1. Sampling Distribution of Differences Between Means

2.  1 =  2  1   2 Null Research Sample 1 Sample 2 Population

3. μ 1 - μ 2 = 0 μ 1 = μ 2 PopulationSampling Distribution

4. Hypothesis: Women earned higher grades than men Choose a statistic: Mean grade point average (GPA) H 1 : μ F –μ M > 0 H 0 : μ F –μ M = 0 α<.05; one tail σ 2 =.10 n F = 25; n M = 20; ;

5. μ F –μ M = 0 z = 1.64,  <.05 Do not reject Null

6. CI 95 : Interval estimate: Point estimate: 0.05 ± 1.96 * 0.09 = -0.1264, 0.2264 -0.12640.2264

7. Hypothesis: GRE Verbal of English speaking students is better than that of Chinese students Choose a statistic: Mean GRE verbal scores H 0 : μ E –μ C = 0 H 1 : μ E –μ C > 0 α<.01; one tail σ = 100 n E = 100; n C = 400; ;

8. μ E –μ C = 0 z =2.33,  <.01 Reject Null

9. Point estimate: Interval estimate: CI 90 : 40 ± 1.64 * 11.18 = 21.66, 58.33 -18.33 18.33

10. t as the sampling distribution of means when  2 is not known and sample variance, s 2, is used to estimate  2. Now you need to estimate the mean but also the variance of the population

11. Variance known; z test Variance unknown, t test

12. df = 2 df = 9 df = 30 0.025 t crit. = -4.30t crit. = 4.30 t crit. = -2.26t crit. = 2.26 z crit. = -1.96z crit. = 1.96 t crit. = 2.04t crit. = - 2.04 Normal

13. Hypothesis: Word acquisition of the two groups of pre school children are different Choose a statistics: Mean word acquisition test scores H 0 : μ E = μ C H 1 : μ E ≠μ C α<.05; two tails s 2 E =5.286; s 2 C =3.671 n E = 15; n C = 20; ; df = n E – 1 + n C – 1 = 15 – 1 + 20 – 1 = 33

14. μ E –μ C = 0

15. t.05 (33)= ±2.04 μ E –μ C = 0 Reject Null

16. Point estimate: Interval estimate: CI 95 : 2.75 ± 2.04 * 0.71 = 1.30, 4.20 -1.451.45

17. Independent vs. Dependent or Paired t-tests PersonMid 1M90 2M65 3M78 4F86 5F77 6F89 7F93 df = n M – 1 n F – 1 df = n – 1 Final D 89 1 68 -3 70 8 97 11 67 -10 99 10 89 -4 +

18. H 0 : μ 1 –μ 2 = 0 H 1 : μ 1 –μ 2 ≠ 0  <0.01/2  2 = 120 n 1 = 12; n 2 = 20; Answers to Homework 3:

19. μ 1 –μ 2 = 0  < 0.01/2 <.005 z critical = 2.58

20. .005 z = -2.58z = 2.58 CI 99 =  1 -  2 = 0 not inside the interval estimates

22.  1 -  2 = 0  0.01/2  0.005 t critical (30)= 2.75  .005 Do not reject Null. t = 2.6

23. .005 t = 2.75t = -2.75  1 -  2 = 0 falls inside the interval estimates CI 99 =