Old Exam Questions Hypothesis Testing 1

Old Exam Questions - Solutions Hypothesis Testing (Chapter 7) 1. First note that this is a claim about a population PROPORTION. Thus we will be using the symbol p. Note that the claim can be written symbolically as p 1 4 The opposite of the claim is p≤ 1 4 Since the claim does NOT contain equality, it becomes the alternative hypothesis. The hypothesis always has  in it. Thus we have H0 : p  1 4 1 H1 : p  4 2. First note that the claim, null & alternative hypotheses are Claim:  ≠ 98. 6 H 0 :   98. 6 H 1 :  ≠ 98. 6 Since the p value (0.045) is less than  (0.05), the conclusion about the null hypothesis would be: Reject the null hypothesis. Using Figure 7-7 on page 346, the final wording about the claim would be: "The sample data support the claim that the mean body temperature of adult humans is not 98.6 degrees." 3. First note that for the sample data we have n  23, x̄  3. 5 years, and s  1. 9 years We do NOT know the population standard deviation and we are told to assume the population distribution is normal. Thus we will use the t distribution to find critical values. Further, note that the claim, null & alternative hypotheses are Claim:   4 H0:   4 H1:   4 Thus we have a LEFT-tail test (we will need to make sure the critical value is negative!). We are told that   0. 1 and note that the degrees of freedom  n - 1  23 - 1  22. Thus look in Table A-3 with   0. 1 (in ONE tail) and df  22. We find that our critical value is -1.321. Note that we had to include the negative sign since this was a LEFT-tail test. Note that the decision test will be that we will reject the null hypothesis if our test statistic is less than -1.321 (T.S.  -1.321). 4. Note we are testing a claim about a population standard deviation. Thus critical values will come from Table A-4 (chi-square distribution). We have a left tail test. Since   0. 05, the area to the RIGHT of the critical value must be 1 - .05  .95. With df  24the critical value is 13.848. The correct choice is (b).

5. Note that this is a claim about a population PROPORTION. We are given that n  603 and p̂  0.43. Claim: p  0.5 H 0 : p  0.5 H 1 : p ≠ 0. 5 We have a two-tail test. Critical Values: 1. 96 (from Table A-2) Decision Test: Reject the null hypothesis if T.S.  -1.96 or if T.S.  1.96. Test Statistic: p̂ − p z  0. 43 − 0. 5 ≈ −3. 44 pq n

0.50.5 603

Null hypothesis Conclusion: Reject H 0 Non-technical restatement: There is sufficient evidence to warrant the rejection of the claim that McDonald’s is preferred by half of all children. 6. To compute the p value for #5, we find the area under the standard normal curve to the left of the test statistic and double it since we have a two-tail test (see Figure 7-6 on page 344). The test statistic was -3.44. Using Table A-2, the area to the left of the test statistic is 0.0003. Thus the p value is 2(0.0003)  0.0006. 7. This is a test of a population MEAN. We have n  35, x̄  6. 3 hours, and s  2. 1 hours Since n  30 and the population standard deviation is unknown, we have a t test. Claim:   7 H0:   7 H1:   7 We have a left-tail test. Thus the critical value is -2.441 (From Table A-3 with df  34 and   0. 01 in one tail). Hence the decision test is to reject the null hypothesis if the test statistic is less than -2.441. The test statistics is x̄ −  t  6. 3 − 7 ≈ −1. 97 s n

2.1 35

Since the test statistic is not in the critical region, the conclusion is: Failure to reject the null hypothesis. Using Figure 7-7 on page 346 (and in your formula card), the wording of the final conclusion is: "There is NOT sufficient sample evidence to support the claim that statistics students spend less than 7 hours studying per week."

8. We are testing a claim about a population mean. We have n  7 and x̄ ≈ 9915. 6 and s ≈ 693. 4 Since we are assuming that the population is normally distributed, we have a t-test. Claim:   9, 000 H 0 :   9, 000 H 1 :   9, 000 We have a right-tail test. Critical Value: 1.943 (t-test with 6 degrees of freedom and   0. 05 in one tail) Decision test: Reject the null hypothesis if T.S.  1.943. Test Statistic: x̄ −  t  9915. 6 − 9000 ≈ 3. 494 s n

693.4 7

H 0 Conclusion: Reject H 0 Non-technical restatement: The sample data SUPPORT the claim that the mean annual consumption amount is more than 9,000 kWh. 9. We are testing a claim about a population standard deviation. We have n  22 and s  0. 0089 The claim is   0. 01 so we have H 0 :   0. 01 H 1 :   0. 01 This is a left-tail test. Critical Value: 11.591 (Table A-4 with df  21 and   0. 05 so that area to right is 1 0.05  0.95) Decision Test: Reject H 0 if TS  11.591. Test Statistic: n − 1s 2 22 − 10. 0089 2  ≈ 16. 634 2  2 0. 01 2 H 0 Conclusion: Fail to reject H 0 Non-technical restatement: There is NOT sufficient sample evidence to support the claim that the standard deviation is less than 0.01 millimeters.