1. Math 113 Chapters 7-8 Review

2. Identify which statement is the null hypothesis and which is the alternative hypothesis. A corpse is dead A corpse is alive The car has gasoline in it The car is out of gas #2

3. #3 Identify which statement is the type I error and which is the type II error. An unnecessary hysterectomy is performed A hysterectomy is not performed when it should be

4. #4 All hypothesis testing is done under the assumption that the null hypothesis is true.

5. #7 A Kolmogorov-Smirnov test is performed on a sample and the following p-values are found. Distributionp-value Normal0.0321 Uniform0.3215 Exponential0.1246 Jones Fictional0.0257 Does the sample appear to be Uniformly distributed? Does the sample appear to have a Jones Fictional Distribution?

6. #9 Identify the test as left-tailed, right-tailed, or two-tailed and give the decision (reject or fail to reject H 0 ) If there are two critical values, it is a two-tail test. If the critical value is less than the mean, it is a left- tail test. If the critical value is greater than the mean, it is a right-tail test.

7. #9 continued The mean of a normal or student’s t distribution is 0. The mean of a chi-squared distribution is its degrees of freedom. The mean of an F distribution is approximately 1. Reject the null hypothesis if the test statistic is more extreme (in the direction of the type of test) than the critical value – that is, if it lies in the critical region.

8. #9 continued again Critical Value:t = -3.215 Test Statistic:t = -2.843 Critical Value:= 35.832 Test Statistic:= 41.721 df= 18

9. #10 Harry claims that there is no agreement (null hypothesis) between the judges of the skating competition. Kendall’s concordance of agreement was computed and the p-value was found to be 0.1275. The decision is to (reject / fail to reject) the null hypothesis. There is (sufficient / insufficient) evidence to (support / reject) the claim that there is no agreement. There is (sufficient / insufficient) evidence to (support / reject) the claim that there is agreement.

10. #11 Give the decision based on the p-value. Reject the null hypothesis when the p-value is less than the level of significance. Fail to reject the null hypothesis when the p-value is greater than the level of significance P-value=0.3158, alpha = 0.3927 P-value=0.1253, alpha = 0.1025

11. #28 Write the null and alternative hypothesis and identify the test as left tailed, right tailed, or two-tailed. 1.24% of Americans smoke 2.The average smoker began smoking before age 16 3.Professional baseball players earn more than professional football players 4.The standard deviation on the last test was 15. 5.More women than men read Cosmopolitan.

12. #28 continued 1.24% of Americans smoke 2.The average smoker began smoking before age 16

13. #28 continued 3.Professional baseball players earn more than professional football players 4.The standard deviation on the last test was 15

14. #28 continued 5.More women than men read Cosmopolitan

15. #30 - #33 Using Microsoft Excel to find critical values The functions and the definitions of those functions are given on the test. Be sure to enter the equal sign at the beginning of the function Be sure you pay careful attention to the definition. Sometimes Excel expects a right tail probability, sometimes a left tail probability, and sometimes a two-tail probability. The critical value notation given on the test is ALWAYS a RIGHT TAIL area.

16. #30-#33 continued Find =CHIINV(prob,df) Returns the critical value from the chi-squared distribution with df degrees of freedom and prob area in the right tail. Since we are looking for a right tail area (always in the critical value notation) and Excel is expecting a right tail area, we can just enter the function as =CHIINV(0.83,18) and Excel gives us the value 12.325 after we hit enter

17. #34 A test statistic from a sample of n=16 items is found to be t=1.531. If this is a right tail test, find the two values that the probability value lies between. One Tail0.0250.050.100.25 Two Tail0.050.100.200.50 df=152.1311.7531.3410.691 Since the test statistic is between 1.753 and 1.341 and it is a one-tail test, the p-value is between 0.05 and 0.10, the one-tail probabilities associated with each of those critical values. 0.05 < p-value < 0.10