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 4886 patients treated with the drug, 129 developed adverse reaction of narusea. Construct a 95% confidence intercal for the proportion of adverse reactions.  a. find the best point estimate of the population proportion p  b. identify the value of the margn of error E c. construct the confidence interval

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
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Chapter10: Statistics
Section10.5: Comparing Sets Of Data
Problem 14PPS
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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 4886 patients treated with the drug, 129 developed adverse reaction of narusea. Construct a 95% confidence intercal for the proportion of adverse reactions. 

a. find the best point estimate of the population proportion p 

b. identify the value of the margn of error E

c. construct the confidence interval 

d. write a statement that correctly interprets the confidence interval. choose the correct answer below 

a. one has 95% confidence that the sample proportion is equal to the population proportion. 

b. there is a 95% chance that the true value of the population proportion will fall between the lower bound and the upper bound.

c. 95% of sample proportions will fall between the lower bound and the upper bound. 

d. one has 95% confidence that the interval from the lowre bound to the upper bound actually does contain the true valie of the population proportion. 

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Step 1

Hello! As you have posted more than 3 sub parts, we are answering the first 3 sub-parts.  In case you require the unanswered parts also, kindly re-post that parts separately.

a)

Among 4886 patients treated with the drug, 129 developed adverse reaction of narusea.

best point estimate of the population proportion p is,

p^=1294886=0.0264

 

Step 2

b)

Confidence level is 95%.

Significance level is 5%

From the standard normal table, the z critical value is 1.96.

Consider, the margin of error is,

E=zp^(1-p^)n  =1.960.0264(1-0.0264)4886 =0.0045

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