain number, you will have to use subtraction and symmetry, as DISTR invNorm( only reports values below, or to the left.

Goals:        Introduce normal curve.  Use TI-83 in place of the standard normal table in the text.

Skills:

¥                Know what a z-score is (standardization).  Sometimes, instead of knowing a variableÕs actual value, we are only interested in how far above or below average it is.  This information is contained in the z-score.  Negative values indicate a below average observation, while positive values are above average.  If the list follows a normal distribution (the familiar Òbell-shapedÓ curve) then it will be relatively rare to have values below -2 or above +2 (only about 5 % of cases).  Even if the list is not normal, surprisingly the z-score still tends to have few values beyond ±2, although this is not guaranteed.

¥                Using the TI-83 to find areas under the normal curve.  When we have a distribution that can be approximated with the bell-shaped normal curve, we can make accurate statements about frequencies and percentages by knowing just the mean and the standard deviation of the data.  Our TI-83 has 2 functions, DISTR normalcdf( and DISTR invNorm( which allow us to calculate these percentages more easily and more accurately than the table in the text.  We use DISTR normalcdf( when we want the percentage as an answer and we use DISTR invNorm( when we already know the percentage but not the value that gives that percentage.

Reading:    Section 1.3.

Day 7

Activity:    Practice normal calculations, using real situations (except 2).

Answer the following questions:

1)    Suppose SAT scores are distributed normally with mean 800 and standard deviation 100.  Estimate the chance that a randomly chosen score will be above 720.  Estimate the chance that a randomly chosen score will be between 800 and 900.  The top 20 % of scores are above what number?  (This is the 80th percentile.)

2)    Find the Inter Quartile Range (IQR) for the standard normal (mean 0, standard deviation 1).   Remember that the first quartile is the 25th percentile and the third quartile is the 75th percentile.  Observe and note whether this IQR is larger or smaller than 1, the standard deviation.

3)    Women aged 20 to 29 have normally distributed heights with mean 64 and standard deviation 2.7.  Men aged 20 to 29 have mean 69.3 with standard deviation 2.8.  What percent of 20 something women are taller than the average 20 something man, and what percent of 20 something men are taller than the average 20 something woman?

4)    Pretend we are manufacturing fruit snacks, and that the average weight in a package is .92 ounces with standard deviation 0.04.  What should we label the net weight on the package so that only 5 % of packages are ÒunderweightÓ?

5)    Suppose that your average commute time to work is 20 minutes, with standard deviation of 2 minutes.  What time should you leave home to arrive to work on time at 8:00?  (You may have to decide a reasonable value for the chance of being late.)

Goals:        Master normal calculations.  Realize that summarizing using the normal curve is the ultimate reduction in complexity, but only applies to data whose distribution is actually bell-shaped.

Skills:

¥                Memorize 68-95-99.7 rule.  While we do rely on our technology to calculate areas under normal curves, it is convenient to have some of the values committed to memory.  Roughly 68 % of data from a normal curve are within one standard deviation of the mean; roughly 95 % of data from a normal curve are within two standard deviations of the mean; and roughly 99.7 % of data from a normal curve are within three standard deviations of the mean.  These values can be used as rough guidelines; if precision is required, you should use the TI-83 instead.  (For example, the area between exactly -2 and exactly +2 is really 95.45 %, not 95 %.)  I will assume you know these three numbers by heart when we encounter the normal numbers again in Chapters 5 through 8.  An important observation to make here is that rarely is a value from the normal curve further than 3 standard deviations from the center of the curve.

¥                Understand that summarizing with just the mean and standard deviation is a special case.  We have progressed from graphs like histograms and quantile plots to summary statistics like medians, means, and standard deviations to finally summarizing an entire list with just two numbers: the mean and the standard deviation.  However, this last step in our summarization only applies to lists whose distribution resembles the bell-shaped normal curves.  If the dataÕs distribution is skewed, or has any other shape, this level of summarization is insufficient.  Also, it is important to realize that these calculations are only approximations; usually in the real world data is only approximately normally distributed.

¥                Interpret a normal quantile plot.  We often want to know if a list of data can be approximated with a normal curve.  While we might try histograms and quantile plots to see if they Òlook normalÓ, it is a difficult task, because we have to match the shape to the very special shape of the normal curve (bell shaped for histograms and the ÒSÓ shape of quantile plots).  One simple alternative graphical method is the normal quantile plot.  This plot is nearly identical to a quantile plot, but instead of graphing the percentiles, we graph the z-scores.  Our TI-83 does this for us: the sixth icon in the STAT PLOT Type.  Be cautious though; the graph, as usual, is unlabeled.  However, we only care if the graph is nearly a straight line or not.  Not surprisingly, this too is a judgment call for us at this point.

Reading:    Sections 2.1 and 2.2.

Day 8

Activity:    Creating scatter plots.  Calculating the correlation coefficient.  Quiz 2 today.

Today we begin the second part of Unit 1.  So far we have dealt with just one list of data at a time.  The graphical and numerical techniques let us know how the data were spread out, whether there was a skew, where it was centered, etc.  With two variables (two lists of data) we will have one graphical display (scatter plots) and one numerical display (correlation).  But we will add a new sort of summary procedure to our list of techniques: regression.  Regression is a summarized description of the pattern in a scatter plot of data.

The correlation coefficient is a unit-less number between -1 and +1.  The two extremes represent data that fall perfectly on a line, with negative or positive slope, respectively.  A correlation of 0 represents data that show no trend whatsoever: your prediction of the y-value is unaffected by the x-value you choose.  These facts can be explored with the following activity.

Using the Arizona Temps data, make a scatter plot of ÒFlagstaff HighÓ versus ÒPhoenix HighÓ.  Then guess what the correlation coefficient might be without using your calculator.  Use the sample diagrams on page 104 to guide you.  Finally, using your calculator, calculate the actual value for the correlation coefficient and compare it to your guess.

Repeat for the variables ÒFlagstaff HighÓ and ÒFlagstaff LowÓ, or for any other pair.

We will have to use the regression function STAT CALC LinReg(ax+b) to calculate correlation, r.  If you type ENTER after the STAT CALC LinReg(ax+b) command, the calculator assumes your lists are in columns L1 and L2; otherwise you will type where they are, for example STAT CALC LinReg(ax+b) L2, L3.  Important note: You will have to have already pressed DiagnosticOn; access this command through the CATALOG (2nd 0).  You will only need to turn the diagnostics on once, unless you happen to have your RAM cleared on your TI-83.

Goals:        Display two variables and measure (and interpret) linear association using the correlation coefficient.

Skills:

¥                Plot data with a scatter plot.  This will be as simple as entering two lists of numbers into your TI-83 and pressing a few buttons, just as for histograms or box plots.  Or, if you are doing plots by hand you will have to first choose an appropriate axis scale and then plot the points.  You should also be able to describe overall patterns in scatter diagrams and suggest tentative models that summarize the main features of the relationship, if any.

¥                Use the TI-83 to calculate the correlation coefficient.  Unfortunately, there isnÕt a quick way to have the TI-83 find the correlation coefficient; we must use the regression function STAT CALC LinReg(ax+b).  However, without the scatter plot, it is a mistake to assume the correlation coefficient automatically is a good summary of a relationship.  We will learn more about this in Section 2.4.  Important note: You will have to have pressed DiagnosticOn; access this command through the CATALOG (2nd 0).

¥                Interpret the correlation coefficient.  You should know the range of the correlation coefficient (-1 to +1) and what a ÒtypicalÓ diagram looks like for various values of the correlation coefficient.  Again, page 104 is your guide.  You should recognize some of the things the correlation coefficient does not measure, such as the strength of a non-linear pattern.  Also, in contrast to the regression results we will learn starting on Day 10, it does not matter in what order the variables are entered; correlation is a fact about a pair of variables.

Reading:    Section 2.2.

Day 9

Activity:    Outlier effects on correlation.  Homework 2 due today.

We are not going to focus on the actual formula used to calculate the correlation coefficient.  However, it is important to note that it involves summation, which is akin to averaging.  Therefore, any ÒproblemsÓ we have with averages will also exist with correlation.  Specifically, extreme values (outliers) will influence the results.  Unfortunately, in two dimensions we encounter a new sort of outlier: it is far from the pattern but is not an outlier in either variable.  Today we will find out what effect a single point that is different from the pattern has on the correlation coefficient.  Important note: TodayÕs activity is one of those we will do for understanding, and is not something you would do in real life to a data set.

The data set we will explore today has 7 data points. 

 

x

y

 

1

4

 

2

3

 

2

5

 

3

4

 

4

8

 

4

6

 

5

9

First new point

3

8

Second new point

7

10

Third new point

7

1

 

First, plot the original seven points and calculate the correlation coefficient.  Now add the eighth point in three different places (making three new data sets, each with 8 points) and for each new data set, calculate the correlation coefficient.  Compare the values, noting where the eighth data point (the outlier) was and how it changed the correlation.  Summarize the effect of outliers in a paragraph.

(You may use any extra time today to discuss Presentation 1 in your groups.)

Goals:        Understand the impact of outliers on correlation.  Gain further practice with the TI-83.

Skills:

¥                Interpret the correlation coefficient.  You should recognize how outliers influence the magnitude of the correlation coefficient.  One simple way to observe the effects of outliers is to calculate the correlation coefficient with and without the outlier in the data set and compare the two values.  If the values vary greatly (this is a judgment call) then you would say the outlier is ÒinfluentialÓ.

Reading:    Section 2.3.

Day 10

Activity:    Using the Olympic data, fit a regression line to predict the 2004 and 2008 race results.

Now that we know how to produce a scatter plot of data, and calculate a correlation coefficient, we will want to further describe the relationship between the two variables.  The chief technique for summarizing a linear relationship is Least Squares Linear Regression.  This technique is also known as Least Squares Regression, Best Fit Regression, Linear Regression, etc.  The important point is that we are going to describe the relationship with a straight line, so if the scatter plot shows some other shape, this technique will be inappropriate.  Your tasks are to 1) come up with a line, either by hand or with technology, that Ògoes throughÓ the data in some appropriate way, 2) to be able to use this model to describe the relationship verbally, and 3) to predict numerically y-values for particular x-values of interest.

Begin by making a scatter plot of the race times.  If you want a rough guess for the slope of the best fitting line through the data, you can connect two points spaced far apart (I will show you the details in class.)

Next, use the TI-83Õs regression features to calculate the best fit.  The command is STAT CALC LinReg(ax+b), assuming the two lists are in L1 and L2.  (L1 will be the horizontal variable, years in this case.)  (We used this command on Day 4 also.  However, for regression it is vital that you get the order correct; the idea here is that you are predicting the vertical variable from knowing the horizontal variable.)

Interpret what your two regression coefficients mean.  Make sure you have units attached to your numbers to help with the meanings.

Have the calculator type this equation into your Y= menu (using VARS Statistics EQ RegEQ), and TRACE on the line to predict the future results.  Specifically, see what your model says the 2008 time should have been.  Then we will look it up and check how predictive our model is.

Here are the data: 100-meter dash winning Olympic times:

1896

Thomas Burke, United States

12 sec

 

 

1900

Francis W. Jarvis, United States

11.0 sec

 

 

1904

Archie Hahn, United States

11.0 sec

 

 

1908

Reginald Walker, South Africa

10.8 sec

 

 

1912

Ralph Craig, United States

10.8 sec

 

 

1920

Charles Paddock, United States

10.8 sec

 

 

1924

Harold Abrahams, Great Britain

10.6 sec

 

 

1928

Percy Williams, Canada

10.8 sec

Elizabeth Robinson, United States

12.2 sec

1932

Eddie Tolan, United States

10.3 sec

Stella Walsh, Poland (a)

11.9 sec

1936

Jesse Owens, United States

10.3 sec

Helen Stephens, United States

11.5 sec

1948

Harrison Dillard, United States

10.3 sec

Francina Blankers-Koen, Netherlands

11.9 sec

1952

Lindy Remigino, United States

10.4 sec

Marjorie, Jackson, Australia

11.5 sec

1956

Bobby Morrow, United States

10.5 sec

Betty Cuthbert, Australia

11.5 sec

1960

Armin Hary, Germany

10.2 sec

Wilma Rudolph, United States

11.0 sec

1964

Bob Hayes, United States

10.0 sec

Wyomia Tyus, United States

11.4 sec

1968

Jim Hines, United States

9.95 sec

Wyomia Tyus, United States

11.0 sec

1972

Valery Borzov, USSR

10.14 sec

Renate Stecher, E. Germany

11.07 sec

1976

Hasely Crawford, Trinidad

10.06 sec

Annegret Richter, W. Germany

11.08 sec

1980

Allen Wells, Britain

10.25 sec

Lyudmila Kondratyeva, USSR

11.6 sec

1984

Carl Lewis, United States

9.99 sec

Evelyn Ashford, United States

10.97 sec

1988

Carl Lewis, United States

9.92 sec

Florence Griffith-Joyner, United States

10.54 sec

1992

Linford Christie, Great Britain

9.96 sec

Gail Devers, United States

10.82 sec

1996

Donovan Bailey, Canada

9.84 sec

Gail Devers, United States

10.94 sec

2000

Maurice Greene, United States

9.87 sec

Marion Jones, United States

10.75 sec

2004

Justin Gatlin, United States

9.85 sec

Yuliya Nesterenko, Belarus

10.93 sec

2008

Usain Bolt, Jamaica

9.69 sec

Shelly-ann Fraser, Jamaica

10.78 sec

2012

?

 

?

 

(a) A 1980 autopsy determined that Walsh was a man.

200-meter dash winning Olympic times:

1900

Walter Tewksbury, United States

22.2 sec

 

 

1904

Archie Hahn, United States

21.6 sec

 

 

1908

Robert Kerr, Canada

22.6 sec

 

 

1912

Ralph Craig, United States

21.7 sec

 

 

1920

Allan Woodring, United States

22 sec

 

 

1924

Jackson Sholz, United States

21.6 sec

 

 

1928

Percy Williams, Canada

21.8 sec

 

 

1932

Eddie Tolan, United States

21.2 sec

 

 

1936

Jesse Owens, United States

20.7 sec

 

 

1948

Mel Patton, United States

21.1 sec

Francina Blankers-Koen, Netherlands

24.4 sec

1952

Andrew Stanfield, United States

20.7 sec

Marjorie, Jackson, Australia

23.7 sec

1956

Bobby Morrow, United States

20.6 sec

Betty Cuthbert, Australia

23.4 sec

1960

Livio Berruti, Italy

20.5 sec

Wilma Rudolph, United States

24.0 sec

1964

Henry Carr, United States

20.3 sec

Edith McGuire, United States

23.0 sec

1968

Tommy Smith, United States

19.83 sec

Irena Szewinska, Poland

22.5 sec

1972

Valeri Borzov, USSR

20.00 sec

Renate Stecher, E. Germany

22.40 sec

1976

Donald Quarrie, Jamaica

20.23 sec

Barbel Eckert, E. Germany

22.37 sec

1980

Pietro Mennea, Italy

20.19 sec

Barbel Wockel, E. Germany

22.03 sec

1984

Carl Lewis, United States

19.80 sec

Valerie Brisco-Hooks, United States

21.81 sec

1988

Joe DeLoach, United States

19.75 sec

Florence Griffith-Joyner, United States

21.34 sec

1992

Mike Marsh, United States

20.01 sec

Gwen Torrance, United States

21.81 sec

1996

Michael Johnson, United States

19.32 sec

Marie-Jose Perec, France

22.12 sec

2000

Konstantinos Kenteris, Greece

20.09 sec

Marion Jones, United States

21.84 sec

2004

Shawn Crawford, United States

19.79 sec

Veronica Campbell, Jamaica

22.05 sec

2008

Usain Bolt, Jamaica

19.30 sec

Veronica Campbell-Brown, Jamaica

21.74 sec

2012

?

 

?

 

Goals:        Practice using regression with the TI-83.  We want the regression equation, the regression line superimposed on the plot, the correlation coefficient, and we want to be able to use the line to predict new values.

Skills:

¥                Fit a line to data.  This may be as simple as ÒeyeballingÓ a straight line to a scatter plot.  However, to be more precise, we will use least squares, STAT CALC LinReg(ax+b) on the TI-83, to calculate the coefficients, and VARS Statistics EQ RegEQ to type the equation in the Y= menu.  You should also be able to sketch a line onto a scatter plot (by hand) by knowing the regression coefficients.

¥                Interpret regression coefficients.  Usually, we want to interpret only slope, and slope is best understood by examining the units involved, such as inches per year or miles per gallon, etc.  Because slope can be thought of as ÒriseÓ over ÒrunÓ, we are looking for the ratio of the units involved in our two variables.  More precisely, the slope tells us the change in the response variable for a unit change in the explanatory variable.  We donÕt typically bother interpreting the intercept, as a zero value for the independent variable is often outside of the range of experimentation.

¥                Estimate/predict new observations using the regression line.  Once we have calculated a regression equation, we can use it to estimate new responses.  The easiest way to use the TI-83 for this is to TRACE on the regression line.  You may need to use up and down arrows to toggle back and forth from the plot to the line.  You may also just use the equation itself by multiplying the new x-value by the slope and adding the intercept.  (This is exactly what TRACE is doing.)  Note: when using TRACE, and the x-value you want is currently outside the window settings (below Xmin or above Xmax) you must reset the window to include your x-value first.

Reading:    Section 2.3.

Day 11

Activity:    Revisit outliers data set, adding regression lines.  Quiz 3 today.

We are not going to focus on the actual formulas used to calculate the regression coefficients.  However, it is important to note that it involves summation (just as it was in correlation), which is akin to averaging.  Therefore, any ÒproblemsÓ we have with averages will also exist with regression.  Specifically, extreme values (outliers) will influence the results.  Unfortunately, in two dimensions we encounter a new sort of outlier: it is far from the pattern but is not an outlier in either variable.  Today we will find out what effect a single point that is different from the pattern has on the regression coefficients.  Important note: TodayÕs activity is one of those we will do for understanding, and is not something you would do in real life to a data set.

The data set we will explore today is the same as that from Day 9 and has 7 data points.  First, plot them and calculate the regression coefficients.  Now add the eighth point in three different places (making three new data sets, each with 8 points) and for each new data set, calculate the regression coefficients.  Compare the values, noting where the eighth data point (the outlier) was and how it changed the values.  Summarize the effect of outliers in a paragraph.

(You may use any extra time today to discuss Presentation 1 in your groups.)

Goals:        Understand the impact of outliers on the regression coefficients.  Gain further practice with the TI-83.

Skills:

¥                Understand the limitations and strengths of linear regression.  Basically, linear regression should only be used with scatter plots that are roughly linear in nature.  That seems obvious by the name.  However, there is nothing that prevents us from calculating the numbers for any data set we can input into our TI-83Õs.  We have to realize what our data looks like before we calculate the regression; therefore a scatter plot is an essential first step.  In the presence of outliers and non-linear patterns, we should avoid drawing conclusions from the fitted regression line.

Reading:    Sections 2.4 and 2.5.

 Day 12

Activity:    Correlation/Regression summary.  U. S. population example.  Alternate regression models.  Homework 3 due today.

Today I want to show you an example similar to the work you will do for your presentation for the state population prediction.  Your TI-83 has many other regression models available to you.  The models that often are adequate for population models are exponential functions and logistic functions.  The polynomial functions (quadratic, cubic, quartic) might fit the data well, yielding small residuals, but they are unlikely to be useful predictive models.  If you zoom out, you will see unrealistic future effects, for example.  One caution with the logistic function on the TI-83: it seems to be finicky.  For some data sets it works extremely well, and for others it wonÕt even produce fitted values.

Another modification you may want to try, instead of different models, is different time periods.  For example, to predict the population the year 2010, data from before 1850 may not be helpful for developing the model, especially if factors affecting population growth have changed substantially.

My presentation of the U. S. population model will not take the entire hour; I suggest using this extra time for your own state model.

Finally, I have listed below some summary statements about correlation and regression that you should be aware of before the first exam.

1)    For correlation, the variables can be entered in any order; correlation is a fact about a pair of variables.  For regression, the order the variables are presented matters.  If you reverse the order, you get a different regression line.

2)    We must have numerical variables to calculate correlation.  For categorical variables, we would use contingency tables, but not in this course.

3)    High correlation does not necessarily mean a straight-line scatter plot.  The U. S. population growth is an example.

4)    Correlation is not resistant to the effects of outliers; the data set from Days 9 and 11 showed that the placement of a single point in the scatter plot could greatly influence the value of the correlation and the coefficients in the regression equation.

Goals:        See scatter plots and correlation in practice.  Understand correlations limitations and features.

Skills:

¥                Recognize the proper use of correlation, and know how it is abused.  Correlation measures straight-line relationships.  Any departures from that model make the correlation coefficient less reliable as a summary measure.  Just as for the standard deviation and the mean, outliers influence the correlation coefficient.  Therefore, it is extremely important to be aware of data that is unusual.  Some two-dimensional outliers are hard to detect with summary statistics; scatter plots are a must then.

¥                Understand that there are competing regression models.  We have focused our attention on the linear regression models, but as we see on our TI-83, there are many other potential models.  If you use an alternate model, be sure to plot the fitted line along with the scatter plot.  Be careful though; our usual measure of fit, r2, will not accurately tell the story.

Reading:    Chapters 1 and 2.

Day 13

Activity:    Presentation 1.  Graphical and Numerical Summaries, Regression and Correlation

Gather 2 to 3 variables on at least 20 subjects; the source is irrelevant, but knowing the data will help you explain its meaning to us.  Be sure to have summarized at least one numerical and at least one categorical variable.  Your goal is to demonstrate that you can summarize one-variable data, both graphically and numerically.  Note that I am not asking you to perform two-variable summaries with this data.

Also, pick one of the 50 states (the data is at the end of this file).  Predict the population in the year 2010 using a regression function (not necessarily linear).  Describe how you decided upon your model, and explain how good you think your prediction is.  Your goal is to convince us to believe your prediction.

This presentation must take less than 7 minutes so that all groups have enough time.  This means that you will have to be organized in how you present your summaries; I will stop you after 7 minutes.

Reading:    Chapters 1 and 2.

Day 14

Activity:    Exam 1.

This first exam will cover graphical summaries (pictures), numerical summaries (summary calculations), normal curve calculations (areas under the bell curve), scatter plots, correlation, and regression (two-variable summaries).  Some of the questions may be multiple choice.  Others may require you to show your worked out solution.  Section reviews are an excellent source for studying for the exams.  DonÕt forget to review your class notes and these on-line notes.

Reading:    Section 3.1.

Day 15

Activity:    History of polls.

Unit 2 is about sampling and probability.  In addition to sampling, we will also explore the related topics of observational studies and experiments.  The underpinnings to Unit 3, statistical inference, rely on knowing precisely how likely a particular outcome is.  If the samples we collect are done with no organized method, and without using probability rules, then we will be unable to assess the likelihood of an outcome.  Thus, studying sampling and probability is crucial to being able to believe statistical inference in scientific studies.

A very high profile use of sampling is presidential elections.  In the early years of the 20th century, many attempts were made to predict elections beforehand.  Below are some consequences of the haphazard Òsamples of convenienceÓ that were conducted.  Because of the lessons learned during these fiascos, we now have very rigorous methods of producing good data.

To 1936:  Selection Bias

Literary Digest calls the 1936 election for Alf Landon (remember him), in an Electoral College landslide.  The poll was performed by sending postcards to people with telephones, magazine subscribers, car owners, and a few people on lists of registered voters.  The first problem was that the sample was biased toward Republican-leaning voters, who could afford magazine subscriptions and telephones during the Great Depression.  The second problem is the response rate error: they received only 2.3 million responses, for a 23 % response rate.  This was compounded by a volunteer error: the respondents are most likely people who wanted change, and Roosevelt is president.  Nevertheless, Franklin Roosevelt, the Democrat, wins in a landslide.  Statistician Jessica Utts (Seeing Through Statistics, Jessica M. Utts, Duxbury Press, Wadsworth Publishing Co., 1996 pages 65-66), who examined issues of the Literary Digest from 1936, says ÒThey were very cocky about George Gallup predicting they would get it wrong.  [Gallup helped make his reputation in polling by correctly calling the race.]  The beauty of something like that is that the winner is eventually known.Ó  [Taken from: http://whyfiles.org/009poll/fiasco.html]

To 1948:  Quota Sampling

Aided by erroneous polling, newspapers prematurely call the presidential election for challenger Thomas Dewey, leading to the famous photograph of an elatedly re-elected Harry Truman.  The problematic polls use then-popular Òquota-samplingÓ techniques.  In other words, they sought out a certain number of men, a certain number of women, and similarly for blacks, whites, and various income levels.  According to statistician Fritz Scheuren, quota sampling in political polling was abandoned after this debacle in favor of random sampling of the population.  [Taken from: http://whyfiles.org/009poll/fiasco.html]

To present:  Random Sampling

Modern sampling techniques use randomness to select participants.  Interviewers are not allowed to choose people; and they are required to read questions from a card.  The purpose of sampling is to gather unbiased information, and each of the techniques we employ is an attempt to keep the information untainted.

I will give you the numbers and we can see how the polls have improved since 1936, and if there are still any biases towards either Republicans or Democrats.

Goals:        Introduce sampling.  Identify biases.  Explore why non-random samples are not trustworthy.

Skills:

¥                Understand the issues of bias.  We seek representative samples.  The ÒeasyÓ ways of sampling, samples of convenience and voluntary response samples, may or may not produce good samples, and because we donÕt know the chances of subjects being in such samples, they are unreliable sampling methods.  Even when probability methods are used, biases can spoil the results.  Avoiding bias is our chief concern in designing surveys.

¥                Huge samples are not necessary.  One popular misconception about sampling is that if the population is large, then we need a proportionately large sample.  This is just not so.  My favorite counter-example is our method of tasting soup.  To find out if soup tastes acceptable, we mix it up, then sample from it with a spoon.  It doesnÕt matter to us whether it is a small bowl of soup, or a huge vat; we still use only a spoonful.  The situation is the same for statistical sampling; we use a small ÒspoonÓ, or sample.  The fundamental requirement though is that the ÒsoupÓ (our population) is Òwell mixedÓ (as in a simple random sample - see Day 17).  In practice, most nation-wide samples have only a few thousand participants.

Reading:    Section 3.1.

Day 16

Activity:    Lurking variables exercises.

Many times, the ÒconclusionÓ drawn is incorrect, because of various other explanations.  We will explore these today through examples.  What you are looking for on these problems are these situations:

1)    Reversed cause and effect.  It may be that instead of X causing Y, Y causes X.  Just because two things occur together does not mean that the first causes the second.

2)    Common response.  These are the ÒlurkingÓ variables.  A third unobserved variable might be causing each of the two observed variables.  Critics of the anti-smoking movement were (correctly) pointing out that there might be a third set of variables (such as genetics) that cause both smoking propensity and cancer susceptibility.

3)    Coincidence.  There may be nothing to the association other than luck.  With large samples, though, we tend to discount the luck notion.

The main conclusion from todayÕs examples is that in observational studies, there are always potential confounding factors.  We may try to control for the other factors, such as by looking at different age categories separately, but critics can always argue that a critical variable that explains the observed association has not been measured.

I have a set of problems from Statistics, by Freedman, Pisani, and Purves, which I think will help us think about looking for alternative explanations than the proffered ÒattractiveÓ conclusion.  Work on each problem for a few minutes; we will discuss each of them in detail before the class is over.  Be prepared to defend your explanations.

Goals:        Explore experimentation ideas.  Discover potential lurking variables and ways to control for them.

Skills:

¥                Examine situations and detect lurking variables.  When searching for lurking variables, it is not enough to suggest variables that might also explain the response variable.  Your potential lurking variable must also be associated with the explanatory variable.  So, for example, suppose you are trying to predict height using weight.  A possible lurking variable might be age, because age tends to be associated with both weight and height.  On the other hand, a variable associated with height that is unlikely to be related to weight (and therefore would not be a lurking variable) is arm span.

¥                Know appropriate ways to attempt to control for lurking variables.  Once a potential lurking variable has been identified, we can make more appropriate comparisons by dividing our subjects into smaller, more homogeneous groups and then making the comparison.  This is why you will see comparisons made by age groups, for example.

¥                Understand that experimentation, done properly, will allow us to establish cause-and-effect relationships.  Observational studies have lurking variables; we can try to control for them by various methods, but we cannot eliminate them.  If the data is collected appropriately through good experimentation, however, the effects of lurking variables can be eliminated.  This is done through randomization, the thinking being that if a sample is large enough, it canÕt realistically be the case that all of one group contains all the large values of a lurking variable, for example.

Reading:    Section 3.2.

Day 17

Activity:    Creating random samples.

The easiest form of random sampling is simple random sampling.  Under this method, each possible sample could be the one selected.  In practice, with samples of 1,000 people out of populations of 100 million, there are an incomprehensibly large number of possible samples.  This means that practically, we do not actually take simple random samples, and other methods must be employed.  Today, though, we will get at the understanding of the procedure.  Important note: TodayÕs activity is one of those we will do for understanding, and is not something you would do in real life.  In practice, you would take one sample and be done.  Today each group will take a total of 60 samples.

We will use three methods of sampling today: dice, Table B in our book, and our calculator.  To make the problem feasible, we will only use a population of size 6.  (I know this is unrealistic in practice, but the point today is to see how randomness works, and hopefully trust that the results extend to larger problems.)  Pretend that the items in our population (perhaps they are people) are labeled 1 through 6.  For each of our methods, you will have to decide in your group what to do with ÒtiesÓ.  Keep in mind the goal of simple random sampling: at each stage, each remaining item has an equal chance to be the next item selected.  Also, remember here we are perhaps interviewing exactly three people.  Keep careful track of which samples you selected; record your results in order, as 125 or 256, for example.  (125 would mean persons 1, 2, and 5 were selected.)

1)     Dice: By rolling dice, generate a sample of three people.  (The number on the die corresponds to one of the items.)  Repeat 20 times, giving 20 samples of size 3.

2)    Table B: Using Table B, and starting at any haphazard location, select three people. (The random digit corresponds to one of the items.)  Repeat 20 times, giving 20 more samples of size 3.  (See page 185 for an example of how to use Table B.)

3)    Calculator: Using your TI-83, select three people. The TI-83 command MATH PRB randInt( 2, 4, 5 ) will produce 5 numbers between 2 and 4, inclusive, for example.  (If you leave off the third number, only one value will be generated.)  Using the proper command, generate 20 samples of size 3.

Your group should have drawn 60 samples at the end. We will pool the results of everyoneÕs work together on the board.

Important note: I want you to be clear what we mean by the word random.  When we refer to ÒrandomnessÓ, we are generally talking about the results of a mathematical model.  Unfortunately in English, random often means something different.  For example, people think random means Òwithout a plan, or without conscious decisionÓ.  In statistics, we actually mean Òwith equal chanceÓ.  I will try to remind you of this the further we get into our probability discussion.

Goals:        Gain practice taking random samples.  Understand what a simple random sample is.  Become familiar with MATH randInt(.  Accept that the calculator is random.

Skills:

¥                Know the definition of a Simple Random Sample (SRS).  Simple Random Samples can be defined in two ways: 1) An SRS is a sample where, at each stage, each item has an equal chance to be the next item selected.  2) A scheme where every possible sample has an equal chance to be the selected sample produces an SRS.

¥                Select an SRS from a list of items.  The TI-83 command MATH randInt( can be used to select numbered items from a list randomly.  If a number selected is already in the list, ignore that number and get a new one.  Remember, as long as each remaining item is equally likely to be chosen as the next item, you have drawn an SRS.

¥                Understand the real world uses of SRSÕs.  In practice, simple random samples are not that common.  It is just too impractical (or impossible) to have a list of the entire population available.  However, the idea of simple random sampling is essentially the foundation for all the other types of sampling.  In that sense then it is very common.

Reading:    Sections 3.3 and 3.4.

Day 18

Activity:    Alternate Sampling Schemes.  Quiz 4 today.

Today we will explore other sampling schemes than the SRS.  In practice, as we have discussed, it is impractical or impossible to collect simple random samples.  This does not mean that we cannot gather useful information, but that we may have to live with some biases.  Of course we will try to eliminate as many biases as we can.  We will look at systematic sampling, stratified sampling, and cluster sampling in todayÕs exercise.  In each method, you will calculate an estimate for the entire U. S. based only on your sample results.  Keep track of your results, as we will pool the class results to see the effects of the different methods.

Using a small population, we will explore alternate sampling schemes.  Our population today is the 50 United States and the variable of interest is the Land Area.

Simple Random Sampling:  Randomly select 10 states.  Find the average Land Area and the standard deviation of Land Area for your 10 states.  To guess the total for the entire U. S., multiply your average by 50.

Systematic Sampling:  Using the alphabetically sorted list, choose a random number between 1 and 5.  Then select every 5th item after that.  In general, you would use a random number between 1 and N/n, and then select every N/nth item.  Find the average Land Area and the standard deviation of Land Area for your 10 states.  To guess the total for the entire U. S., multiply your average by 50.

Stratified Sampling:  From the large states (Alaska, Texas, California, Montana, New Mexico, Arizona, Nevada, Colorado, Wyoming, and Oregon), randomly select 4 states.  From the small states (Rhode Island, Delaware, Connecticut, Hawaii, New Jersey, Massachusetts, New Hampshire, Vermont, and Maryland), randomly select 1 state.  From the remaining 31 states, randomly select 5 states.

To guess the total Land Area for the entire U. S., guess the totals for the three strata separately.  That is, for the small states, multiply your average for them by 9.  For the large states, multiply by 10.  For the other states, multiply by 31.  Then add the 3 sub-totals together.  In addition, calculate the standard deviation for your 10 states chosen.

Cluster Sampling:  Using the following breakdown of the states (North, Southwest, Central, Southeast, and Northeast), randomly select two regions.  From each of the two regions, randomly select 5 states.

North:  Alaska, Washington, Oregon, Idaho, Montana, Wyoming, N. Dakota, S. Dakota, Nebraska, and Minnesota.

Southwest:  Hawaii, California, Nevada, Utah, Arizona, Colorado, New Mexico, Kansas, Oklahoma, and Texas.

Central:  Iowa, Missouri, Arkansas, Wisconsin, Michigan, Illinois, Indiana, Ohio, Kentucky, and Tennessee.

Southeast:  Louisiana, Mississippi, Alabama, Georgia, Florida, S. Carolina, N. Carolina, Virginia, W. Virginia, and Maryland.

Northeast:  Pennsylvania, Delaware, New Jersey, New York, Vermont, New Hampshire, Massachusetts, Connecticut, Rhode Island, and Maine.

Find the average Land Area and the standard deviation of Land Area for your 10 states.  To guess the total for the entire U. S., multiply your average by 50.

For our entire class, which method produced the most reliable results?  I will summarize the benefits and drawbacks of each method, despite whether we actually saw these effects in our simulation today.

 

State

Land Area

 

State

Land Area

 

State

Land Area

1

Alabama

50,744

18

Louisiana

43,562

35

Ohio

40,948

2

Alaska

571,951

19

Maine

30,862

36

Oklahoma

68,667

3

Arizona

113,635

20

Maryland

9,774

37

Oregon

95,997

4

Arkansas

52,068

21

Massachusetts

7,840

38

Pennsylvania

44,817

5

California

155,959

22

Michigan

56,804

39

Rhode Island

1,045

6

Colorado

103,718

23

Minnesota

79,610

40

South Carolina

30,109

7

Connecticut

4,845

24

Mississippi

46,907

41

South Dakota

75,885

8

Delaware

1,954

25

Missouri

68,886

42

Tennessee

41,217

9

Florida

53,927

26

Montana

145,552

43

Texas

261,797

10

Georgia

57,906

27

Nebraska

76,872

44

Utah

82,144

11

Hawaii

6,423

28

Nevada

109,826

45

Vermont

9,250

12

Idaho

82,747

29

New Hampshire

8,968

46

Virginia

39,594

13

Illinois

55,584

30

New Jersey

7,417

47

Washington

66,544

14

Indiana

35,867

31

New Mexico

121,356

48

West Virginia

24,078

15

Iowa

55,869

32

New York

47,214

49

Wisconsin

54,310

16

Kansas

81,815

33

North Carolina

48,711

50

Wyoming

97,100

17

Kentucky

39,728

34

North Dakota

68,976

 

 

 

 

Goals:        Understand the differences (good and bad) between various sampling schemes.

Skills:

¥                Systematic Sampling.  In systematic sampling, we decide on a fraction of the population we would like to sample, and then randomly select a number between 1 and N/n, where N is the population size and n is the sample size.  Then from the list of items, we take every N/nth item.  This scheme works best when we have a list of items, and it is simple to select items periodically from that list.  For example, we may have decided to take every 20th name, and there are 20 names per page.  It would be thus very convenient to take the 11th item from each page, for example.

¥                Stratified Sampling.  In stratified random sampling, we draw a simple random sample from several sub-populations, called strata.  The sample size may differ by stratum; in fact, this leads to the most efficient use of stratified sampling.  This scheme works best when there is variability between strata.  For example, one stratum may be more homogeneous than another stratum.  Then, just a few items are needed from the stratum that is homogenous, since all the items in that stratum are basically the same.  Alternatively, from a stratum that is quite diverse, you should sample more heavily.  In our example in class today, the large states are quite different from one another in areas, so we took 40 % of them.  The small states are very similar in area, so we took only 1 (10 % of them).  The remaining 31 states are in between, so we sampled 16 % of them.

¥                Cluster Sampling.  In cluster sampling, we select several groups of items.  Within each selected group, we then repeat by selecting from several smaller groups.  This process continues until we have our ultimate sampled units.  This scheme works best when we do not have a list of all the items in a population, but we have lists of groups of items, for example states or counties.  One drawback of this method is that the ultimate clusters are generally quite similar (all people living on a block in a town are generally more similar than different), so the effective sample size is lower than it appears.  To compensate, cluster samples typically have larger sample sizes.

Reading:    Section 4.1.

Day 19

Activity:    What is Randomness?  Homework 4 due today.

Our formal probability theory is based on the Òlong runÓ, but our everyday lives are dominated by the Òshort runÓ.  This contradiction will follow us throughout our discussions for the rest of the course.  I believe it is an important issue, and I will try to convince you of that too.  To begin our probability material, we will explore the most basic idea we have of randomness: coin flipping.  I believe that if we understand coin flipping, we can understand the other models of randomness we will encounter.

Today we will look at some everyday sequences to see if they exhibit ÒrandomÓ behavior.

Coin experiment 1:  Write down a sequence of HÕs and TÕs representing head and tails, pretending you are flipping a coin.  Then flip a real coin 50 times and record these 50 HÕs and TÕs.  Without knowing which list is which, in most cases I will be able to identify your real coin.

Basketball teams:  In sports you often hear about the Òhot handÓ.  We will pick a team, look at their last few games, and see if flipping a coin will produce a simulation that resembles their real performance.  Then we will examine whether we could pick out the simulation without knowing which was which.

Coin experiment 2:  Spin a penny on a flat surface, instead of tossing it into the air.  Record the percentage of heads.

Coin experiment 3:  Balance a nickel on its edge on a flat surface.  Jolt the surface enough so that the nickel falls over, and record the percentage of heads.

Goals:        Observe some real sequences of random experiments.  Develop an intuition about variability.

Skills:

¥                Recognize the feature of randomness.  In statistics, random does not mean haphazard, or without pattern.  It means, Òequal chancesÓ, from trial to trial.  While we are unable to predict what will happen on a single toss of a coin, we are able to predict what will happen in 1,000 tosses of a coin.  This is the hallmark of a random process: uncertainty in a small number of trials, but a predictable pattern in a large number of trials.

¥                Resist the urge to jump to conclusions with small samples.  Typically our daily activities do not involve large samples of observations.  Therefore our ideas of Òlong runÓ probability theory are not applicable.  You need to develop some intuition about when to believe an observed experiment, and when to doubt the results.  We will hone this intuition as we develop our upcoming inference methods.  For now, understand that you may be jumping to conclusions by just believing the simulationÕs observed value.

Reading:    Section 4.2.

Day 20

Activity:    Sample Spaces, Simulation.

If it is possible to list the entire sample space for an experiment, you can answer any probability question asked by just counting.  For example, with two dice, to find the chance of rolling doubles, look at the diagram from class and count how many of the 36 elements are doubles.  Then express the answer as a ratio (out of 36 in this case).  We will do examples like this in class before you do the three exercises below.  Sometimes, however, listing the entire sample space is a daunting task, especially if the sample space is quite large.  In those cases, it may be more appropriate to approximate the answer by simulating the experiment; that is actually perform the experiment a large number of times.  The difficulty with simulation is that we often will need an extremely large number of trials before the results are trustworthy.

Today we will try both methods.

Using both complete sampling spaces (theory) and simulation, find (or estimate) these chances:

1)     Roll two dice, one colored, one white.  Find the chance of the colored die being less than the white die.  For the simulation, try 100 dice rolls.  If you would rather use the calculator instead of actually rolling two dice 100 times, that would be fine.  One way to compare dice rolls with the calculator is to use ( randInt( 1, 6 ) < randInt( 1, 6 ) ).  (The < key is found in the TEST menu.)  This command will record a Ò1Ó if the result is true and a Ò0Ó if it is false.  Using this clever trick of the calculator, we can essentially convert dice roll numbers into yes/no responses.  Note that by continuing to hit ENTER the command will be repeated; this way you do not have to retype for every trial.  However, a more efficient method is to use seq(, found in the LIST OPS menu.  The command is seq( ( randInt( 1, 6 ) < randInt( 1, 6 ) ), X, 1, 100 ) -> L1.  The simulated results have been stored in L1, and you can now find out how many times the first die was larger than the second die by using 1-Var Stats and looking at the average, which will be a percentage now.  However, see the skill note below for simulations; you may need to do this too many times to make it worthwhile.

2)    Roll three dice and find the chance that the largest of the three dice is a 6.  (If two or more dice are tied for the largest, use that value; for example, the largest value when 6, 6, 4 is rolled is 6.)  To simulate this one, you can keep track of what the largest number is on three rolls using max( randInt( 1, 6, 3 ) ).  max is found in the LIST MATH menu.  You will still use seq, but substitute the expression with max for the expression with the < from 1.  (I underlined them for you).  After the simulation stores the results in a list, we can make a histogram to see how many 6Õs (or any other number) were produced.  Again, see the skill note below for simulations; you may need to do this too many times to make it worthwhile.

3)    Roll three dice and find the chance of getting a sum of less than 8.  You do not need to draw the entire sample space for three dice, but maybe by drawing a portion of it you will be able to see what you need to pay attention to.  I will give you hints in class when you get to this one if you ask me.  The simulation expression for this one is sum( randInt( 1, 6, 3 ) ).  sum is in the same menu as max.  Once again, see the skill note below for simulations; you may need to do this too many times to make it worthwhile.

Goals:        Create sample spaces.  Use simulation to estimate probabilities.

Skills:

¥                List simple sample spaces.  Flipping coins and rolling dice are common events to us, and listing the possible outcomes lets us explore probability distributions.  We will not delve deeply into probability rules; rather, we are more interested in the ideas of probability and I think the best way to accomplish this is by example.

¥                Know the probability rules and how to use them.  We have three rules we use primarily: the complement rule, the addition rule for disjoint events, and the multiplication rule for independent events.  The complement rule is used when the calculation of the complement turns out to be simpler than the event itself.  For example, the complement of Òat least oneÓ is ÒnoneÓ, which is a simpler event to describe.  The addition rule for disjoint events is used when we are asking about the chances of one event or another occurring.  If the two events are disjoint (they have no elements in common) we can find the chance of their union (one event or the other) by adding their individual probabilities.  The multiplication rule for independent events is used when we have a question about the intersection of two sets, which can be phrased as a question about two events occurring simultaneously.  Both sets occurring at once can be phrased as Òone event and the other event occurringÓ, so the multiplication rule is used for ÒandÓ statements.

¥                Simulation can be used to estimate probabilities, but only for a very large number of trials.  If the number of repetitions of an experiment is large, then the resulting observed frequency of success can be used as an estimate of the true unknown probability of success.  However, a ÒlargeÓ enough number of repetitions may be more than we can reasonably perform.  For example, for problem 1 today (and using some calculations from Chapter 8), a sample of 100 will give results between 32/100 and 51/100 (.32 to .51) 95 % of the time.  That may not be good enough for us.  But even with 500 rolls, the range is 187/500 to 230/500 (.374 to .460).  Eventually the answers will converge to a useful percentage; the question is how soon that will occur.  We will have answers to that question after Section 8.1.

Reading:    Section 4.3.

Day 21

Activity:    Continue coins and dice.  Introduce Random Variables and Probability Histograms.  Quiz 5 today.

To be successful at probability models, we need to understand Random Variables and Probability Distributions.  Our goal with a probability model is to predict real-world behavior.  If we have chosen a good model and know how it works, we can then make calculations that match observed results.  This will allow us to make conclusions theoretically instead of having to rely on simulations, which as we have seen may require huge sample sizes and therefore may be prohibitively costly.

Today we will continue to explore the sample spaces for the sum when rolling dice.  (We started these sample spaces in yesterdayÕs exercises.)  In addition to the sample space, we will include random variables, which are numbers assigned to each element.  For example, the sum of the two values on the dice is a random variable.  Another example is the larger of the two values on the dice.  If we keep track of the possible values (for example 2 to 12 for the sum on two dice) then we can tabulate the probabilities in a chart, which we call the probability distribution.  We will see this idea again and again as we continue towards statistical inference (Unit 3).  The most common discrete probability distribution we will encounter is the binomial distribution of Day 24.  As an introduction to this distribution, we can use PascalÕs triangle.

We will finish up the problems from Day 20, and also examine PascalÕs triangle, which is a way of figuring binomial probabilities (chances on fair coins).  Also in our tables, we will include random variables.

Goals:        Understand that variables may have values that are not equally likely.

Skills:

¥                Understand sampling distributions and how to create simple ones.  We have listed sample spaces of equally likely events, like dice and coins.  Events can further be grouped together and assigned values.  These new groups of events may not be equally likely, but as long as the rules of probability still hold, we have valid probability distributions.  PascalÕs Triangle is one such example, though you should realize that it applies only to fair coins.  We will work with Òunfair coinsÓ (proportions) later, in Chapter 8.  Historical note: examining these sampling distributions led to the discovery of the normal curve in the early 1700Õs.  We will copy their work and ÒdiscoverÓ the normal curve for ourselves too using dice.

Reading:    Section 4.4.

Day 22

Activity:    Means and Variances Rules.  Using the frequency option on STAT CALC 1-Var Stats to calculate mx and sx.  Simulating data to see the rules in action.

We can use the Frequency option of the TI-83 to find mx and sx for discrete distributions.  This option allows different weights for the data entered.  When we used STAT CALC 1-Var Stats L1 before, the values in L1 were each representing one data value each.  In probability distributions, we donÕt have actual data, but possible values, the values of the Random Variable, and their associated probabilities.  To find the mean and standard deviation of the distribution, then, we let the probabilities be the weights.  We will practice with the dice rolling results we created on Day 21.

Our second exploration today will be about the formulas for means and variances of linear combinations. Basically, we are talking about combining information by adding and subtracting, and multiplying by constants.  An example of a linear transformation is the Fahrenheit/Centigrade conversion F = (9/5) C + 32.  An example of a linear combination is total profit for a company that makes $10 each for X chairs and $30 each for Y tables; their profit would be 10 X + 30 Y.  The main example I will use is the average of a sample of measurements.  This derivation will lead to the means and variances that are critical to the Central Limit Theorem.  The main assumption we need is independence.

Goals:        Explore the rules we have for means and variances using one particular simulation of normal data.  Use the TI-83 to calculate the mean and variance for a discrete distribution.

Skills:

¥                Know how to use the TI-83 to calculate the mean and variance for a discrete distribution.  By including a variable of weights or frequencies, the TI-83 will calculate mx and sx for a discrete distribution.  The syntax is STAT CALC 1-Var Stats L1, L2, where the x-values are entered in L1 and the weights (probabilities) are entered in L2.

¥                Know how to combine Expected Value and Variance for Linear Combinations of Random Variables.  We often are interested in linear functions of two random variables.  For instance, we may know the mean and variance of times for workers traveling to a job site.  We may also know the mean and variance of the length of the job.  What we really want to know is the total length of time the workers are gone from the main plant, which is a sum of the two variables.

Reading:    Section 4.5.

Day 23

Activity:    Constructing probability trees.  Demonstrating BayesÕ formula with the rare disease problem.  Homework 5 due today.

Consider a card trick where two cards are drawn sequentially off the top of a shuffled deck.  (There are 52 cards in a deck, 4 suits of 13 ranks.)  We want to calculate the chance of getting hearts on the first draw, on the second draw, and on both draws.  We will organize our thoughts into a tree diagram, much like water flowing in a pipe.  On each branch, the label will be the probability of taking that branch; thus at each node, the exiting probabilities (conditional probabilities) add to one.

On the far right of the tree, we will have the intersection events.  Their probability is found by multiplying all the probabilities on the branches in the tree leading to this end result.  At each branch, we have a conditional probability, which is a probability calculation based on what has already occurred to get to where we are in the tree.  The notation we will use is P(A|B) which reads Òthe probability of event A given that event B has occurredÓ.  The multiplication formula we use in trees is P(A‚B) = P(A) P(B|A).  Note that this can also be written with the roles of A and B reversed: P(A‚B) = P(B) P(A|B).  Combining these two formulas gives us the basis of BayesÕ formula:  P(A) = P(B) P(A|B) / P(A).  I will demonstrate how to use these conditional formulas using the rare disease problem today.  Essentially, BayesÕ Theorem allows us to reverse the conditions in a conditional probability.  This commonly occurs when data or information is available in one direction or order of events, but the interesting questions involve the reverse direction or order.

For the 2-cards problem, calculate the chances of:

1)     Drawing a heart on the first card.

2)    Drawing two hearts.

3)    Drawing a heart on the second card.

4)    Drawing at least one heart.

5)    Drawing a heart on the second draw given that a heart was drawn first.

6)    Drawing a heart on the first draw given that a heart was drawn second.

Now we will do this work for the rare disease problem.

Goals:        Be able to express probability calculations as tree diagrams.  Be able to reverse the events in a probability tree, which is what BayesÕ formula is about.

Skills:

¥                Know how to use the multiplication rule in a probability tree.  Each branch of a probability tree is labeled with the conditional probability for that branch.  To calculate the joint probability of a series of branches, we multiply the conditional probabilities together.  Note that at each branching in a tree, the (conditional) probabilities add to one, and that overall, the joint probabilities add to one.

¥                Recognize conditional probability in English statements.  Sometimes the key word is ÒgivenÓ.  Other times the conditional phrase has ÒifÓ.  But sometimes the fact that a statement is conditional is disguised.  For example:  ÒAssuming John buys the insurance, what is the chance he will come out aheadÓ is equivalent to ÒIf John buys insurance, then what is the chance he will come out aheadÓ.

¥                Be able to use the conditional probability formula to reverse the events in a probability tree.  The key here is the symmetry of the events in the conditional probability formula.  We exchange the roles of A and B, and tie them together with our formula for Pr(A and B).  This reversal is the essence of BayesÕ formula.

¥                Know the definition of independence.  Independence is a fact about probability, not about sets.  Contrast this to ÒdisjointÓ which is a property of sets.  In particular, independent events are by definition not disjoint.  Independence is important later as an assumption as it allows us to multiply individual probabilities together without having to worry about conditional probability.

Reading:    Section 5.1.

Day 24

Activity:    Coin flipping.  Quiz 6 today.

The binomial distribution is a common example of probability in the real world.  Many processes satisfy the conditions, and it is a quite useful model.  You need to become familiar with the conditions, and the ways we calculate probabilities.  Also, you must be able to recognize situations where a binomial model would be appropriate.

Coin flipping is a good way to understand randomness, but because most coins have a probability of heads very close to 50 %, we donÕt get the true flavor of the binomial distribution.  Today we will simulate the flipping of an unfair coin; that is, a binomial process with probability not equal to 50 %.

Experiment 1:  Our unfair ÒcoinÓ will be a die, and we will call getting a 6 a success.  Roll your die 10 times and record how many sixes you got.  Repeat this process 10 times each.  Your group should have 40 to 50 trials of 10 die rolls.  Pool your results and enter the data into a list on your calculator.  We want to see the histogram (be sure to make the box width reasonable) and calculate the summary statistics, in particular the mean and variance.  Also produce a quantile plot.  Compare the simulated results with theory.  (Of course, you may use the TI-83 to ÒrollÓ your die if you like.  See Day 20.)

Experiment 2:  Your calculator will generate binomial random variables for you, but it is not as illuminative as actually producing the raw data yourself.  Still, we can see the way the probability histogram looks (if we generate enough cases; this is an application of the law of large numbers).  I suggest 100 at a minimum.  Again be sure to make your histogram have an appropriate box width.  The command is MATH PRB randBin( n, p, r ), where n is the sample size, p is the probability of success, and r is the number of times to repeat the experiment.

Using DISTR binompdf( and/or DISTR binomcdf(, calculate these probabilities:

1)     What is the chance of 4 or fewer successes with n = 10 and p = .4?

2)    What is the chance of 3 to 10 successes, inclusive, with n = 20 and p = .7?

3)    What is the chance of more than 6 successes with n = 15 and p = .2?

Goals:        Become familiar with the binomial distribution and its applications.

Skills:

¥                Know the four assumptions.  The binomial distribution requires four assumptions:  1) There must be only two possible outcomes to each trial of the experiment, commonly called ÒsuccessesÓ and ÒfailuresÓ.  2) Trials must be independent of one another.  3) There must be a fixed sample size, chosen ahead of time and not determined while the experiment is running.  4) The probability of success is the same from trial to trial.  Number 2 is the most difficult to check, so it is usually simply assumed.  Number 4 rules out finite populations, where the success probability depends on what is left in the pool.  Number 3 rules out situations where the experiment continues until some event happens, such as repeating a coin flipping experiment until the 3rd head occurs.

¥                Know how to calculate binomial probabilities.  Using DISTR binompdf( n, p, x ), we can calculate the chance of getting exactly x successes in n trials with constant probability p.  Using DISTR binomcdf( n, p, x ), we can calculate the chance of getting at most x successes in n trials with constant probability p.  The second command adds up repeated results of the first command, starting at x and working down to zero.  Because binomcdf only calculates x successes or less, when we want to know about y successes or more, we must use the complement rule of probability.  To find the chance of y successes or more, use y - 1 successes or less.  For example, the chance of 70 or more successes is found this way:  Pr(X ³ 70) = 1 - Pr(X ² 69).

¥                Mean and Variance.  With algebra, we can show that the mean of a binomial random variable is n p, and the variance is n p (1 - p ).  You should memorize these two facts, but donÕt worry about how they are proven.

¥                Continuity correction.  We can use the normal curve to approximate binomial probabilities.  But because the binomial is a discrete distribution and the normal is continuous, we need to adjust our endpoints by ½ unit.  I recommend drawing a diagram with rectangles to see which way the ½ unit goes.  For example, calculating the chance of getting between 40 and 50 heads inclusive on 100 coin flips entails using 39.5 to 50.5.

Reading:    Section 5.2.

Day 25

Activity:    Central Limit Theorem exploration.

Our last topic in Unit 2 is the important Central Limit Theorem.  In your everyday lives, you have already seen the evidence of this famous theorem, because most of the normal distributions youÕve encountered are results of this theorem.  There are two main features of the Central Limit Theorem material that will be important to us.  One is that averages are more reliable than individual measurements.  This says that the average for a bunch of items in a sample is likely to be closer to the true population average than a single measurement.  The other main conclusion is that while the sample average is getting closer to the population average, it is also getting closer in an organized way; in fact, the normal curve describes the sampling distribution of the average quite well for large samples.  Many processes in nature are the result of sums, so the normal distribution is quite common in nature.

One of the problems students have with the Central Limit Theorem is recognizing that the problem at hand is about sample averages instead of individual measurements.  It is critical for you to have this ability.  One way to think of the situation is that all problems are really Central Limit Theorem problems, but some of them are just samples of size 1 (n = 1).  All of our problems earlier in Section 1.3 were of this n = 1 nature.  The calculation for the spread of an average (s = s/Ãn ) always works for the appropriate standard deviation for an average, even when n = 1.

Another problem I have noticed is dealing with sums rather than averages.  There are two approaches, and we will look at both in 3) below.  The first method is to label the normal curve with both the scale for the average and the scale for the sum.  You first find the ÒaverageÓ scale using the standard deviation for the average, as we have been doing.  Then the ÒsumÓ scale is the ÒaverageÓ scale multiplied by n.  The second method is to calculate the standard deviation for the sum directly, using ssum = s Ãn.  Notice, for standard deviations, that for the average, you divide by the square root of n whereas for the sum you multiply by it.  That is how I remember which is which: sums are larger than averages.  These new standard deviations (for the sum or for the average) are called Òstandard errorsÓ, a new term to keep our various standard deviations straight.

Important Note: We have many standard deviations we have been referring to.  It is important for you to keep them all straight.  We have the standard deviation in a population of items, s.  We have the standard deviation in a sample from a population, s.  We have the standard deviation of a series of trials, like sums, or averages, and we call these Òstandard errorsÓ.  We have the standard deviation of a probability model, which is much like the population standard deviation s.  I know all these terms are confusing, using the same word to describe them, but that is why your practicing problems and having discussions with me are important.

In addition to coins and dice, MATH PRB rand on your calculator is another good random mechanism for exploring Òsampling distributionsÓ.  These examples will give you some different views of sampling distributions.  The important idea is that each time an experiment is performed, a potentially different result occurs.  How these results vary from sample to sample is what we seek.  You are going to produce many samples, and will therefore see how these values vary.

1)     Sums of two items: Each of you will roll two dice.  Record the sum on the dice.  Repeat this 100 times, generating 100 sums.  Make a histogram and a QUANTILE plot of your 100 sums.  Compare to the graphs of people around you, particularly noting the shape.  Sketch the graphs you made and compare to the theoretical results.  A simulation on the calculator uses seq ( sum( randInt( 1, 6, 2 ) ), X, 1, 100 ) -> L1.  The 100 two-dice sums are stored in L1.  (seq is found in the LIST OPS menu.)

2)    Averages of 4 items: Each of you will generate 4 random numbers on your calculator, add them together, average them, and record the result; repeat 100 times.  The full command is seq ( rand + rand + rand + rand, X, 1, 100 ) / 4 -> L2, which will generate 100 four-number averages and store them in L2.)  Again, make a graph of the distribution.  (seq is found in the LIST OPS menu.)

3)    Sums of 12 items: Each of you generate 12 random normal numbers on your calculator using MATH PRB randNorm( 65, 5, 12).  Add them together and record the result; repeat 100 times.  Use seq (sum ( randNorm( 65, 5, 12 ) ), X, 1, 100 ) -> L3.)  Again, make a graph of the distribution.  (This example could be describing how the weight of a dozen eggs is distributed.)  (sum is found in the LIST MATH menu.)

For all the lists you generated, calculate the standard deviation and the mean.  We will find these two statistics to be immensely important in our upcoming discussions about inference.  It turns out that these means and standard deviations can be found through formulas instead of having to actually generate repeated samples.  These means depend only on the mean and standard deviation of the original population (the dice or rand or randNorm in this case) and the number of times the dice were rolled or rand was pressed (called the sample size, denoted n).

Goals:        Examine histograms or quantile plots to see that averages are less variable than individual measurements.  Also, the shape of these curves should get closer to the shape of the normal curve as n increases.

Skills:

¥                Understand the concept of sampling variability.  Results vary from sample to sample.  This idea is sampling variability.  We are very much interested in knowing what the likely values of a statistic are, so we focus our energies on describing the sampling distributions.  In todayÕs exercise, you simulated samples, and calculated the variability of your results.  In practice, we only do one sample, but calculate the variability with a formula.  In practice, we also have the Central Limit Theorem, which lets us use the normal curve in many situations to calculate probabilities.

Reading:    Section 5.2.

Day 26

Activity:    Practice Central Limit Theorem (CLT) problems.  Homework 6 due today.

The Central Limit Theorem is remarkable in that it allows us to use the normal curve to find probabilities for distributions that are not normal, but only if we are talking about a sum or an average from a large enough sample.  Unfortunately, we sometimes forget that we cannot use the normal curve on small samples.  We should check to see how close the actual data is to the normal curve, though, because if the original data is reasonably symmetric, with no outliers or extreme skewness, then samples of 20 or 30 may be large enough to use the normal curve approximation.  I admit that this determination is a judgment call; there are fancier techniques to determine this that we are skipping over.  I donÕt expect you to always know if the sample size is large enough; I do expect that you know about the issue, and that you have looked at the data to at least check for outliers.

We will have examples of non-normal data and normal data to contrast the diverse cases where the CLT applies. 

1)     People staying at a certain convention hotel have a mean weight of 180 pounds with standard deviation 35.  The elevator in the hotel can hold 20 people.  How much weight will it have to handle in most cases?  Do we need to assume weights of people are normally distributed?

2)    Customers at a large grocery store require on average 3 minutes to check out at the cashier, with standard deviation 2.  Because checkout times cannot be negative, they are obviously not normally distributed.  Can we calculate the chance that 85 customers will be handled in a four-hour shift?  If so, calculate the chance; if not, what else do you need to know?

3)    Suppose the number of hurricanes in a season has mean 6 and standard deviation Ã6.  What is the chance that in 30 years the average has been less than 5.5 hurricanes per year?  What assumptions must you make?

4)    Suppose the students at UWO take an average of 15 credits each semester with a standard deviation of 3 credits.  What is the chance that the average in a sample of 20 randomly chosen UWO students is between 14.5 and 15.5?

5)    The number of boys in a 4-child family can be modeled reasonably well with the binomial distribution.  If ten such families live on the same street, what is the chance that the total number of boys is 24 or more?  What assumptions must you make?

Goals:        Use normal curve with the CLT.

Skills:

¥                Recognize how to use the CLT to answer probability questions concerning sums and averages.  The CLT says that for large sample sizes, the distribution of the sample average is approximately normal, even though the original data in a problem may not be normal.  Sums are simply averages multiplied by the sample size, so any question we can answer about averages, we can also answer about the corresponding sums.

¥                For small samples, we can only use the normal curve if the actual distribution of the original data is normally distributed.  It is important to realize when original data is not normal, because there is a tendency to use the CLT even for small sample sizes, and this is inappropriate.  When the CLT does apply, though, we are armed with a valuable tool that allows us to estimate probabilities concerning averages.  A particular example is when the data is a count that must be an integer, and there are only a few possible values, such as the number of kids in a family.  Here the normal curve wouldnÕt help you calculate chances of a single family having 3 kids.  However, we could calculate quite accurately the number of kids in 100 such families.

Reading:    Chapters 3 to 5.

Day 27

Activity:    Catch up day and Review of Probability.  Quiz 7 today.

Reading:    Chapters 3 to 5.

Day 28

Activity:    Presentation 2.  Sampling and Probability (Chapters 3 and 4).

Sample 20 students from UWO.  For each student, record the number of credits they are taking this semester, what year they are in school, and whether or not they are graduating this semester.  Summarize each variable for us, using both graphical and numerical summaries, as appropriate.  At the very least, report to us the average credit load and the standard deviation of credit loads.  Make your sample as representative as you can, but you must have some sort of a probability sample to get full credit.  Discuss the biases your sample has and what you did to try to avoid bias.

Your sample has a particular number of men.  Using whatever resources you feel appropriate, calculate and present to us the chance that your sample had the number of males it had.  Be sure to justify your assumptions.

Reading:    Chapters 3 to 5.

Day 29

Activity:    Exam 2.

This second exam is on sampling, experiments, and probability, including sampling distributions.  Some of the questions may be multiple choice.  Others may require you to show your worked out solution.  Section reviews are an excellent source for studying for the exams.  DonÕt forget to review your class notes and these on-line notes.

Reading:    Section 6.1.  (Skip Bootstrap.  Skip Choosing the Sample Size.)

Day 30

Activity:    Guess m&mÕs percentage.

Unit 3 is about statistical inference, where we collect a sample of data and try to make statements about the entire population the sample came from.  The classic situation is where we have an actual population of individuals from which we have chosen a sample, for example pre-election polls.  A more prevalent use of statistical inference though is in the situation where we extend our results from the current individuals to the entire population of individuals, even though we donÕt have a random sample.  An example of this kind of inference is when we use subjects in an experiment and then make claims about the whole population.  If it turns out our subjects were not representative in some important way, then our calculations will be ridiculous.  For example, does it matter to us that saccharin causes cancer in laboratory rats?  Are we claiming that rat physiology is like human physiology?  I do not know the answer to that question, but it is at the heart of the inference.

There are two main tools of statistical inference, confidence intervals and hypothesis testing.  The first one we will examine is confidence intervals, where we ÒguessÓ the population value using the sample information.  We know our guess isnÕt right though; what is the probability that our example is perfectly representative!  So, to convey our belief that we are close but not exact, we report an interval of guesses.  Ideally this interval would be narrow; then we would be saying our guess is very close to the truth.  In practice though it is hard to get narrow intervals, because as we will see the narrow intervals are associated with large samples and large samples are expensive.

What fraction of m&mÕs are blue or green?  Is it 25 %?  33 %?  50 %?  We take samples today to find out.  Technically the samples we take today will not be SRSÕs; we are not labeling the m&mÕs individually and then randomly selecting them by number.  However, if the population is sufficiently mixed, I believe we can use our results and have reasonable faith that they make sense.  (Note that this situation differs from samples of convenience such as asking students in Reeve about campus life.)

Each of you will sample from my jar of m&mÕs, and you will all calculate your own confidence interval.  Of course, not everyone will be correct, and in fact, some of us will have ÒlousyÓ samples.  But that is the point of the confidence coefficient, as we will see when we jointly interpret our results.

It has been my experience that confidence intervals are easier to understand if we talk about sample proportions instead of sample averages.  Thus I will use techniques from Chapter 8.  Each of you will have a different sample size and a different number of successes.  In this case the sample size, n, is the total number of m&mÕs you have selected, and the number of successes, x, is the total number of blue or green m&mÕs in your sample.  Your guess is simply the ratio x/n, or the sample proportion.  We call this estimate p-hat or .  Use STAT TEST 1-PropZInt with 70 % confidence for your interval here today.  However, in practice you will rarely see anyone using less than 90 % confidence.

When you have calculated your confidence interval, record your result on the board for all to see.  We will jointly inspect these confidence intervals and observe just how many are ÒcorrectÓ and how many are ÒincorrectÓ.  The percentage of correct intervals should match our chosen level of confidence.  This is in fact what is meant by confidence.

Goals:        Introduce statistical inference - Guessing the parameter.  Construct and interpret a confidence interval.

Skills:

¥                Understand how to interpret confidence intervals.  The calculation of a confidence interval is quite mechanical.  In fact, as we have seen, our calculators do all the work for us.  Our job is then not so much to calculate confidence intervals as it is to be able to understand when one should be used and how best to interpret one.

¥                Know what the confidence level measures.  We would like to always make correct statements, but in light of sampling variability we know this is impossible.  As a compromise, we use methods that work most of the time.  The proportion of times our methods work is expressed as a confidence coefficient.  Thus, a 95 % confidence interval method produces correct statements 95 % of the time.  (By Òcorrect statementÓ we mean one where the true unknown value is contained in our interval, that is, it is between our two numbers.)

Reading:    Section 6.1.  (Skip Bootstrap.  Skip Choosing the Sample Size.)

Day 31

Activity:    Changing confidence levels and sample sizes.

We want narrow confidence intervals.  What effect do the confidence level and the sample size have on interval width?  Remember, what we are using this technique for is to guess the true parameter, in these cases population percentage (the first chart) or population average (the second chart).  We have used and interpreted the command for the first chart.  For the second chart, we will use a new command today, the ZInterval.  We interpret this command the same as the other: it gives a range of guesses, and our confidence is expressed as a percentage telling us how often the technique produces correct answers.

Today we will explore how changing confidence levels and sample sizes influence CIÕs.  Complete the following table, filling in the confidence interval width in the body of the table.  Use STAT TEST 1-PropZInt but in each case make x close to 50 % of n.  (The calculator will not let you use non-integers for x; round off if needed.)

Confidence Level =======> Sample Size

70 %

90 %

95 %

99 %

99.9 %

10

 

 

 

 

 

20

 

 

 

 

 

50

 

 

 

 

 

100

 

 

 

 

 

1000

 

 

 

 

 


We will try to make sense of this chart, keeping in mind the meaning of confidence level, and the desire to have narrow intervals.

Now repeat the above table using STAT TEST ZInterval, with s = 15 and  = 100.

Confidence Level =======> Sample Size

70 %

90 %

95 %

99 %

99.9 %

10

 

 

 

 

 

20

 

 

 

 

 

50

 

 

 

 

 

100

 

 

 

 

 

1000

 

 

 

 

 

Goals:        See how the TI-83 calculates our CIÕs.  Interpret the effect of differing confidence coefficients and sample sizes.

Skills:

¥                Understand the factors that make confidence intervals believable guesses for the parameter.  The two chief factors that make our confidence intervals believable are the sample size and the confidence coefficient.  The key result is larger confidence makes wider intervals, and larger sample size makes narrower intervals.

¥                Know the details of the Z Interval.  When we know the population standard deviation, s, our method for guessing the true value of the mean, m, is to use a z confidence interval.  This technique is unrealistic in that you must know the true population standard deviation.  In practice, we will estimate this value with the sample standard deviation, s, but a different technique is appropriate (See Day 35).

Reading:    Sections 6.2 and 6.4.

Day 32

Activity:    Argument by contradiction and the Scientific method.  Type I and Type II error diagram.  Courtroom terminology.  Homework 7 due today.

We have been introduced to the first of our two statistical inference techniques.  Today we will begin our introduction to the other technique: statistical hypothesis testing.  In this technique, we again collect a sample of data, but this time instead of guessing the value of the parameter, we check to see if the sample is close to some pre-specified value for the parameter.  This pre-specified value is the statistical hypothesis, a statement of equality.  The result of our test will be either ÒI believe the hypothesis is reasonableÓ or Òthe hypothesis looks wrongÓ.  Just as in confidence intervals, we will give a percentage that can be interpreted as the success rate of our method.

Some terminology:

Null hypothesis.  The null hypothesis is a statement about a parameter.  The null hypothesis is always an equality or a single claim (like two variables are independent).  We assume the null hypothesis is true in our ensuing calculations, so it is important that the null be a specific value or fact that can be assumed.

Alternative hypothesis.  The alternative hypothesis is a statement that we will believe if the null hypothesis is rejected.  The alternative does not have to be the complement of the null hypothesis.  It just has to be some other statement.  It can be an inequality, and usually is.

One- and Two-Tailed Tests.  A one-tailed test is one where the alternative hypothesis is in only one direction, like Òthe mean is less than 10Ó.  A two-tailed test is one where the alternative hypothesis is indiscriminate about direction, like Òthe mean is not equal to 10Ó.  When a researcher has an agenda in mind, he will usually choose a one-tailed test.  When a researcher is unsure of the situation, a two-tailed test is appropriate.

Rejection rule.  To decide between two competing hypotheses, we create a rejection rule.  ItÕs usually as simple as ÒReject the null hypothesis if the sample mean is greater than 10.  Otherwise fail to reject.Ó  We always want to phrase our answer as Òreject the null hypothesisÓ or Òfail to reject the null hypothesisÓ.  We never want to say, ÒAccept the null hypothesisÓ.  The reasoning is this: rejecting the null hypothesis means the data have contradicted the assumptions weÕve made (assuming the null hypothesis was correct); failing to reject the null hypothesis doesnÕt mean weÕve proven the null hypothesis is true, but rather that we havenÕt seen anything to doubt the claim yet.  It could be the case that we just havenÕt taken a large enough sample yet.

Type I Error.  When we reject the null hypothesis when it is in fact true, we have made a Type I error.  We have made a conscious decision to treat this error as a more important error, so we construct our rejection rule to make this error rare.

Type II Error.  When we fail to reject the null hypothesis, and in fact the alternative hypothesis is the true one, we have made a Type II error.  Because we construct our rejection rule to control the Type I error rate, the Type II error rate is not really under our control; it is more a function of the particular test we have chosen.  The one aspect we can control is the sample size.  Generally, larger samples make the chance of making a Type II error smaller.

Significance level, or size of the test.  The probability of making a Type I error is the significance level.  We also call it the size of the test, and we use the symbol a to represent it.  Because we want the Type I error to be rare, we usually will set a to be a small number, like 0.05 or 0.01 or even smaller.  Clearly smaller is better, but the drawback is that the smaller a is, the larger the Type II error becomes.  Remember, we often express this error rate as a percentage, so 0.05 is also 5 %.

P-value.  There are two definitions for the P-value.  Definition 1: The P-value is the alpha level that will cause us to just reject our observed data.  Definition 2: The P-value is the chance of seeing data as extreme or more extreme than the data actually observed.  Using either definition, we calculate the P-value as an area under a tail in a distribution.  Caution: the P-value calculation will depend on whether we have a one- or a two-tailed test.

Power.  The power of a test is the probability of rejecting the null hypothesis when the alternative hypothesis is true.  We are calculating the chance of making a correct decision.  Because the alternative hypothesis is usually not an equality statement, it is more appropriate to say that power is a function rather than just a single value.

We will examine these ideas using the z-test.  The TI-83 command is STAT TEST ZTest.  The command gives you a menu of items to input.  It assumes your null hypothesis is a statement about a mean m.  You must tell the assumed null value, m0, and the alternative claim, either the two-sided choice, or one of the one-sided choices.  You also need to tell the calculator how your information has been stored, either as a list of raw DATA or as summary STATS.  If you choose CALCULATE the machine will simply display the test statistic and the P-value.  If you choose DRAW, the calculator will graph the P-value calculation for you.  You should experiment to see which way you prefer.

Goals:        Introduce statistical inference - Hypothesis testing.

Skills:

¥                Recognize the two types of errors we make.  If we decide to reject a null hypothesis, we might be making a Type I error.  If we fail to reject the null hypothesis, we might be making a Type II error.  If it turns out that the null hypothesis is true, and we reject it because our data looked weird, then we have made a Type I error.  Statisticians have agreed to control this type of error at a specific percentage, usually 5 %.  On the other hand, if the alternative hypothesis is true, and we fail to reject the null hypothesis, we have also made a mistake.  We generally do not control this second type of error; the sample size is the determining factor.

¥                Understand why one error is considered a more serious error.  Because we control the frequency of a Type I error, we feel confident that when we reject the null hypothesis, we have made the right decision.  This is how the scientific method works; researchers usually set up an experiment so that the conclusion they would like to make is the alternative hypothesis.  Then if the null hypothesis (usually the opposite of what they are trying to show) is rejected, there is some confidence in the conclusion.  On the other hand, if we fail to reject the null hypothesis, the most useful conclusion is that we didnÕt have a large enough sample size to detect a real difference.  We arenÕt really saying we are confident the null hypothesis is a true statement; rather we are saying it could be true.  Because we cannot control the frequency of this error, it is a less confident statement.

Reading:    Section 6.2.

Day 33

Activity:    Practice problems on hypothesis testing.  Introduce z-test as an example.  Quiz 8 today.

One difficulty I see students having with hypothesis testing is setting up the hypotheses.  Today we will work some examples so you become familiar with the process and the interpretations.  Before we begin the examples, though, I want to mention several ideas.

Researchers generally have agendas in their research.  After all, they are trying to show their ideas are true.  But we require a statement of equality in our statistical inference, so the strategy generally goes this way: assume the opposite of what you are trying to prove.  When the data donÕt look right under that assumption, you have ÒprovenÓ the equality is untrue, or very unlikely.  You can then believe your original idea was the right one.  For example, a researcher thinks adding fertilizer to plants will increase growth.  She will assume for the sake of argument that the fertilizer doesnÕt work.  Her null hypothesis is that the mean growth will be exactly zero.  After she collects some data using fertilizer on plants, she will be able to say the data agree with the Òno-growthÓ claim or that they disagree with the claim, in which case she will happily assume that fertilizer instead helps plants grow.

The notation we use for hypotheses is shorthand.  Instead of saying ÒThe null hypothesis is that the population mean is 20Ó we say ÒH0: m = 20Ó.  Because it is a sentence, you need a subject and a verb.  The subject is H0.  The verb is the colon.  m = 20 is the clause.  It would be incorrect to write just H0: = 0.  That would be like saying ÒThe null hypothesis is that equals zero.Ó

We have two interpretations of P-values, as explained on Day 32.  To perform a statistical hypothesis test using P-values, we compare the calculated P-value to the pre-chosen significance level, often 5 %.  If the P-value is small (at or below 5 %) we reject the claim, stating that the data contradict the null hypothesis and we then conclude the alternate hypothesis is more believable.  If the P-value is large (above 5 %) we do not have any reason to think the null hypothesis is incorrect.  The data ÒagreeÓ with the claim.  We Òfail to rejectÓ the null hypothesis and conclude that it could indeed be true.  Notice the weak language:  ÒcouldÓ instead of Òis definitelyÓ.  When we reject the null hypothesis, we are saying with high confidence that the null isnÕt right.  When we fail to reject the null hypothesis we are saying it is plausible, but we are not saying it is for sure true.  We will explore this particular phenomenon more on Day 34.

The second interpretation for a P-value is that it is the chance of seeing more extreme results than those observed.  It is our measure of how unusual our current results are.  If our P-value is 0.25, or 25 %, we say, Òthere is a 25 % chance of seeing data like this or worse when the null hypothesis really is trueÓ.  That doesnÕt seem so unusual then.  Conversely, if the P-value is 0.03, or 3 %, we say there is only a 3 % chance of seeing data this extreme or worse.  That seems too unusual to us, and we therefore conclude that the claim on which this calculation was based is wrong.  Specifically, we assume the alternate hypothesis to be the correct sentence.

Work on the following problems.  Compare answers with your neighbor or your group.  We will summarize before the end of class.

1)    An educational researcher wants to show that the students in his school district score better than the national average, (which is 65) on a standardized test.  He collects a sample of 20 test scores, and finds the sample average is 66.  If s is 3, what is his conclusion?  Set up the hypothesis and perform the test.  Use a = 0.05.  What assumptions is the researcher making?

2)    A downhill skier makes 5 timed runs on a course with these times in seconds: 52, 56, 55, 58, and 51.  In the past, this skierÕs times have had a standard deviation of 2, so you can assume s is 2.  To qualify for a local competition, a skier must beat 55 seconds.  Is there evidence that our skierÕs true ability is enough to qualify for the local competition?  Set up the hypothesis and perform the test.  Use a = 0.05.  What assumptions must we make about the skierÕs times?

3)    You work for a company that makes nuts for bolts.  You have to calibrate a machine to put 100 nuts in a bag.  (The label on the bag says, ÒContains 100 nutsÓ.)  Suppose you run the machine 10 times, and find the average is 98.  Assume the standard deviation is 1.  Is the machine working well enough or do you have to adjust it?

Goals:        Practice contradiction reasoning, the basis of the scientific method.

Skills:

¥                Become familiar with Òargument by contradictionÓ.  When researchers are trying to ÒproveÓ a treatment is better or that their hypothesized mean is the right one, they will usually choose to assume the opposite conclusion as their null hypothesis.  For election polls, they assume the candidate has 50 % of the vote, and hope to show that is an incorrect statement.  For showing that a local population differs from, say, a national population, they will typically assume the national average applies to the local population, again with the hope of rejecting that assumption.  In all cases, we formulate the hypotheses before collecting data; therefore, you will never see your sample average in either a null or alternative hypothesis.

¥                Understand why we reject the null hypothesis for small P-values.  The P-value is the probability of seeing a sample result ÒworseÓ than the one we actually saw.  In this sense, ÒworseÓ means even more evidence against the null hypothesis, or even more evidence favoring the alternative hypothesis.  If this probability is small, it means either we have observed a rare event, or that we have made an incorrect assumption, namely the value in the null hypothesis.  Statisticians and practitioners have agreed that 5 % is a reasonable cutoff between a result that contradicts the null hypothesis and a result that could be argued to be in agreement with the null hypothesis.  Thus, we reject our claim only when the P-value is a small enough number.

Reading:    Section 6.2.

Day 34

Activity:    Testing Simulation.

Today I want to give you practice performing hypothesis tests, and interpreting results.  This is a day for understanding, not something you would do in practice.  Work in pairs; one of you will generate some data for the other person.  The other person will conduct a test and decide whether the first person chose 20 or not for the mean.  Then you will repeat the procedure a number of times.  Each time, the second person will not know what value the first person chose, but the second person will always answer with ÒYou could have used 20Ó (Failing to reject the null hypothesis) or ÒI believe you did not use 20Ó (Rejecting the null hypothesis).

In this experiment, you will work in pairs and generate data for your partner to analyze.  Your partner will come up with a conclusion (either reject the null hypothesis or fail to reject the null hypothesis) and you will let them know if they made the right decision or not.  Keep careful track of the success rates.  Record the sample averages and the P-values so we can check our work.

For each of these simulations, let the null hypothesis mean be 20, n = 10, and s = 5.  You will let m change for each replication.

1)     Without your partner knowing, choose a value for m from this list: 16, 18, 20, 22, or 24.  Then use your calculator and generate 10 observations.  Use MATH PRB randNorm( M, 5, 10 ) -> L1 where M is the value of m you chose for this replication.  Clear the screen (so your partner canÕt see what you did) and give them the calculator.  They will perform a hypothesis test using the 0.05 significance level and tell you their decision.  Remember, each test uses the same null hypothesis:  ÒThe mean is 20.Ó vs. the alternative ÒThe mean is different from 20Ó.  The decision will be either ÒReject the null hypothesisÓ or ÒFail to reject the null hypothesisÓ.  Finally, compare the decision your partner made to the true value of m that you chose.

2)    Repeat step 1 until you have each done at least 10 hypothesis tests; it is not necessary to have each value of m exactly twice, but try to do each one at least once.  Do m = 20 at least twice each.  (We need more cases for 20 because weÕre using a small significance level.)

3)    Keep track of the results you got (number of successful decisions and number of unsuccessful decisions) and report them to me so we can all see the combined results.

Goals:        Interpret significance level.  Observe the effects of different values of the population mean.  Recognize limitations to inference.  Realize the potential abuses of hypothesis tests.

Skills:

¥                Interpret significance level.  Our value for rejecting, usually .05, is the percentage of the time that we falsely reject a true null hypothesis.  It does not measure whether we had a random sample; it does not measure whether we have bias in our sample.  It only measures whether random data could look like the observed data.

¥                Understand how the chance of rejecting the null hypothesis changes when the population mean is different than the hypothesized value.  When the population mean is not the hypothesized value, we expect to reject the null hypothesis more often.  This is reasonable, because rejecting a false null hypothesis is a correct decision.  Likewise, when the null hypothesis is in fact true, we hope to seldom decide to reject.  If we have generated enough replications in class, we should see a power curve emerge that tells us how effective our test is for various values of the population mean.

¥                Know the limitations to confidence intervals and hypothesis tests.  Section 6.3 has examples of when our inference techniques are inappropriate.  The main points to watch for are non-random samples, misinterpreting what Òrejecting the null hypothesisÓ means, and misunderstanding what error the margin of error is measuring.  Be sure to read the examples in Section 6.3 carefully as I will not go over them in detail in class.

Reading:    Section 6.3.

Day 35

Activity:    Gosset Simulation.  Homework 8 due today.

Today, a little history into the problem discovered 100 years ago with small samples.  A century ago, William Sealy Gosset noticed that for small samples, his confidence intervals were not as correct as predicted.  For example, he was supposed to capture the parameter in his intervals 95 % of the time and found he was only accomplishing this 70 % of the time.  The problem was not knowing the true population standard deviation.  At the time, the standard practice was to just use the sample standard deviation as if it were the unknown population standard deviation.  This turns out to be the wrong way to do things, and Gosset guessed what the true distribution should be.  A few years later Sir R. A. Fisher proved that GossetÕs guess was correct, and the statistical community accepted the t distribution, as it came to be known.  Gosset was unable to publish his results under his own name (to protect trade secrets), so he used the pseudonym ÒA. StudentÓ.  You will therefore sometimes see the t distribution referred to as ÒStudentÕs t distributionÓ.

In practice, we seldom know the true population standard deviation, so we need to select the t procedures instead of the z procedures in our TI-83.  Our interpretations of confidence intervals and hypothesis tests are exactly the same.

Take samples of size 5 from a normal distribution.  Use s instead of s in the standard 95 % confidence z-interval.  Repeat 100 times to see if the true coverage is 95 %.  (My program GOSSET accomplishes this.)  We will pool our results to see how close we are to 95 %. 

While we will use the TI-83 to calculate confidence intervals, it will be helpful to know the formulas in addition.  All of the popular confidence intervals are based on adding and subtracting a margin of error to a point estimate.  This estimate is almost always an average, although in the case of proportions it is not immediately clear that it is an average.

Goals:        Introduce t-test.  Understand how the z-test is inappropriate in most small sample situations.

Skills:

¥                Know why using the t-test or the t-interval when s is unknown is appropriate.  When we use s instead of s and do not use the correct t distribution, we find that our confidence intervals are too narrow, and our hypothesis tests reject H0 too often.  We pay a ÒpenaltyÓ for not knowing the true population standard deviation.

¥                Realize that the larger the sample size, the less serious the problem.  When we have larger sample sizes, say 15 to 20, we notice that the simulated success rates are much closer to the theoretical.  Thus the issue of t vs. z is a moot point for large samples.  I still recommend using the t-procedures any time you do not know s.

Reading:    Section 7.1.

Day 36

Activity:    Matched Pairs vs. 2-Sample.  Quiz 9 today.

Matched Pairs problems are really one sample data sets disguised as two sample data sets because two measurements on the same subject are taken.  Sometimes ÒsubjectÓ is a person; other times it is less recognizable, such as a year.  The key issue is that two measurements have been taken that are related to one another.  One quick way to tell if you have a two-sample problem is whether the lists are of different lengths.  Obviously if the lists are of different lengths, they are not paired together.  Naturally the tricky situation is when the lists are of the same length, which occurs often when researchers assign the same number of subjects to each of treatment and control groups.  Another way to assess that you have matched pairs is to imagine rearranging one of the lists.  Would we still be able to analyze the data or would the rearranging make the results meaningless?  Do the first items in each list ÒbelongÓ to each other?

Once you realize that a sample is a matched pairs data set and that the difference in the two measurements is the important fact, the analysis proceeds just like one-sample problems, but you use the list of differences.  In this respect, there is nothing new about the matched pairs situation.

I will generate some data for you in class where there is a small difference in means between the two lists, but a real difference, in that for each case the difference is about the same.  However, by including variability within each list, we will have the situation where analyzing the data incorrectly as two-sample data will yield insignificant results.  This is the danger; failing to reject a significant difference, and thereby perhaps wasting your research resources.

Goals:        Recognize when matched-pairs applies.

Skills:

¥                Detect situations where the matched pairs t-test is appropriate.  The nature of the matched pairs is that each value of one of the variables is associated with a value of the other variable.  The most common example is a repeated measurement on a single individual, like a pre-test and a post-test.  Other situations are natural pairs, like a married couple, or twins.  In all cases, the variable we are really interested in is the difference in the two scores or measurements.  This single difference then makes the matched pairs test a one-variable t-test.

Reading:    Section 7.2.

Day 37

Activity:    Finish 2-sample work.

Two sample problems involve comparing whether the means from two populations are the same.  Our null hypothesis is that the means are equal, and the alternate is that they differ.  In the case of the one-tailed test, we claim that one populationÕs mean is smaller than the otherÕs.  The test statistic is a t-statistic.  We will skip over the details of the formula in favor of the interpretation of computer output, in our case the TI-83.  However, the basic strategy is the same:  formulate a null hypothesis (in this case, we usually assume the means are equal, or that their difference equals zero), calculate a test statistic, and make a decision about which hypothesis to believe.  The interpretation of the P-value is the same as before.  The interpretation of failing to reject is the same.  The only difference is the calculation of the test statistic.

We will look at several of the two-sample problems from the text today.

Goals:        Complete 2-sample t-test.

Skills:

¥                Know the typical null hypothesis for 2-sample hypothesis tests.  The typical null hypothesis for 2-sample problems, both matched and independent samples, is that of Òno differenceÓ.  For the matched pairs, we say H0: m = 0, and for the 2 independent samples we say H0: m1 = m2.  As usual, the null hypothesis is an equality statement, and the alternative is the statement the researcher typically wants to end up concluding.  In both 2-sample procedures, we interpret confidence intervals as ranges for the difference in means, and hypothesis tests as whether the observed difference in means is far from zero.

Reading:    Section 8.1.

Day 38

Activity:    Proportions.  Homework 9 due today.

We revisit the m&m type problem from Day 30.

Proportions: What are the true batting averages of baseball players?  Do we believe results from a few games?  A season?  A career?  We can use the binomial distribution as a model for getting hits in baseball, and examine some data to estimate the true hitting ability of some players.  Keep in mind as we do this the four assumptions of the binomial model, and whether they are truly justifiable.

For a typical baseball player, we can look at confidence intervals for the true percentage of hits he gets.  On the calculator, the command for the one-sample problem is STAT TEST 1-PropZInt.

Technical note:  the Plus 4 Method will give more appropriate confidence intervals.  As this method is extraordinarily easy to use (add 2 to the numerator, and 4 to the denominator), I recommend you always use it when constructing confidence intervals for proportions.  For two sample problems, divide the 2 and 4 evenly between the two samples; that is, add 1 to each numerator and 2 to each denominator.  Furthermore, the Plus 4 Method seems to work even for very small sample sizes, which is not the advice generally given by textbooks for the large sample approximation.  The Plus 4 Method advises that samples as small as 10 will have fairly reliable results; the large sample theory requires 5 to 10 cases in each of the failure and success group.  Thus, at least 20 cases are required, and that is only when p is close to 50 %.

Goals:        Introduce proportions.

Skills:

¥                Detect situations where proportions z-test is correct.  We have several conditions that are necessary for using proportions.  We must have situations where only two outcomes are possible, such as yes/no, success/failure, live/die, Rep/Dem, etc.  We must have independence between trials, which is typically simple to justify; each successive measurement has nothing to do with the previous one.  We must have a constant probability of success from trial to trial.  We call this value p.  And finally we must have a fixed number of trials in mind beforehand; in contrast, some experiments continue until a certain number of successes have occurred.

¥                Know the conditions when the normal approximation is appropriate.  In order to use the normal approximation for proportions, we must have a large enough sample size.  The typical rule of thumb is to make sure there are at least 10 successes and at least 10 failures in the sample.  For example, in a sample of voters, there must be at least 10 Republicans and at least 10 Democrats, if we are estimating the proportion or percentage of Democrats in our population.  (Recall the m&mÕs example: when you each had fewer than 10 blue or green m&mÕs, I made you take more until you had at least 10.)

¥                Know the Plus 4 Method.  A recent study (1998) from statistical research suggested that the typical normal theory failed in certain unpredictable situations.  Those researchers recommend a convenient fix: pretend there are 4 additional observations, 2 successes and 2 failures.  By adding these pretend cases to our real cases, the resulting confidence intervals almost magically capture the true parameter the stated percentage of the time.  Because this ÒfixÓ is so simple, it is the recommended approach in all confidence interval problems.  Hypothesis testing procedures remain unchanged.

Reading:    Section 8.2.

Day 39

Activity:    2-Sample Proportions.  Quiz 10 today.

To be added later.

To be decided.

Goals:        2-Sample proportions.

Skills:

¥                Detect situations where the 2-proportion z-test is correct.  Description.

Reading:    Chapters 6 to 8.

Day 40

Activity:    Review Statistical inference.

To be added later.

To be decided.

Goals:        Conclude course topics.  Know everything.

Skills:

¥                Be able to correctly choose the technique from among the z-test, the t-test, the matched pairs t-test, the two-sample t-test, and z-tests for proportions.  To effectively use statistical inference, you must be able to select the appropriate procedure.  The issues are sample sizes, knowledge of the population standard deviation, proper sampling schemes, etc.

Reading:    Chapters 6 to 8.

Day 41

Activity:    Presentation 3.  Statistical Inference (Chapters 6 to 8).  Homework 10 due today.

Make a claim (a statistical hypothesis) and perform a test of your claim.  Gather appropriate data and choose an appropriate technique to test your claim.  (I do not want you to just do a problem from the book; be creative, but see me if you are stuck.)  Discuss and justify any assumptions you made.  Explain why your test is the appropriate technique.

Reading:    Chapters 6 to 8.

Day 42

Activity:    Exam 3.

This last exam covers the z- and t- tests and intervals in Chapters 6 and 7, and the z tests and intervals for proportions in Chapter 8.  Some of the questions may be multiple choice.  Others may require you to show your worked out solution.  Section reviews are an excellent source for studying for the exams.  DonÕt forget to review your class notes and these on-line notes.

 

Arizona Temps Dataset:

 

 

Flagstaff

 

 

 

Phoenix

 

 

 

 

Date

Temperature (¡F)

Humidity (%)

Wind (mph)

Temperature (¡F)

Humidity (%)

Wind (mph)

 

high

low

high

low

avg

high

low

high

low

avg

 

 

 

 

 

 

 

 

 

 

 

5/1/06

66

35

89

29

3

99

69

31

8

7

5/2/06

72

35

85

11

7

96

69

25

7

8

5/3/06

66

36

49

16

10

95

69

25

7

10

5/4/06

65

31

41

15

13

93

65

24

8

10

5/5/06

59

28

82

26

6

87

65

26

11

7

5/6/06

65

27

92

16

6

91

67

34

12

6

5/7/06

69

34

64

14

4

94

67

37

9

7

5/8/06

71

36

59

18

11

95

69

29

7

7

5/9/06

70

34

62

20

6

94

69

31

10

6

5/10/06

72

32

82

12

6

99

70

31

8

5

5/11/06

74

31

41

10

6

100

73

31

7

6

5/12/06

79

38

40

9

7

100

71

27

6

6

5/13/06

78

44

40

11

6

101

73

20

7

8

5/14/06

80

40

43

14

3

105

74

31

5

7

5/15/06

79

46

50

20

8

101

74

24

7

7

5/16/06

75

44

83

27

6

104

76

31

12

10

5/17/06

76

37

92

17

4

104

78

38

11

7

5/18/06

78

45

60

16

6

104

77

36

8

9

5/19/06

79

44

65

15

8

105

78

28

8

7

5/20/06

78

45

58

14

10

105

78

28

7

6

5/21/06

77

39

44

18

8

105

78

23

7

10

5/22/06

63

32

70

24

14

84

73

36

9

12

5/23/06

72

27

85

17

5

92

66

40

11

5

5/24/06

81

32

61

9

6

102

71

31

7

5

5/25/06

79

43

40

9

8

103

74

25

6

8

5/26/06

74

39

43

11

12

103

76

22

5

9

5/27/06

68

50

46

14

23

96

72

19

10

12

5/28/06

63

33

65

6

13

88

70

41

4

10

5/29/06

67

27

33

7

7

92

65

26

5

6

5/30/06

74

29

29

7

4

97

68

26

4

5

5/31/06

79

30

36

6

4

102

70

23

4

5

6/1/06

81

33

41

5

7

108

73

19

3

5

6/2/06

84

39

28

8

6

110

80

17

6

7

6/3/06

84

46

37

10

6

112

83

27

6

6

6/4/06

86

53

30

9

7

112

83

23

6

9

6/5/06

87

53

35

8

7

108

84

27

9

11

6/6/06

86

53

34

9

7

105

86

26

12

10

6/7/06

75

19

62

34

9

106

82

32

14

10

6/8/06

75

50

87

28

5

106

81

49

12

9

6/9/06

77

50

80

12

12

105

81

28

8

9

6/10/06

78

42

58

9

10

104

78

25

6

7

6/11/06

80

40

43

8

7

107

76

25

2

7

6/12/06

80

35

33

5

8

107

74

21

3

9

6/13/06

83

36

32

14

10

111

79

17

6

7

6/14/06

82

57

36

5

17

106

84

20

4

9

6/15/06

75

43

30

11

10

101

76

25

7

10

6/16/06

77

35

64

13

6

102

77

21

7

7

6/17/06

84

37

48

8

5

106

77

23

6

6

6/18/06

87

42

37

7

7

109

80

26

6

6

 

Populations for the 50 states, DC, and the USA, by decade. (in thousands)

 

AL

AK

AZ

AR

CA

CO

CT

DE

DC

FL

GA

HI

ID

IL

IN

IA

KS

KY

1790

 

 

 

 

 

 

238

59

 

 

83

 

 

 

 

 

 

74

1800

1

 

 

 

 

 

251

64

8

 

163

 

 

 

6

 

 

221

1810

9

 

 

1

 

 

262

73

16

 

252

 

 

12

25

 

 

407

1820

128

 

 

14

 

 

275

73

23

 

341

 

 

55

147

 

 

564

1830

310

 

 

30

 

 

298

77

30

35

517

 

 

157

343

 

 

688

1840

591

 

 

98

 

 

310

78

34

54

691

 

 

476

686

43

 

780

1850

772

 

 

210

93

 

371

92

52

87

906

 

 

851

988

192

 

982

1860

964

 

 

435

380

34

460

112

75

140

1057

 

 

1712

1350

675

107

1156

1870

997

 

10

484

560

40

537

125

132

188

1184

 

15

2540

1680

1194

364

1321

1880

1263

33

40

803

865

194

623

147

178

269

1542

 

33

3078

1978

1625

996

1649

1890

1513

32

88

1128

1213

413

746

168

230

391

1837

 

89

3826

2192

1912

1428

1859

1900

1829

64

123

1312

1485

540

908

185

279

529

2216

154

162

4822

2516

2232

1470

2147

1910

2138

64

204

1574

2378

799

1115

202

331

753

2609

192

326

5639

2701

2225

1691

2290

1920

2348

55

334

1752

3427

940

1381

223

438

968

2896

256

432

6485

2930

2404

1769

2417

1930

2646

59

436

1854

5677

1036

1607

238

487

1468

2909

368

445

7631

3239

2471

1881

2615

1940

2833

73

499

1949

6907

1123

1709

267

663

1897

3124

423

525

7897

3428

2538

1801

2846

1950

3062

129

750

1910

10586

1325

2007

318

802

2771

3445

500

589

8712

3934

2621

1905

2945

1960

3267

226

1302

1786

15717

1754

2535

446

764

4952

3943

633

667

10081

4662

2758

2179

3038

1970

3444

303

1775

1923

19971

2210

3032

548

757

6791

4588

770

713

11110

5195

2825

2249

3221

1980

3894

402

2717

2286

23668

2890

3108

594

638

9747

5463

965

944

11427

5490

2914

2364

3660

1990

4040

550

3665

2351

29760

3294

3287

666

607

12938

6478

1108

1007

11430

5544

2777

2478

3685

2000

4447

627

5131

2673

33872

4301

3406

784

572

15982

8186

1294

1294

12419

6080

2926

2688

4042

 

 

LA

ME

MD

MA

MI

MN

MS

MO

MT

NE

NV

NH

NJ

NM

NY

NC

ND

OH

1790

 

97

320

379

 

 

 

 

 

 

 

142

184

 

340

394

 

 

1800

 

152

342

423

 

 

8

 

 

 

 

184

211

 

589

478

 

45

1810

77

229

381

472

5

 

31

20

 

 

 

214

246

 

959

556

 

231

1820

153

298

407

523

9

 

75

67

 

 

 

244

278

 

1373

639

 

581

1830

216

399

447

610

32

 

137

140

 

 

 

269

321

 

1919

736

 

938

1840

352

502

470

738

212

 

376

384

 

 

 

285

373

 

2429

753

 

1519

1850

518

583

583

995

398

6

607

682

 

 

 

318

490

62

3097

869

 

1980

1860

708

628

687

1231

749

172

791

1182

 

29

7

326

672

94

3881

993

 

2340

1870

727

627

781

1457

1184

440

828

1721

21

123

42

318

906

92

4383

1071

2

2665

1880

940

649

935

1783

1637

781

1132

2168

39

452

62

347

1131

120

5083

1400

37

3198

1890

1119

661

1042

2239

2094

1310

1290

2679

143

1063

47

377

1445

160

6003

1618

191

3672

1900

1382

694

1188

2805

2421

1751

1551

3107

243

1066

42

412

1884

195

7269

1894

319

4158

1910

1656

742

1295

3366

2810

2076

1797

3293

376

1192

82

431

2537

327

9114

2206

577

4767

1920

1799

768

1450

3852

3668

2387

1791

3404

549

1296

77

443

3156

360

10385

2559

647

5759

1930

2102

797

1632

4250

4842

2564

2010

3629

538

1378

91

465

4041

423

12588

3170

681

6647

1940

2364

847

1821

4317

5256

2792

2184

3785

559

1316

110

492

4160

532

13479

3572

642

6908

1950

2684

914

2343

4691

6372

2982

2179

3955

591

1326

160

533

4835

681

14830

4062

620

7947

1960

3257

969

3101

5149

7823

3414

2178

4320

675

1411

285

607

6067

951

16782

4556

632

9706

1970

3645

994

3924

5689

8882

3806

2217

4678

694

1485

489

738

7171

1017

18241

5084

618

10657

1980

4206

1125

4217

5737

9262

4076

2521

4917

787

1570

801

921

7365

1303

17558

5880

653

10798

1990

4220

1228

4781

6016

9295

4375

2573

5117

799

1578

1202

1109

7730

1515

17990

6629

639

10847

2000

4469

1275

5296

6349

9938

4919

2845

5595

902

1711

1998

1236

8414

1819

18976

8049

642

11353

 

 

OK

OR

PA

RI

SC

SD

TN

TX

UT

VT

VA

WA

WV

WI

WY

USA

1790

 

 

434

69

249

 

36

 

 

85

692

 

56

 

 

3929

1800

 

 

602

69

346

 

106

 

 

154

808

 

79

 

 

5308

1810

 

 

810

77

415

 

262

 

 

218

878

 

105

 

 

7240

1820

 

 

1049

83

503

 

423

 

 

236

938

 

137

 

 

9638

1830

 

 

1348

97

581

 

682

 

 

281

1044

 

177

 

 

12866

1840

 

 

1724

109

594

 

829

 

 

292

1025

 

225

31

 

17069

1850

 

12

2312

148

669

 

1003

213

11

314

1119

1

302

305

 

23192

1860

 

52

2906

175

704

5

1110

604

40

315

1220

12

377

776

 

31443

1870

 

91

3522

217

706

12

1259

819

87

331

1225

24

442

1055

9

38558

1880

 

175

4283

277

996

98

1542

1592

144

332

1513

75

618

1315

21

50189

1890

259

318

5258

346

1151

349

1768

2236

211

332

1656

357

763

1693

63

62980

1900

790

414

6302

429

1340

402

2021

3049

277

344

1854

518

959

2069

93

76212

1910

1657

673

7665

543

1515

584

2185

3897

373

356

2062

1142

1221

2334

146

92228

1920

2028

783

8720

604

1684

637

2338

4663

449

352

2309

1357

1464

2632

194

106022

1930

2396

954

9631

687

1739

693

2617

5825

508

360

2422

1563

1729

2939

226

123203

1940

2336

1090

9900

713

1900

643

2916

6415

550

359

2678

1736

1902

3138

251

132165

1950

2233

1521

10498

792

2117

653

3292

7711

689

378

3319

2379

2006

3435

291

151326

1960

2328

1769

11319

859

2383

681

3567

9580

891

390

3967

2853

1860

3952

330

179323

1970

2559

2092

11801

950

2591

666

3926

11199

1059

445

4651

3413

1744

4418

332

203302

1980

3025

2633

11865

947

3121

691

4591

14226

1461

511

5347

4132

1950

4706

470

226542

1990

3146

2842

11882

1003

3487

696

4877

16987

1723

563

6187

4867

1793

4892

454

248710

2000

3451

3421

12281

1048

4012

755

5689

20852

2233

609

7079

5894

1808

5364

494

281422

 

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Last updated August 15, 2009