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Statistics B Chapter 8 Practice Test 1. Which of the following correctly expresses the null and alternative hypotheses for the situation: A psychologist claims that more than 6% of the population suffers from professional problems due to extreme shyness. a. H0: p = 0.06, H1: p > 0.06 b. H0: p = 0.06, H1: p < 0.06 c. H0: p > 0.06, H1: p ≤ 0.06 d. H0: p < 0.06, H1: p ≥ 0.06 2. Which of the following correctly expresses the null and alternative hypotheses for the situation: The manufacturer of a certain type of refrigeration unit claims it keeps the mean temperature at no more than 48°. a. H0: = 48, H1: ≠ 48 c. H0: ≠ 48, H1: = 48 b. H0: = 48, H1: > 48 d. H0: ≤ 48, H1: ≥ 48 3. The “2” critical value for a sample size of 31 with a = 0.05 for a right-tail test is 43.773. 4. The “z” critical value for a sample size more than 50 with a = 0.01 for a two-tail test is ±2.575. 5. The “t” critical value for a sample size of 46 with a = 0.10 for a right-tail test is 1.301 6. In a hypothesis test, the original claim is the alternative hypothesis, H1. a. always true b. sometimes true c. never true 7. If a claim is to be rejected, the claim will be worded so that it becomes the alternative hypothesis, H1. a. always true b. sometimes true c. never true (when the claim is H1, the claim is either supported or not supported) 8. The P-value for a test statistic z = 2.44 for a two-tail test is 0.0146. 9. The P-value for a test statistic t = –1.529 (n = 15) for a left-tail test is 0.05 < P < 0.10. 10. The P-value for a test statistic 2 = 25.213 (n = 21) for a right-tail test is P > 0.10. 11. Which of the following is the correct final conclusion: The owner of a brewery claims that the true mean temperature of the refrigeration units for beer kegs is not equal to 45°F, which is the ideal temperature for its German-style pilsner. Assume that a hypothesis test was performed and the initial conclusion was to reject the null hypothesis. c. The sample data support the claim that the true mean temperature of the refrigeration units for beer kegs is not equal to 45°F. 12. Which of the following is the correct final conclusion: A researcher claims that the amounts of acetaminophen in cold tables have a standard deviation of = 3.3 mg. Assume that a hypothesis test was performed and the initial conclusion is to not reject the null hypothesis. b. There is not sufficient evidence to warrant rejection of the claim that the amounts of acetaminophen in cold tablets have a standard deviation of = 3.3 mg For each hypothesis test problem, use whichever method you prefer, and clearly show all the steps. 13. A light bulb manufacturer advertises that the average life for its light bulb is 900 hours. A random sample of 15 light bulbs resulted in the data below (life span in hours). At a significance level of 0.10, test the claim that the mean life for the company’s bulbs differs from what’s advertised. 995 917 664 Step 1: H0: = 900 *H1: ≠ 900 590 571 693 510 555 708 Step 2: Test Stat: t 724.1 900 156.9 15 t 4.342 ▪ NEVER use “” from calc screen for sample, always use “s” ▪ because this is a sample, it is a “t” problem ▪T.S. is more extreme than C.V., that’s why reject ▪P-value is smaller than , that’s why reject ▪ original claim is H1, so SUPPORT in final answer 539 916 887 739 728 849 Step 3: Step 4: Critical value: Initial Conclusion: t=±1.761 Reject H0 P-value: P < 0.01 Step 5: There is enough evidence to SUPPORT the claim that the mean life for the company’s bulbs differs from what’s advertised 14. According to a recent poll, 53% of residents stated that they would vote for the incumbent politician. If a random sample of 100 people resulted in 45 who stated they would vote for the incumbent, test the claim that the actual percentage really is 53%. Use a 0.05 significance level. Step 1: *H0: p = 0.53 H1: p ≠ 0.53 Step 2: Step 3: Test Stat: z .45 .53 .53 .47 100 1.60 Step 4: Critical value: Initial Conclusion: z = ±1.96 Do not reject H0 P-value: ▪ claim: becomes null hypothesis H0 ▪ that’s why “REJECT” is in last box ▪ two-tail test because of “≠” ▪ that’s why critical value has “±” ▪ P-value was number found in Table A-2, then doubled ▪ P-value greater than , that’s why do not reject ▪ T.S. not more extreme than CV, that’s why do not reject 0.1096 Step 5: There is not enough evidence to REJECT the claim that the actual percentage really is 53% 15. A random sample of 100 pumpkins gave a mean circumference of 40.5 centimeters. Assuming that the population standard deviation is known to be 1.6 centimeters, use a 0.05 significance level to test the claim that the mean circumference of all such pumpkins is at least 40 centimeters. Step 1: *H0: = 40 H1: < 40 Step 2: Step 3: Test Stat: z 40.5 40 1.6 100 z 3.13 ▪ claim: “at least” “≥” becomes H0 ▪ opposite of claim is “<” ▪ “<” means left-tail test ▪ this is a “z” problem since “” is known ▪T.S. not more extreme than C.V., that’s why do not reject ▪P-value is way bigger than , that’s why do not reject Step 4: Critical value: Initial Conclusion: z=–1.645 Do not reject H0 P-value: P = .9991 Step 5: There is not enough evidence to REJECT the claim that the mean circumference of all such pumpkins is at least 40 centimeters 16. The standard deviation of math test scores at one high school is 16.1. A teacher claims that the standard deviation of girls’ test scores is actually smaller than 16.1. A random sample of 22 girls results in scores with a standard deviation of 14.3. Use a 0.01 significance level to test the teacher’s claim. Step 1: H0: = 16.1 *H1: < 16.1 Step 2: Test Stat: 21 14.32 16.12 2 16.567 2 Step 3: Critical value: Initial Conclusion: 2=8.897 Do not reject H0 P-value: ▪ “<” means left-tail test ▪ that’s why look in the “0.99” column for C.V. ▪T.S. not more extreme than C.V., that’s why do not reject ▪P-value is bigger than , that’s why do not reject Step 4: P > 0.10 Step 5: There is not enough evidence to SUPPORT the claim that the standard deviation of girl’s test scores is actually smaller than 16.1