1. 381 Hypothesis Testing (Testing with Two Samples-II) QSCI 381 – Lecture 31 (Larson and Farber, Sect 8.2)

2. 381 Comparing Two Means z-testt-test Samples Must be independent Distribution and sample size Both samples must have at least 30 members or the populations must be normal with known populationstandard deviations. The populations must be normal (n 1 or n 2 can be < 30)

3. 381 Comparing Means with “Small” Sample Sizes (Conditions for use) To use a t-test for small (independent) samples, the following conditions must be met: The samples must be selected randomly. The samples must be independent. The data for each population must be normally distributed.

4. 381 Comparing Means with “Small” Sample Sizes (The two-sample t-test) A is used to test the difference between two population means  1 and  2 when the sample size for at least one population is less than 30.The standardized test statistic is:

5. 381 Comparing Means with “Small” Sample Sizes (Standard Error Specification) If the population variances are equal, then: d.f. = If the population variances are not equal then: d.f. = smaller of n 1 -1 and n 2 -1.

6. 381 A small survey includes two strata. The results of the survey are summarized below. Test the hypothesis that the density is the same in the two strata using  =0.05. Assume the populations are normally distributed and the population variances are not equal. Example-A-I Stratum 1Stratum 2 s 1 = 8s 2 = 5 n 1 =11n 2 =12

7. 381 Example-A-II 1. H 0 :  1 =  2 ; H a :  1   2. 2. The level of significance is 0.05, the variances are not equal so the d.f. is 11-1=10. The rejection region is therefore |t|>2.228. 3. The variances are not equal so: 4. The standardized test statistic is: 5. The null hypothesis cannot be rejected because t is not in the rejection region.

8. 381 Example-B-I Two areas are surveyed. One area is fished and another is in a marine reserve. It is claimed (before the data are collected) that the density in the marine reserve will be higher than in the fished area. Assume that: a)  =0.01, b) the populations are normally distributed and, c) the variances are equal. Stratum 1Stratum 2 s 1 = 9s 2 = 10 n 1 =11n 2 =15

9. 381 Example-B-II 1. H 0 :  1  2 ; H a :  1 <  2. 2. The level of significance is 0.01, the variances are equal so the d.f. is 11+15-2=24. The rejection region is therefore t>2.492. 3. The variances are equal so: 4. The standardized test statistic is: 5. The null hypothesis should be rejected because t is in the rejection region.

10. 381 Confidence Intervals for Differences Between Means-I If the sampling distribution for is a t- distribution and the populations have equal variances, you can construct a c-confidence interval for  1 -  2 using the equation: d.f. =

11. 381 Confidence Intervals for Differences Between Means-II If the sampling distribution for is a t- distribution and the populations have unequal variances, you can construct a c-confidence interval for  1 -  2 using the equation: d.f. = smaller of and

12. 381 Example-I Find a 99% confidence interval for the difference in density between the fished area and marine reserve in example B.