Economics 173 Business Statistics Lecture 10b © Spring 2002, Professor J. Petry

2 Organizational Note Today’s lecture notes are not in your packet. They are posted in a note on web-board for those of you that do not already have them. The mid-term is 1 week from Thursday on February 28 th, from 7:00-9:00. The rooms are organized by lab section. Look on the web for your lab’s exam location. Ask during this week’s lab if you have any questions. The mid-term will cover the material through lecture 12 in your course- packet as well as lecture 10b—today’s material. There are two practice exams on line. One was the actual mid-term from last semester, and the other was the practice for last semester’s mid-term. You can take the Mallard homeworks over as many times as you would like, it keeps your best score—so there is no penalty for doing them again. Wednesday we do Lecture 11, a week from today, we do Lecture 12. We have no lecture a week from Wednesday, review in lab on Thurs, 2/28. We will assign Project 1 shortly after your mid-term.

3 F-test, ratio of variances We can have either a one-tailed test or a two-tailed test. Excel only produces output based on a one-tail test. Excel provides the critical value on the same side of the distribution as the test statistic. If the test statistic is above 1 (which is the middle of the distribution), it gives you the right tail as the rejection region. If the test statistic is below 1, it gives you the left tail as the rejection region. You need to be able to manipulate what excel gives you to get what you want. Draw the distributions for examples 12.5, 12.2 and from lecture 9.

4 F-test, ratio of variances

Matched Pairs Experiment Designing the experiment in the most effective way is an important aspect of statistical analysis. Understanding the implications of experimental design can also provide insight into the statistical testing process. We have already shown you three tests which can be done when comparing the difference in population means. Which are? There is a fourth way, however. It is called “matched pairs”. To illustrate, we will take an example and first do conduct it as a difference in means test with independent samples, then we will conduct it as a matched pairs experiment.

6 Example 12.3 To determine whether a new steel-belted radial tire lasts longer than an existing model, the manufacturer designs the following experiment. –A pair of newly designed tires are installed on the rear wheels of 20 randomly selected cars. –A pair of existing-model tires are installed on the rear wheels of another randomly selected 20 cars. –Drivers drive in their usual way until the tires worn out. –The number of miles driven by each driver were recorded.

7 Solution Compare two populations of quantitative data. The parameter is  1 -  2 11 22 The hypotheses are: H 0 : (  1 -  2 ) = 0 H 1 : (  1 -  2 ) > 0 Mean distance driven before worn out occurs for the new design tires Mean distance driven before worn out occurs for the existing design tires

8 Which difference in means test do we use? –We have sample data for variance, so the only question is which t-test. –We have to do the F-test, ratio of variances test to determine whether we can assume the variances are unequal, or should assume that the variances are unequal.

9 The hypotheses are H 0 :  1 -  2 = 0 H 1 :  1 -  2 > 0 The test statistic is We run the t test, and obtain the following Excel results. We conclude that there is insufficient evidence to reject H 0 in favor of H 1.

More New design More Existing design While the sample mean of the new design is larger than the sample mean of the existing design, the variability (noise) within each sample is large enough for the sample distributions to overlap and cover about the same range. It is therefore difficult to argue that one expected value is different than the other. What does “noise” mentioned above refer to?

11 Example 12.4 –Is all that variability necessary? Can we eliminate some of the noise so we can take a closer look at the means? –to eliminate variability among observations within each sample the experiment was redone. –One tire of each type was installed on the rear wheel of 20 randomly selected cars. –The number of miles until wear-out was recorded

12 Solving –Calculate the difference for each x i –Calculate the average differences and the standard deviation of the differences –Build the statistics as follows: –Run the hypothesis test using t distribution with n D - 1 degrees of freedom.

13 –The hypotheses test for this problem is H 0 :  D = 0 H 1 :  D > 0 The statistic is The rejection region is: t > t  with d.f. = 20-1 = 19. If  =.05, t.05,19 = Since > 1.729, there is sufficient evidence in the data to reject the null hypothesis in favor of the alternative hypothesis. Conclusion: At 5% significance level the new type tires last longer than the current type.

14

15 Estimating the mean difference

16 Checking the required conditions for the paired observations case The validity of the results depends on the normality of the differences.

17 Matched Pairs Usage Since matched pairs got rid of excess variability should it always be used to compare difference in means? Are there any disadvantages to its use? Under what conditions should you not use matched pairs?