A drug is used to help prevent blood clots in certain patients. In clinical trials, among4918patients treated with the drug,163developed the adverse reaction of nausea. Construct a99%confidence interval for the proportion of adverse reactions. Use the sample data and confidence level given below to complete parts (a) through (d).A drug is used to help prevent blood clots in certain patients. In clinical trials, among 4918 patients treated with the drug, 163 developed the adverse reaction of nausea. Construct a 99% confidence interval for the proportion of adverse reactions.a) Find the best point estimate of the population proportion p.[ ] (Round to three decimal places as needed.)b) Identify the value of the margin of error E.E = [ ] (Round to three decimal places as needed.)c) Construct the confidence interval.[ ] < p < [ ] (Round to three decimal places as needed.)d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below.O A. There is a 99% chance that the true value of the population proportion will fall between the lower bound and the upper bound.O B. One has 99% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.O C. 99% of sample proportions will fall between the lower bound and the upper bound.O D. One has 99% confidence that the sample proportion is equal to the population proportion.

Name: A drug is used to help prevent blood clots in certain patients. In cli
Uploaded: 2024-03-20T07:07:00.000Z
Channel: Bartleby
Description: A drug is used to help prevent blood clots in certain patients. In clinical trials, among4918patients treated with the drug,163developed the adverse reaction of nausea. Construct a99%confidence interval for the proportion of adverse reactions. Use

The number of patients who suffered adverse reactions is 163 in the sample of 4918 patients. Therefore, the sample proportion of patients who experienced adverse reactions = 163/4918 = 0.03 Part a: The sample proportion is the best unbiased estimate of the population proportion. Therefore, the point estimate of the population proportion, p^ = 0.033 Part b: The sampling distribution of the sample proportion is normal for large samples. Therefore, at a 0.01% level of significance, the margin of error E is E = 2.58×p^(1-p^)/n = 2.58×0.033×0.967/4918 = 0.007 Part c: A 99% confidence interval for the true population proportion of patients who suffered adverse reactions is given by p^ ± E= 0.033 ± 0.007= (0.026, 0.040) Part d: There is a 99% chance that the value of the population proportion will fall between 0.026 to 0.04.