Question 7, 6.3.13 HW Score: 72.59%, 13.07 of 18 Homework: Sec 6.3 Part 1 of 2 points O Points: 0 of 1 Save In a survey of 3324 adults, 1490 say they have started paying bills online in the last year. Question list K Construct a 99% confidence interval for the population proportion. Interpret the results. Question 6 A 99% confidence interval for the population proportion is () (Round to three decimal places as needed.) O Question 7 O Question 8 Question 9 Question 10 Question 11 Question 12Question 8, 6.3.16-T HW Score: 72.59%, 13.07 of 18 E Homework: Sec 6.3 Part 1 of 2 points O Points: 0 of 1 Save Next question In a survey of 3974 adults, 721 oppose allowing transgender students to use the bathrooms of the opposite biological Question list K sex. Construct a 99% confidence interval for the population proportion. Interpret the results. Question 6 A 99% confidence interval for the population proportion is (,). (Round to three decimal places as needed.) O Question 7 Question 8 O Question 9 Question 10 Question 11 Question 12. 0 Question 9, 6.3.17-T HW Score. 72.591), 13.07 of 13 @ Homework: Sec 6.3 points Par\" f 3 0 Points: 0 of 1 _ _ A researcher wishes to estimate, with 95% condence, the population proportion of adults who think Congress is doing a Questlon \"St '6 good or excellent job. Her estimate must be accurate within 3% of the true proportion. (a) No preliminary estimate is available. Find the minimum sample size needed. (b) Find the minimum sample size needed, using a prior study that found that 20% of the respondents said they think Congress is doing a good or excellent job. (0) Compare the results from parts (a) and (b). 0 Question 6 E) 0 Question 7 (a) What is the minimum sample size needed assuming that no prior information is available? n = Cl (Round up to the nearest whole number as needed.) 0 Question 8 0 Question 9 7/ Question 10 '2' Question 11 7/ Question 12 Homework: Sec 7.2 Homework Overview Question list 0 Question 6 0 Question 7 0 Question 8 7; Question 9 \\x Question 10 \\x Question 11 g\\" Question 12 I? HW Score:66.72%, 12.68 of 19 points 9 Points: 0.4 of 1 Question 12, 7.2.34T Part 4 of 4 A nutritionist claims that the mean tuna consumption by a person is 3.8 pounds per year. A sample of 70 people shows that the mean tuna consumption by a person is 3.5 pounds per year. Assume the population standard deviation is 1.22 pounds. At 0: = 0.06, can you reject the claim? H0:ps;5.u H02p>d.a ' H0:p=;5.ts Ha: p>3.8 Ha: \"53.5 Ha: was H01p53.5 H0:u>3.8 H0:p3.5 Ha: p.>3.5 Ha: 1.53.8 Ha: p=3.5 (b) Identify the standardized test statistic. z= - 2.06 (Round to two decimal places as needed.) (c) Find the Pvalue. 0.039 (Round to three decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. (E) A. Fail to reject H0. There is sufcient evidence to Z . Reject H0. There is not sufcient evidence to reject reject the claim that mean tuna consumption is the claim that mean tuna consumption is equal to equal to 3.8 pounds. 3.8 pounds. (:3 C. Reject Ho- There is sufcient evidence to reject the C) D. Fail to reject H0. There is not sufcient evidence to claim that mean tuna consumption is equal to 3.8 reject the claim that mean tuna consumption is pounds. equal to 3.8 pounds