Ch18-HTforP (2)

Practice Problems – Chapter 18 Hypothesis Test for a Proportion Chapter 18 Exercises, #1, #5 Better than aspirin? A very large study showed that aspirin reduced the rate of first heart attacks by 44%. A pharmaceutical company thinks they have a drug that will be more effective than aspirin, and plans to do a randomized clinical trial to test the new drug. a) What is the null hypothesis the company will use? b) Is the alternative to the null hypothesis more naturally one-sided or two-sided? Explain. c) The P-value from a clinical trial testing the hypothesis is 0.0028. What do you conclude? d) What would you have concluded if the P-value had been 0.28? Chapter 18 Exercises, #7 Hispanic origin According to the 2010 Census, 16% of the people in the United States are of Hispanic or Latino origin. One county supervisor believes her county has a different proportion of Hispanic people than the nation as a whole. She looks at their most recent survey data, which was a random sample of 437 county residents, and found that 44 of those surveyed are of Hispanic origin. a) State the hypotheses. b) Name the model and check appropriate conditions for a hypothesis test. c) Draw and label a sketch, and then calculate the test statistic and P-value. d) State your conclusion using a 1% significance level. Chapter 18 Exercises, #15 Hypotheses Write the null and alternative hypotheses you would use to test each of the following situations: a) A governor is concerned about his “negatives”— the percentage of state residents who express disapproval of his job performance. His political committee pays for a series of TV ads, hoping that they can keep the negatives below 30%. They will use follow- up polling to assess the ads’ effectiveness. b) Is a coin fair? c) Only about 20% of people who try to quit smoking succeed. Sellers of a motivational tape claim that listening to the recorded messages can help people quit. Chapter 18 Exercises, #17 Negatives After the political ad campaign described in Exercise 15, part a, pollsters check the governor’s negatives. They test the hypothesis that the ads produced no change against the alternative that the negatives are now below 30% and find a P-value of 0.22. Which conclusion is appropriate? Explain. a) There’s a 22% chance that the ads worked. b) There’s a 78% chance that the ads worked. c) There’s a 22% chance that their poll is correct. d) There’s a 22% chance that natural sampling vari ation could produce poll results like these if there’s really no change in public opinion.

Chapter 18 Exercises, #25 Dowsing In a rural area, only about 30% of the wells that are drilled find adequate water at a depth of 100 feet or less. A local man cl aims to be able to find water by “dowsing”— using a forked stick to indicate where the well should be drilled. You check with 80 of his customers and find that 27 have wells less than 100 feet deep. What do you conclude about his claim? a) Write appropriate hypotheses. b) Check the necessary assumptions and conditions. c) Perform the mechanics of the test. What is the P-value? d) Explain carefully what the P-value means in context. e) What’s your conclusion? Chapter 18 Exercises, #37 Women executives A company is criticized because only 13 of 43 people in executive-level positions are women. The company explains that although this proportion is lower than it might wish, it’s not a surprising value given that only 40% of all its employees are women. What do you think? Test an appropriate hypothesis using 𝛼 = 0.05 and state your conclusion. Be sure the appropriate assumptions and conditions are satisfied before you proceed. Other questions: 1. When Gregor Mendel conducted his famous hybridization experiments with peas, one such experiment resulted in offspring consisting of 430 peas with green pods and 158 peas with yellow pods. According to Mendel’s theory, one fourth of all the offspring peas should have yellow pods. a. Write a set of hypotheses for this test. b. Match the items listed to the correct letters in the image below. i. Null distribution ii. Standardized test statistic iii. P-value iv. Mean of the null distribution c. At a significance level of 0.05, test the claim that the proportion of peas with yellow pods is one fourth.

Key Chapter 18 Exercises, #1, #5 a) Let p = the rate of reduction in heart attacks. The null is that the new drug is not more effective than aspirin (the same effectiveness or less effective). That is, ? 0 : ? ≤ 0.44 b) One-sided (the rate of reduction would be greater than that of aspirin): ? 𝐴 : ? > 0.44 c) There is convincing evidence to conclude that the new drug is better than aspirin. d) There is not convincing evidence that the new drug is more effective than aspirin. Chapter 18 Exercises, #7 a) ? 0 : ? = 0.16, ?? ? 𝐴 : ? ≠ 0.16 b) This is a one-proportion z-test. The sample is random and ?? = 437 ∗ 0.16 = 69.9 ≥ 10 ??? ?? = 437 ∗ 0.84 = 367.1 ≥ 10 c) ?̂ = 44 437 = 0.101 𝑧 ∗ = 0.101 − 0.16 √ 0.16 ∗ 0.84 437 = −3.37 ? − 𝑉???? = 2 ∗ ?(? < −3.37) = 2(0.004) = 0.008 d) Reject the null hypothesis (p-value is less than 0.01). There is sufficient evidence at the 1% significance level to suggest that the Hispanic/Latino population in this county differs from that of the nation as a whole. Chapter 18 Exercises, #15 a) ? 0 : ? ≥ 0.30, ?? ? 𝐴 : ? < 0.30 b) ? 0 : ? = 0.5, ?? ? 𝐴 : ? ≠ 0.5 c) ? 0 : ? ≤ 0.20, ?? ? 𝐴 : ? > 0.20 Chapter 18 Exercises, #17 Statement d is correct. It talks about the probability of seeing the data, not the probability of the hypotheses.

Chapter 18 Exercises, #25 a) ? 0 : ? ≤ 0.30, ?? ? 𝐴 : ? > 0.30 b) The sample is probably random and ?? = 80 ∗ 0.30 = 17.2 ≥ 10 ??? ?? = 80 ∗ 0.70 = 25.8 ≥ 10 c) ?̂ = 27 80 = 0.3375 𝑧 ∗ = 0.3375 − 0.30 √ 0.30 ∗ 0.70 80 = 0.73 ? − 𝑉???? = ?(? > 0.73) = 1 − ?(? < 0.73) = 1 − 0.7673 = 0.2327 d) If his dowsing is no different from standard methods, there is a 23.27% chance of seeing results as good as those of the dowser’s, or better, by random chance. e) Since the p-value is large, we fail to reject the null hypothesis. There is not strong enough evidence to suggest tha t the dowser’s chance of finding water is any better than normal drilling. Chapter 18 Exercises, #37 Let p be the proportion of women in the company in executive-level positions. Is this proportion smaller than the proportion of women at the company (40%)? ? 0 : ? ≥ 0.40, ?? ? 𝐴 : ? < 0.40 Data are for all executives in this company and may not be able to be generalized to all companies; but the sample is random and ?? = 43 ∗ 0.40 = 24 ≥ 10 ??? ?? = 43 ∗ 0.86 = 56 ≥ 10 ?̂ = 13 43 ≈ 0.3023 𝑧 ∗ = 13 43 − 0.40 √ 0.40 ∗ 0.60 43 = −1.31 ? − 𝑉???? = ?(? < −1.31) = 0.0951 If the proportion of women in executive positions is truly 40% or more, the probability of finding a random sample whose proportion is 30.23% or smaller is 0.0951 (not extremely small) Since the p-value is larger than 5%, we fail to reject the null hypothesis. There is not enough evidence to suggest that the proportion of women executives is less than the 40% of women in the company in general. Other questions: 1. a. Hypotheses: ? 0 : ? = 1 4 (???𝑖?) ? 𝐴 : ? ≠ 1 4 b. Match the items listed to the correct letters in the image below. i. Null distribution = c ii. Standardized test statistic = b iii. P-value = d iv. Mean of the null distribution = a

e. No; The p-value is not the probability that the null hypothesis is true. It is the probability that we observe a sample proportion as extreme or more extreme than the one in our sample, under the assumption that the null hypothesis is true. f. Since the p-value is 0.064, it is greater than the significance level ( 𝛼 = 0.05 ), so we Fail to Reject the null hypothesis. g. CI for p: ?̂ = 0.33 ? ̂ = 0.67 ? = 100 ??𝑖?𝑖??? ?????: 𝑧 0.025 = 1.96 𝑀?𝐸 = 1.96 ∗ √ 0.33 ∗ 0.67 100 = 0.0922 95% 𝐶? ??? ?: (0.238, 0.422) Since this interval contains ? = 0.25 , it is plausible that the true proportion of red paper wasps in the state may be 25%. Our interval does not contradict the entomologist’s claim. This is the same conclusion we reach when we fail to reject the null hypothesis above (that p = 0.25). We were unable to reject the entomologist’s claim. These two methods agree in conclusion!