1. Chapter 11 – Student’s t Tests Continuation of hypothesis testing. When population standard deviation is unknown, one can not use normal distribution. In such cases, one has to depend on sample standard deviation and a new kind of distribution. This is the t distribution. The t distribution is also bell-shaped, symmetrical, but it is more spread out compared to the standard normal distribution. See figure 11-1 on page 273.

2. Large sample size and t distribution When sample size becomes large, there is no difference between sample standard deviation and population standard deviation, and the t distribution becomes the standard normal distribution.

3. Shape of the t distribution curve Depends on a quantity, called degree of freedom. Denoted by the Greek letter, υ (pronounced as “nu”). The degree of freedom is defined by sample size minus one (n-1). For example, if the sample size is 10, then the degree of freedom is 10-1 = 9.

4. The t distribution The test statistic for t tests is: _ (x -  ) T = --------- s / √ n formula 11-1 p. 272 Critical t values are available in Appendix E (page 480). Exercises from book problem 7, page 276 problem 11, page 276

5. Problem 7, pg 276 (a) given: υ = 10, α = upper 0.025, t=? –From Appendix E, t = 2.228 (b) given: υ = 10, α = lower 0.05, t=? –From Appendix E, t = - 1.812

6. Problem 11, pg 276 Given: t = 2.624, α = 0.01 H 1 : μ > μ 0 What is n? Degrees of freedom = 14, so n = 15

7. Testing for population mean The testing procedure is very similar to what we did in chapter 10. The appendix to be used is different. Let’s illustrate with 2 examples. Example problem 11-2 on page 277 H 0 :  = 500 H 1 :  < 500 _ X = 485, s = 45, n = 25 Decision rule: Reject H 0 if T < - 1.711

8. Decision: Do not reject H 0 Interpretation: The average profit margin per car of the dealer is not significantly lower than the national average at 5% level of significance. Example problem 11-3 on page 277-278. H :  = 8 H :   8 0 1 α = 0.01, υ  two-tailed test _ n = 16, X = 7.5, s = 0.8

9. Problem 11-3, cont. Decision rule: Reject Ho if T 2.947 T = (7.5 – 8)/(0.8/√16) = -2.5 Decision: Do not reject Ho Interpretation: The vending machine is operating properly at 1% significance level.

10. Problem 2, pg 279 Ho:  = 30,000; H 1 :  > 30,000 _ N = 16, X = 31,000, s = 2000 α = 0.025 Decision rule: Reject Ho if T > 2.131 T = (31,000 – 3000)/(2000/√16) = 2 Decision: Do not reject Ho Interpretation: The vendor’s claim that his/her tires average more than 30,000 miles is not supported at 2.5% level of significance.

11. Testing for the difference between two means These tests are for: Independent samples and dependent samples Independent samples We assume that the two populations have the same variance (or standard deviation). The test statistic is T in Formula 11-5 (page 282)

12. Decision rules stay the same as before Example problem 11-5 on page 282. Effects of two training methods on learning Given: Sample Size n = 18 n = 12 1 2 -- -- Averages X = 85 X = 80 1 2 2 2 Variances s = 36 s = 34 1 2

13. Two-tailed test     =   H 1 :     Critical t at α of 0.01, degree of freedom of 28 for a two tailed test is ± 2.763 

14. Decision: Do not reject H 0 Interpretation: The two training methods have the same effects on learning at alpha of 1%. At alpha of 5%, critical t is ± 2.048. Thus, the same hypothesis is rejected at 5% level of significance. That means, the two training methods have different (unequal) effects on learning.