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
0.50.5 603
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 − 1s 2 22 − 10. 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.