A medical researcher says that less than 82% of adults in a certain country think that healthy children should be required to be vaccinated. In a random sample of 300 adults in that country, 78% think that healthy children should be reguired to be vaccinated. At a =0.10, is there enough evidence to support the researcher's claim? Complete parts (a) through (e) below. Hip< 0.82 OF. Hg: ps H:p= H:ps OE. Ho: p= O D. Ho: p< H: p2 Hy:p> (b) Find the critical value(s) and identify the rejection region(s). Identify the critical value(s) for this test. Z,= -1.28 (Round to two decimal places as needed. Use a comma to separate answers as needed.) Identify the rejection region(s). Select the correct choice below and fill in the answer box(es) to complete your choice. (Round to two decimal places as needed.) O A. The rejection regions are z and z> O B. The rejection region is I test statistic z. Fail to reject laces as needed.) Reject ect or fail to reject the null hypothesis and (e) interpret the decision in the context of the original claim. the null hypothesis. There Y enough evidence to V the researcher's claim.

The coach of a very popular men’s basketball team claims that the average distance the fans travel to the campus to watch a game is 35 miles. The team members feel otherwise. A sample of 16 fans who travel to games was randomly selected and yielded a mean of M= 36 miles and s= 5 miles. Test the coach’s claim at the 5% (.05) level of significance.    one-tailed or two-tailed test:  State the hypotheses: df= tα or t value for the critical region =  sM =    t (test statistic)= Decision:

A newsletter publisher believes that less than 36%36% of their readers own a laptop. For marketing purposes, a potential advertiser wants to confirm this claim. After performing a test at the 0.010.01 level of significance, the advertiser failed to reject the null hypothesis.   What is the conclusion regarding the publisher's claim?

Consider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 6052 patients treated with this drug, 149 developed the adverse reaction of nausea. Use a 0.05 significance level to test the claim that 3% of users develop nausea. Does nausea appear to be a problematic adverse reaction? Identify the null and alternative hypotheses for this test. Choose the correct answer below. O A. Ho: p=0.03 H;:p>0.03 O B. Ho: p#0.03 H;:p= 0.03 OC. Ho: p= 0.03 H4: p+0.03 O D. Ho: p= 0.03 H:p<0.03 Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is (Round to two decimal places as needed.) Identify the P-value for this hypothesis test. The P-value for this hypothesis test is. (Round to three decimal places as needed.) Identify the conclusion for this hypothesis test.

The data in the accompanying table give the weights (in g) of randomly selected quarters that were minted after 1964. The quarters are supposed to have a median weight of 5.670 g. Use the sign test and a 0.01 significance level to test the claim that the median is equal to 5.670 g. Do quarters appear to be minted according to specifications? Click the icon to view the data. What are the null and alternative hypotheses? O A. Ho: Median weight = 5.670 g H₁: Median weight 5.670 g More Info OB. Ho: Median weight ≤ 5.670 g H₁: Median weight = 5.670 g O D. Ho: Median weight = 5.670 g H₁: Median weight # 5.670 g 0 Post-1964 Quarters 5.5863 5.5764 5.6582 5.5649 5.5844 5.7402 5.7367 5.6057 5.5995 5.6165 5.6448 5.5974 5.6609 5.6363 5.6352 5.5986 5.5797 5.5501 5.6238 5.6462 5.6423 5.7021 5.5481 5.7078 5.6055 5.5324 5.7417 5.6776 5.5924 5.5691 5.5666 5.7287 5.6312 5.6017 5.6626 5.6436 5.6634 5.6132 5.5441 5.6363 -

A magazine reported that at the top 50 business schools in a region, students studied an average of 13.3 hours. Set up a hypothesis test to try to prove that the mean number of hours studied at your school is different from the reported 13.3 hour benchmark. Complete parts (a) through (c) below. O E. H X#13.3 H: X= 13.3 P. Ho p=13.3 H, p#13.3 O G. H, p= 13.3 H, X+13.3 OH. H, p#13.3 H, p=13.3 OL H a 13.3 H: a = 13.3 O J. Ho a= 13.3 H a# 13.3 O K. H, p# 13.3 H1: X= 13.3 OL Ho B#13.3 H B= 13.3 b. What is a Type I error for your test? A. Concluding that the mean number of hours studied at your school is different from the reported 13.3 hour benchmark when in fact it is not different O B. Concluding that the mean number of hours studied at your school is different from the reported 13.3 hour benchmark when in fact it is different C. Concluding that the mean number of hours studied at your school is not different from the reported 13.3 hour benchmark when in fact it is different c. What is a…

A magazine reported that at the top 50 business schools in a​ region, students studied an average of 14.4 hours. Set up a hypothesis test to try to prove that the mean number of hours studied at your school is different from the reported 14.4 hour benchmark. Complete parts​ (a) through​ (c) below.       Question content area bottom Part 1 a. State the null and alternative hypotheses. Choose the correct answer below.     A. H0​: μ=14.4 H1​: X≠14.4     B. H0​: α≠14.4 H1​: α=14.4     C. H0​: α≠14.4 H1​: β=14.4     D. H0​: β=14.4 H1​: β≠14.4     E. H0​: μ≠14.4 H1​: X=14.4     F. H0​: μ=14.4 H1​: μ≠14.4     G. H0​: α=14.4 H1​: β≠14.4     H. H0​: β≠14.4 H1​: β=14.4     I. H0​: X≠14.4 H1​: X=14.4     J. H0​: α=14.4 H1​: α≠14.4     K. H0​: μ≠14.4 H1​: μ=14.4     L. H0​: X=14.4 H1​: X≠14.4   Part 2 b. What is a Type I error for your​ test?     A. Concluding that the mean…