A>Consider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 5765 patients treated with this drug. 152 developed the adverse reaction of nausea. Use a 0.01significance level to test the claim that 3% of users develop nausea. Does nausea appear to be a problematic adverse reaction?WSIdentify the null and alternative hypotheses for this test. Choose the correct answer below.OA. Ho: p*0.03H₁: p=0.03ew an example Get more help.OC. Ho: p=0.03H₁: p>0.03B. Ho: p=0.03H₁: p=0.03DOD. Ho: p=0.03H₁: p<0.03двandIdentify the test statistic for this hypothesis test.The test statistic for this hypothesis test is(Round to two decimal places as needed.)XMonroe's Temp....docxRemoved#380F3EDC$4aRFSAMPLE Persuas.pdf%5VTG^6ABMacBook AirY5373255 (1) (1).docxH&78FUN*8JDIIIM(95373255 (1).docx exKO<IH)OLPClear allPcommandV:;ISFIT{+[optionCheck answer?111Show All4191delete

n=5765x=152 Thus, sample proportion p^=1525765=0.0264

Answered: Consider a drug that is used to help…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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2. Consider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 6079 patients treated with this drug, 153 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 O C. Ho: p=0.03 H1: p<0.03 O D. Ho: p= 0.03 H1: 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. O A. Reject Ho. There is sufficient evidence to warrant rejection of the claim that 3% of users develop nausea. O B. Reject Ho. There is not…
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The Gallup Poll asked 1015 U.S. adults whether they believed that people should pay sales tax on items purchased over the internet. Of these, 477 said they supported such a tax. Does the survey provide convincing evidence that less than 42% of U.S. adults favor an internet sales tax? Use the 0.05 level of significance. 1. STATE THE NULL AND ALTERNATIVE HYPOTHESIS: 2. T STATISTIC/ P VALUE 3.REJECT OR FAIL THE NULL HYPOTHESIS 4. IS THE CLAIM SUPPORTED AT THE 0.05 LEVEL OF SIGNIFICANCE?
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Consumer Reports did a survey of 1100 adults and found that 565 were unhappy that their food did not look like the picture in the menu. Use a 0.05 significance level to test the claim that the majority of adults are annoyed by food that does not look like the picture in the menu. Identify the null and alternate hypotheses. TTTArial 3 (12pt) v T- = - E·V
Consider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 6225 patients treated with this drug, 164 developed the adverse reaction of nausea. Use a 0.01 significance level to test the claim that 3% of users develop nausea. Does nausea appear to be a problematic adverse reaction? 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. O A. Reject Hn. There is sufficient evidence to warrant rejection of the claim that 3% of users develop nausea. O B. Reject Ho. There is not sufficient evidence to warrant rejection of the claim that 3% of users develop nausea. OC. Fail to reject Ho. There is sufficient evidence to warrant rejection…
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Test the claim that the proportion of people who own cats is smaller than 60% at the 0.025 significance level. The null and alternative hypothesis would be: Ho: p=0.6 Ho:μ ≥ 0.6 Ho: p 0.6 The test is: left-tailed right-tailed two-tailed Based on a sample of 600 people, 53% owned cats The test statistic is: The p-value is: Based on this we: Ho:p>0.6 Ho: 0.6 Ho: μ 0.6 Reject the null hypothesis O Fail to reject the null hypothesis (to 2 decimals) (to 2 decimals) =
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