1. Section 8.3 Testing the Difference Between Means (Dependent Samples)

2. Section 8.3 Objectives Perform a t-test to test the mean of the differences for a population of paired data

3. The test statistic is the mean of these differences.  t - Test for the Difference Between Means To perform a two-sample hypothesis test with dependent samples, the difference between each data pair is first found:  d = x 1 – x 2 Difference between entries for a data pair Mean of the differences between paired data entries in the dependent samples

4. t - Test for the Difference Between Means Three conditions are required to conduct the test. 1.The samples must be randomly selected. 2.The samples must be dependent (paired). 3.Both populations must be normally distributed. If these requirements are met, then the sampling distribution for is approximated by a t-distribution with n – 1 degrees of freedom, where n is the number of data pairs. -t0-t0 t0t0 μdμd

5. Symbols used for the t - Test for μ d The number of pairs of data The difference between entries for a data pair, d = x 1 – x 2 The hypothesized mean of the differences of paired data in the population n d Symbol Description

6. Symbols used for the t - Test for μ d Symbol Description The mean of the differences between the paired data entries in the dependent samples The standard deviation of the differences between the paired data entries in the dependent samples sdsd

7. t - Test for the Difference Between Means The test statistic is The standardized test statistic is The degrees of freedom are d.f. = n – 1.

8. t - Test for the Difference Between Means (Dependent Samples) 1.State the claim mathematically and verbally. Identify the null and alternative hypotheses. 2.Specify the level of significance. 3.Determine the degrees of freedom. 4.Determine the critical value(s). State H 0 and H a. Identify α. Use Table 5 in Appendix B if n > 29 use the last row (∞). d.f. = n – 1 In WordsIn Symbols

9. t - Test for the Difference Between Means (Dependent Samples) 5.Determine the rejection region(s). 6.Calculate and 7. Find the standardized test statistic. In WordsIn Symbols

10. t - Test for the Difference Between Means (Dependent Samples) 8.Make a decision to reject or fail to reject the null hypothesis. 9.Interpret the decision in the context of the original claim. If t is in the rejection region, reject H 0. Otherwise, fail to reject H 0. In WordsIn Symbols

11. Example: t - Test for the Difference Between Means Student1234567 Errors (before)1510128549 Errors (after)11965109 A teacher claims that a grammar seminar will help students reduce the number of grammatical errors made when writing essays. The table shows the number of errors made before and after participating in the seminar. Assuming the populations are normally distributed, is there enough evidence to support the claim at α = 0.01?

12. Example: t - Test for the Difference Between Means Hypotheses: claim: mean_before > mean_after mean_before – mean_after > 0 µ before - µ after > 0 µ 1 - µ 2 > 0 µ d > 0 Hypotheses: H 0 :µ d ≤ 0 H a : µ d > 0 (claim)

13. Example: t - Test for the Difference Between Means Student1234567sum Errors before 1510128549 Errors after11965109 d (x 1 – x 2 )416344022 d2d2 16136916 094

14. Example: t - Test for the Difference Between Means

15. Athletes12345678 Height (old)2422252835323027 Height (new)2625 2933343530 A shoe manufacturer claims that athletes can increase their vertical jump heights using the manufacturer’s new Strength Shoes ®. The vertical jump heights of eight randomly selected athletes are measured. After the athletes have used the Strength Shoes ® for 8 months, their vertical jump heights are measured again. The vertical jump heights (in inches) for each athlete are shown in the table. Assuming the vertical jump heights are normally distributed, is there enough evidence to support the manufacturer’s claim at α = 0.10?

16. Example: t - Test for the Difference Between Means Hypotheses: claim: increases vertical jump mean_before < mean_after mean_before – mean_after < 0 µ before - µ after < 0 µ 1 - µ 2 < 0 µ d < 0 Hypotheses: H 0 :µ d ≥ 0 H a : µ d < 0 (claim)

17. Solution: Two - Sample t - Test for the Difference Between Means H 0 : H a : α = d.f. = Rejection Region: 0.10 8 – 1 = 7 μ d ≥ 0 μ d < 0 (claim) d = (jump height before shoes) – (jump height after shoes)

18. Solution: Two - Sample t - Test for the Difference Between Means BeforeAfterdd2d2 2426–24 2225–39 25 00 2829–11 3533 24 3234–24 3035–525 2730–39 Σ = –14Σ = 56 d = (jump height before shoes) – (jump height after shoes)

19. Solution: Two - Sample t - Test for the Difference Between Means H 0 : H a : α = d.f. = Rejection Region: Test Statistic: 0.10 8 – 1 = 7 µ d ≥ 0 μ d < 0 (claim) Decision: d = (jump height before shoes) – (jump height after shoes) At the 10% level of significance, there is enough evidence to support the shoe manufacturer’s claim that athletes can increase their vertical jump heights using the new Strength Shoes ®. Reject H 0. t ≈ –2.333

20. Section 8.3 Summary Performed a t-test to test the mean of the difference for a population of paired data