1. Lecture 8 Matched Pairs Review –Summary –The Flow approach to problem solving –Example

2. Announcements Extra office hours: –Today, after class – 1:00. –Sunday, 2-5 –Monday 9-11:30. I also should be in my office most of the afternoon after 2:15

3. Example 13.04 –It was suspected that salary offers were affected by students’ GPA, (which caused S 1 2 and S 2 2 to increase). –To reduce this variability, the following procedure was used: 25 ranges of GPAs were predetermined. Students from each major were randomly selected, one from each GPA range. The highest salary offer for each student was recorded. –From the data presented can we conclude that Finance majors are offered higher salaries? The Matched Pairs Experiment

4. Adminstrative Info for Midterm Location: Steinberg Hall-Dietrich Hall 351 Time: Monday (Feb. 10 th ), 6:00-8:00 p.m. Closed book. Allowed one double-sided 8.5 x 11 note sheet Bring calculator.

5. Matched Pairs => One-Sample Test After taking differences of observations within each pair, one can continue with a one-sample test with

6. Solution 13.04 (by hand) –The parameter tested is  D (=  1 –  2 ) –The hypotheses: H 0 :  D = 0 H 1 :  D > 0 –The t statistic: Finance Marketing The matched pairs hypothesis test Degrees of freedom = n D – 1 The rejection region is t > t.05,25-1 = 1.711

7. Solution 13.04 –Calculate t The matched pairs hypothesis test

8. Conclusion: There is sufficient evidence to infer at 5% significance level that the Finance MBAs’ highest salary offer is, on the average, higher than that of the Marketing MBAs. The matched pairs hypothesis test

9. Matched Pairs vs. Independent Samples Potential advantage of matched pairs: The variance of can be much less than the variance of if the variable(s) on which the pairs are matched account for much of the variation within each group. However, if the variance of is close to the variance of, the independent samples design will produce a more powerful test because it has almost twice as many degrees of freedom.

10. Recognizing Matched Pairs Experiments Does there exist some natural relationship between the first pairs of observations that makes it more appropriate to compare the first pair than the first observation in group 1 and the second observation in group 2? Before and after designs Example: A researcher for OSHA wants to see whether cutbacks in enforcement of safety regulations coincided with an increase in work related accidents. For 20 industrial plants, she has number of accidents in 1980 and 1995.

11. Examples Example: A scientist claims that tomatoes will grow large if they are played soft, soothing music. To test the claim, 10 tomatoes grown without music and 10 tomatoes grown with music are randomly selected. The tomatoes are weighed in ounces. Example: Many Americans use tax preparation companies to prepare their taxes. In order to investigate whether there are are any differences between companies, an experiment was conducted in which two of the largest companies were asked to prepare the tax returns of a sample of 55 taxpayers. The amounts of tax payable for each taxpayer at the two companies were recorded.

12. Caveat Keep in mind the difference between observational and experimental data in terms of what can be inferred. –Does the study of finance/marketing majors in Example 13.4 show that students educated in finance are more attractive to prospective employers?

13. Additional Example-Problem 13.75 Tire example contd. (Problem 13.49) Suppose now we redo the experiment: On 20 randomly selected cars, one of each type of tire is installed on the rear wheels and, as before, the cars are driven until the tires wear out. The number of miles(in 1000s) is stored in Xr13- 75. Can we conclude that the new tire is superior?

15. A Review of Chapters 12 and 13 Summary of techniques seen Here (Chapter 14) we build a framework that helps decide which technique (or techniques) should be used in solving a problem. Logical flow chart of techniques for Chapters 12 and 13 is presented next.

16. Summary of statistical inference: Chapters 12 and 13 Problem objective: Describe a population. –Data type: Interval Descriptive measurement: Central location –Parameter: –Test statistic: –Interval estimator: –Required condition: Normal population

17. Summary - continued Problem objective: Describe a population. –Data type: Interval Descriptive measurement: Variability. –Parameter: s 2 –Test statistic: –Interval estimator: –Required condition: normal population.

18. Summary - continued Problem objective: Describe a population. –Data type: Nominal –Parameter: p –Test statistic: –Interval estimator: –Required condition:

19. Summary - continued Problem objective: Compare two populations. –Data type: Interval Descriptive measurement: Central location –Experimental design: Independent samples »population variances: »Parameter: m 1 - m 2 »Test statistic: Interval estimator: »Required condition: Normal populations d.f. = n 1 + n 2 -2

20. Summary - continued Problem objective: Compare two populations. –Data type: Interval. Descriptive measurement: Central location –Experimental design: Independent samples »population variances: »Parameter: m 1 - m 2 »Test statistic: Interval estimator: »Required condition: Normal populations

21. Summary - continued Problem objective: Compare two populations. –Data type: Interval. Descriptive measurement: Central location –Experimental design: Matched pairs »Parameter: m D »Test statistic: Interval estimator: »Required condition: Normal differences d.f. = n D - 1

22. Problem objective? Describe a populationCompare two populations Data type? Interval Nominal Interval Nominal Type of descriptive measurement? Type of descriptive measurement? Z test & estimator of p Z test & estimator of p Z test & estimator of p 1 -p 2 Z test & estimator of p 1 -p 2 Central location Variability Central location Variability t- test & estimator of  t- test & estimator of    - test & estimator of  2   - test & estimator of  2 F- test & estimator of   2 /   2 F- test & estimator of   2 /   2 Experimental design? Continue

23. t- test & estimator of  1 -  2 (Unequal variances) t- test & estimator of  1 -  2 (Unequal variances) Population variances? t- test & estimator of  D t- test & estimator of  D t- test & estimator of  1 -  2 (Equal variances) t- test & estimator of  1 -  2 (Equal variances) Independent samplesMatched pairs Experimental design? Unequal Equal

24. Identifying the appropriate technique Example 14.1 –Is the antilock braking system (ABS) really effective? –Two aspects of the effectiveness were examined: The number of accidents. Cost of repair when accidents do occur. –An experiment was conducted as follows: 500 cars with ABS and 500 cars without ABS were randomly selected. For each car it was recorded whether the car was involved in an accident. If a car was involved with an accident, the cost of repair was recorded.

25. Example – continued –Data 42 cars without ABS had an accident, 38 cars equipped with ABS had an accident The costs of repairs were recorded (see Xm14-01).Xm14-01 –Can we conclude that ABS is effective? Identifying the appropriate technique

26. Solution –Question 1: Is there sufficient evidence to infer that the cost of repairing accident damage in ABS equipped cars is less than that of cars without ABS? –Question 2: How much cheaper is it to repair ABS equipped cars than cars without ABS? Identifying the appropriate technique

27. Question 2: Compare the mean repair costs per accident Solution Problem objective? Describing a single populationCompare two populations Data type? Interval Nominal Type of descriptive measurements? Central location Variability Cost of repair per accident

28. Equal Solution - continued Population variances equal? Independent samplesMatched pairs Unequal Experimental design? Central location t- test & estimator of  1 -  2 (Equal variances) t- test & estimator of  1 -  2 (Equal variances) Run the F test for the ratio of two variances. Equal Question 2: Compare the mean repair costs per accident

29. Solution – continued –  1 = mean cost of repairing cars without ABS  2 = mean cost of repairing cars with ABS –The hypotheses tested H 0 :  1 –  2 = 0 H 1 :  1 –  2 > 0 –For the equal variance case we use Question 2: Compare the mean repair costs per accident

30. Solution – continued –To determine whether the population variances differ we apply the F test –From JMP we have (Xm14-01)Xm14-01 Do not reject H 0. There is insufficient evidence to conclude that the two variances are unequal. Question 2: Compare the mean repair costs per accident

31. Solution – continued –Assuming the variances are really equal we run the equal-variances t-test of the difference between two means At 5% significance level there is sufficient evidence to infer that the cost of repairs after accidents for cars with ABS is smaller than the cost of repairs for cars without ABS. Question 2: Compare the mean repair costs per accident

32. Checking required conditions The two populations should be normal (or at least not extremely nonnormal)

33. Question 3: Estimate the difference in repair costs Solution –Use Estimators Workbook: t-Test_2 Means (Eq-Var) worksheet We estimate that the cost of repairing a car not equipped with ABS is between $71 and $651 more expensive than to repair an ABS equipped car.