1. 381 Hypothesis Testing (Decisions on Means) QSCI 381 – Lecture 27 (Larson and Farber, Sect 7.2)

2. 381 Using p-values to Make Decisions (Recap-I) To use a p-value (the probability of obtaining a test statistic as extreme or more extreme than that actually obtained, given that the null hypothesis is true) to draw a conclusion in a hypothesis test, compare the p-value to  : If p  , then reject H 0. If p > , then fail to reject H 0. Therefore, if p=0.0761 what conclusions do you draw for  =0.1, 0.05 and 0.01?

3. 381 Using p-values to Make Decisions (Recap-II) After determining the test statistic: p = Area in left tail for a left-tailed test. p = Area in right tail for a right-tailed test. p = 2 x area in tail of the test statistic for a two- tailed test. For a right-tailed test (i.e. the population parameter  some value), the test statistic is 2.31. What is the p-value to use? Would you reject the hypothesis for  =0.01 or 0.05?

4. 381 Z-test for a Mean The is a statistical test for a population mean. The test statistic is the sample mean and the standardized test statistic is: Notes: The z-test can be used when the population is normal and  is known, or for any population when the sample size n is at least 30. When n  30, you can use the sample standard deviation s in place of .

5. 381 Overview – Large sample sizes-I 1. State the claim mathematically and verbally. Identify the null and alternative hypotheses. H 0 = ?; H a = ? 2. Specify the level of significance.  =? 3. Determine the standardized test statistic: If n  30, use  s. 4. Find the area that corresponds to z.

6. 381 Overview – Large sample sizes-II Find the P-value: For a left-tailed test, p = (area in left tail). For a right-tailed test, p = (area in right tail). For a two-tailed test, p = 2*(area in the tail of test statistic). Make decision whether or not to reject H 0. Reject H 0 if the p-value  . Fail to reject H 0 if the p-value > . Interpret the results.

7. 381 Worked Example-I The mean output from an age reader in a lab is known to be 30 otoliths per day. A new age reader joins the team and reads 795 otoliths (standard deviation 10 per day) in his first 30 days. Is the performance of the new age reader sufficiently poor that he should be fired?

8. 381 Worked Example-II 1. H 0 is   30; H a < 30. 2. We had better use  =0.01 – we don’t want to fire this person incorrectly! 3. Calculate the test statistic: 4. The area corresponding to this value of z is 0.027 ( =NORMDIST(- 1.92,0,1,TRUE) in EXCEL)

9. 381 Worked Example-III 5. This is a left-tailed test. 6. The p-value is greater than 0.01 so we don’t reject the null hypothesis. 7. Notes: 1. Had we assumed  =0.05 we may have fired our new age reader! 2. We could have calculated the area more directly using the EXCEL statement: =NORMDIST(795/30,30,10/SQRT(30),TRUE)

10. 381 Another Example It is known that students taking MATH 481 sleep on average 3 hours per night. Some changes are made to Windows and we sample sleep time for 60 students and find it to be 3.3 hours (standard deviation 1 hour). Has the new version of Windows changed the amount of sleep – use  =0.05.

11. 381 Rejection Regions and Critical Values-I A (or critical region) of the sampling distribution is the range of values for which the null hypothesis is not probable. If a test statistic falls in this region, the null hypothesis is rejected. A z 0 separates the rejection region from the non- rejection region.

12. 381 Rejection Regions and Critical Values-II 1. Specify the level of significance. 2. Decide whether the test is left-tailed, right- tailed or two-tailed. 3. Find the critical value, z 0 : 1. Left-tailed, find the z-score that corresponds to an area of . 2. Right tailed, find the z-score that corresponds to an area of 1- . 3. Two-tailed, find the z-scores that correspond to areas of ½  and 1-½ .

13. 381 Rejection Regions and Critical Values-III do not reject do not reject do not reject Reject H 0 Calculate the standardized test statistic and compare it with the rejection region.

14. 381 Worked Example again This was a left-tailed test. The test statistic was -1.92 The critical value is -2.33. We do not reject the null hypothesis.  =0.01  =0.05

15. 381 Another example The target escapement for salmon rivers in a given watershed is 2,000,000 fish per river per annum. You believe that the actual escapement differs from this so you sample 40 rivers. You find the escapement to be 2,200,000 (s.d. 500,000). You are likely to be an expert witness in a hearing so choose your level of significance carefully.