The final weeks

Clarification: The AP exam exemption policy says that you must be passing the class and have fewer than 6 absences to exempt the final exam. Anyone who is not passing must take the exam.

If you are part of the crowd who has to take the final, then take those practice exams that you used to prepare for the real exam. Your exam will be multiple choice.

Getting ready for the exam

I have sent out emails to 3rd period teachers asking for you to be released in time to eat (except for JROTC and pers fitness. Please let them know!!!).

Meet at the outdoor classroom between 11 and 11:30.
Bring pencils, calculator, pen, smile, sweater, mittens, scarf, hand warmers. . .

Just kidding a little about the stuff to keep you warm. Reports from Monday's exams were that it was FREEZING in the gym. Bundle up in layers.

Some sites to visit:
for fun
http://www.youtube.com/watch?v=Ooa8nHKPZ5k
for real
http://tinyurl.com/gwcmq


Somethings to remember:
You CANNOT discuss the MC problems at all--not in person, not on the phone, not on the web.

You can discuss the free response problems after 4:00 on Thursday.

Bring pencils and a pen to the test. Do not bring your cell phones, i-pods, etc. You can leave them in my room. I will lock them up.

Know your assumptions/conditions.

What questions do you have?

Chapter 13 Inferences for regression

Here are the answers from today's activities:
Please forgive formatting. Note that the values of SECoef, T, and P for the constant are not used in our inference calculations.


Regression Analysis: C6 versus C5
The regression equation is
C6 = 26.7 + 57.0 C5

Predictor Coef SECoef T P
Constant 26.75 19.44 1.38 0.263
C5 57.00 42.17 1.35 0.269

S = 9.42956 R-Sq = 37.8% R-Sq(adj) = 17.1%

Regression Analysis: C9 versus C8
The regression equation is
C9 = 32.4 + 37.5 C8

Predictor Coef SECoef T P
Constant 32.38 25.87 1.25 0.299
C8 37.50 60.19 0.62 0.577

S = 14.1877 R-Sq = 11.5% R-Sq(adj) = 0.0%


Regression Analysis: C12 versus C11
The regression equation is
C12 = 17.5 + 90.3 C11

Predictor Coef SECoef T P
Constant 17.50 10.45 1.68 0.192
C11 90.31 14.96 6.04 0.009

S = 8.52974 R-Sq = 92.4% R-Sq(adj) = 89.9%
--------------------------------------------
Regression Analysis: C15 versus C14
The regression equation is
C15 = 52.7 - 14.0 C14

Predictor Coef SECoef T P
Constant 52.67 18.74 2.81 0.067
C14 -14.00 42.50 -0.33 0.764

S = 7.76030 R-Sq = 3.5% R-Sq(adj) = 0.0%
-------------------------------------------

Descriptive Statistics: C19, C20
Variable N N* Mean SE Mean StDev
C19 5 0 0.5000 0.0791 0.1768
C20 5 0 61.00 8.34 18.64


Regression Analysis: C20 versus C19
The regression equation is
C20 = 9.00 + 104 C19

Predictor Coef SECoef T P
Constant 9.000 5.279 1.70 0.187
C19 104.00 10.07 10.33 0.002

S = 3.55903 R-Sq = 97.3% R-Sq(adj) = 96.4%

--------------------------------------------

Now, can you generate a confidence interval for the slope of the REAL regression line from one of your estimates? What does your (large) interval tell you about the strength of the relationship between x and y?

What were the three types of evidence you used to answer the questions about the model? Match the evidence to the question online at the quizplace. http://www.proprofs.com/quiz-school/quizview.php?id=968

Chapter 13 Inferences using Chi-square procedures

Hang in there; we're getting close to the end of the race.

This chapter introduces us to chi-square procedures. These methods are generally used to analyze tables of counts from samples which are separated into CELLS based on one or more categorical variables. The advantage of these methods is that you can perform many comparisons at once, instead of just two as in our previous procedures using z and t. Most students like chi-square procedures better than z and t procedures because we will be using counts rather than continuous data and our tests are automatically two-tailed.

For instance, we might analyze the COUNTS of M&Ms of each of the six usual colors in a bag or the distribution of the COUNTS of teachers at each combination of YEARS OF EXPERIENCE and HIGHEST DEGREE ATTAINED. Each element counted must be placed in exactly one CELL. We will compare the OBSERVED counts from a sample or samples to the EXPECTED counts in a way that will quantify the likelihood of this size error so we can make inferences.

There are two common versions of this test, one for situations where there is a set of guidelines or percentages that your sample data should match and one where the observations themselves determine the expected counts using the independence principle. In order to make an inference about the population or populations involved, the samples used must be SRS.

Another condition that must be met is that each expected count must be at least one. Furthermore, at least 80% of the expected counts must be at least 5. Although the observed counts must be integer values, the expected counts (just like expected values) do not need to be integers. [Error alert: Many students INCORRECTLY use the observed counts instead of the expected counts to determine whether the test is appropriate.]

Depending on the type of test we are performing there will be one of two different methods for calculating the expected counts. For both types of tests, once you have found the expected counts, you calculate chi-square components for each pair of observed and expected counts:

chi-square component = (observed - expected)^2/expected. (Of course, these are all non-negative.)


You add up all of the chi-square components to get the chi-square statistic, X^2.

You compare this X^2 value to the chi-square distribution with the appropriate number of degrees of freedom to find the p-value, or probability that you could get a X^2 value at least this large, randomly, when the null hypothesis is true. If this p-value is small, we reject the null hypothesis. If the p-value is large, we do not have sufficient evidence to conclude that the alternative is preferred.

So, I haven't addressed the hypotheses. . ..

Chi-square Goodness of Fit Test (GOF)

This is the test you use to compare a sample set of observed counts to a model that is defined somewhere else by a higher authority. Some examples:

Comparing your bag of M&Ms to the distribution of colors posted on the M&M/Mars website.

Comparing your bag of M&Ms to a uniform distribution by color (1/6th of the bag / color).

Comparing the age distribution of your town to the U.S. Census proportions.

Comparing the number of students at your school making 1, 2, 3, 4, or 5 on the AP exam compared to the global distribution.


Your null hypothesis states that the distribution matches the expected distribution. The alternative is that the distributions do not match. It is important that you write the first statement in context.

If the null hypothesis says something like p1 = p2 = p3, the alternative hypothesis SHOULD NOT be "p1 is not equal to . . . " because some of the pairs of proportions could still be equal yet the numbers do not match the distribution you wanted. Instead, use verbal descriptions like the distribution does not match the model.

To find each expected count, you take the proportions from the higher authority and multiply them by the total of all observations. You will generally get non-integer values.

Check the expected counts to make sure that all of them are at least one and check a second time to make sure that at least 80% are 5 or more.

Perform the calculations described, computing the chi-square components, adding them up to get the chi-square statistic, using that statistic to find the p-value, and making a decision in the context of the problem. If you choose to reject the null hypothsis, go back through the components to find the greatest contributor to the high chi-square statistic and cite that in your decision.


Chi-square Tests of Association: Independence and Homogeneity

When you have two or more samples from one population or two or more samples from two or more populations that you are comparing against each other with respect to categorical variables you will generally perform a chi-square test of association on the two-way table that you create to summarize the samples.

Use the words of the problem to generate the Ho and Ha for this test. The null hypothesis will customarily follow the pattern there is no association between [characteristic one] and [characteristic two].

The method for finding the expected values is different from the method described for the goodness of fit test. Otherwise, the tests are virtually the same.

To find each expected value for the cells of the two-way table, multiply the row total by the column total for that cell and divide by the grand total. Again, you will likely get non-integer numbers. Check the expecteds to see if they are at least 1 and at least 5 as described above. Calculate the components and chi-square statistic using the same formulas as in the goodness of fit test and evaluate the statistic in the same way.

Setting up the hypotheses

One of the hardest problems for students seems to be figuring out what the null and alternative hypotheses should be. Consider the test itself. Whenever the observed count matches the expected count you get a chi-square component equal to zero--something that does not contribute anything to our chi-square statistic. If ALL of the numbers matched, then our statistic would be zero and it would be graphed on the far left end of our distribution, leaving 100% of the probability to the right--a p-value of 1. (Fail to reject the null!!!!)


On the other hand, if our observed values are far from the expecteds, then the chi-square components will contribute to a larger statistic and, ultimately, a smaller p-value. (If p is small enough, reject the null!!!!!)

How does this help us to generate our hypotheses? For our null hypothesis, our observeds must be close to our expecteds. When does that happen? When our idea of what should have happened actually DID happen, for instance, when we expected the distribution to be practically uniform and it was.


This is just a little trickier when we are talking about association. The null hypothesis is that the characteristics listed along the top of the two-way table have nothing to do with the characteristics listed on the side of the table. If we proposed that video-gaming and gender were independent, then we would expect the same proportion of boys to be gamers as the girl gamers. Even though the wording of the problem may be ambiguous (Are gaming and gender independent? vs Is there a relationship between gaming and gender?), the test is still the same. The comparison that you make is between the observed counts and what the counts should be if the two characteristics are independent.

Reviewing concepts

The problem for Thursday's HW:
P-hat is 0.3, Ho: p = .25, Ha: p > .25

Your rival thinks that the sample indicates over 25% support for his program. He found 12/40 customers liked the idea. Write an email to the boss to enlighten him/her.


**********
The card problem:
Three cards are in a hat. One is white on both sides, one is red on both sides, and one has one white face and one red face. The cards are mixed and one is drawn from the hat and placed face down on the table without showing the underside. If the face up is red, what is the probability that the other face is also red?

Chapter 12 Inference for proportions

Statistics in action. . .

Here's the basketball video.
http://viscog.beckman.uiuc.edu/grafs/demos/15.html

NCAA Brian's out in front with no way for anyone to catch up (I think). Pretty amazing. http://linnerstats.mayhem.sportsline.com/e
You'll need the password, which tells you who I think will win: gogators


Please try out this quiz and let me know how it works for you.
http://www.proprofs.com/quiz-school/quizview.php?id=567 :Basic stuff quiz

http://www.proprofs.com/quiz-school/quizview.php?id=585 :Which test do we do?

Cool sites for playing with proportions:
http://http://www.ltcconline.net/greenl/java/Statistics/HypTestProp/HypTestProp.htm

http://www.math.csusb.edu/faculty/stanton/m262/proportions/proportions.html
List of top engineering schools for recruiting as discussed in class (not in any particular order):
Cal Poly, Penn State, Penn, MIT, Florida A&M, Florida, RPI, Morgan State, Maryland, UCLA, Virginia, VA Tech, Iowa State, GA Tech, Howard, Colorado, Arizona, Cal – Berkley, North Carolina A&T, Puerto Rico, Michigan, Carnegie Mellon, Ohio State, Purdue, Illinois, Cornell, Texas, Texas A&M, Stanford, USC

This chapter is more of the same methods we saw in the last two chapters. You perform hypothesis tests and confidence intervals for proportions and for differences between proportions.

The tricky bits: (1) you have to keep track of which version of the proportion you will use for testing assumptions and for calculating standard deviations/std errors. Simply use the "best" information available. (2) Recognize when the inference is about proportions and when it is about measurements (chapter 11 methods). If you use X when you should have used p you let the reader know that you are confused.

When you have a 1 proportion hypothesis test, you have a hypothesized value for p that you use for both checking assumptions (conditions) and calculating the std dev.

When you are constructing a 1 proportion confidence interval, use the best info you have--the sample proportion. This is the lucky case where you just record the number of successes and the number of failures when you are checking the conditions. Because the estimator is used, we call the sqrt(p-hat(1- p-hat)/n) the standard error. Estimate------>>>>std error.

When you have a 2 proportion hypothesis test and you are testing to see if the two proportions are the same, well, doesn't that mean that the two proportions that you use in the std error calculation should be the same? In this case you generate a "pooled" estimator (Pooled sample proportion = sum of x / sum of n)to use for condition checking and for std error calculations. When checking conditions, use the pooled proportion * each value of n and (1 - the pooled proportion) * each value of n and make sure that each product is 5 or more.

On the other hand, when you are creating a 2 proportion confidence interval for the difference, you are not assuming that the proportions are the same, so the proportions must be checked separately and the formula for the std error resembles the formulas for two-sample conf interval std errors from Ch 11 a little bit. Checking conditions: for each sample check p-hat for that sample * sample size and (1-p-hat for that sample) * sample size.

Chapter 11 Inference for Distributions

To understand this chapter you have to understand the processes of Chapter 10.


The t-distribution is a lot like the normal (z) distribution. It is much more forgiving (look for the references in the book to robustness) than the normal and we use it mostly when we have only a sample to work from--no population standard deviation.

The formulas involving t start out a lot like the z formulas.

t-statistic = (x-bar - mean)/(sample std dev/sqrt n)

and t-interval boundaries are x-bar +/- t* (sample std dev/sqrt n)

We use n-1 degrees of freedom because we "lost " one when we used x-bar to create the estimator s.

The sample std dev / sqrt n is called the standard error of the mean.


The value we use for t*, in fact the line of the table we use when considering probabilities, is based on the number of degrees of freedom (df). You can't use a line with a df = some number if you don't have at least that number of degrees of freedom. It's kind of like buying stuff. If you don't have the money, you can't buy the product. Do you realize what this means??? If you have 990 degrees of freedom and the closest choices in the text are 100 and 1000, you are supposed to select the conservative number, the one you can afford, 100 df. Now, if you can get a closer number from your calculator, use it.

How can you get the value from your calculator? (1) Use the Inv T program or function. Ti-84s with system 2.41 have it. If you have an '84, upgrade your system. If you have something else, get the program.
(2) Use the trick we demonstrated in class: Use T-INT with x-bar = 0, sx = sqrt of n, and n = n. The upper bound of the interval you generate is the estimate for t*.


Paired t-test

This is a routine t-test that is done on matched-pairs data. When you can load the first data set into L1 and the second into L2 and the following two conditions hold, you are looking at a matched-pairs design. (1) Each row of the data has to be naturally linked, as in data coming from the same person--and a different person from the rest of the rows. The two lists are DEFINITELY NOT independent of each other. (2) The variable of interest is the difference between the two values, like L1 - L2. The null hypothesis is usually mu(of the differences) = 0.

To perform the test, just do the regular t-procedures on the column of differences. DF still equals n-1.

If the two sets of data are two independent samples, that's something different. . ..

Two-sample tests

Note: The t-statistic for the difference betwen two means IS NOT t-distributed, but it is pretty close under most conditions.

We use two-sample procedures when we are looking at two separate, independent samples and trying to make an inference about the difference between the two population means.

While most of the procedure is intuitive, the standard error and the number of degrees of freedom require a little explanation.

Std Error of the difference of the means:
Do you remenber how we can't add std deviations? And how the variance of the difference of two variables is the sum of the variances? Put it together for this problem.

Find each sample variance--(s/sqrt(n))^2. Add the two sample variances together. Take the square root. In these formulas, s1 is the sample std dev for the first sample, n1 is the size of the firs sample, etc.

Then the std error of the difference = sqrt( (s1^2/n1) + (s2^2/n2) ).

Degrees of freedom:
For the number of degrees of freedom, either use the number that the calculator or the computer calculates for you or use the more conservative minimum of n1-1 or n2-1.

Hypothesis:
Ho: mu1 = mu 2 which is equivalent to Ho: mu1 - mu2 = 0

Other than these little changes, the procedures are similar to those you've already practiced.



Pooled vs unpooled

This refers to the situations when you believe that the variances of the two populations should really be equal. Using a concept similar to our Law of Large Numbers, combining the standard deviations from the samples in a clever way creates an even stronger estimate for the ONE estimated standard deviation. This is pooling of variances.

Just because the means are the same we cannot assume that the variances are equal also.

We almost never pool variances of X-bar. You can generally leave your calculator set on UNPOOLED and forget about memorizing the formula. You can only pool variances if you are really sure that the variances are equal.