B) identify the value of the margin of error (E)C)Construct the confidence interval ### Understanding Confidence Intervals and Proportions in Clinical Trials#### Problem Statement: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 4939 patients treated with the drug, 166 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.)*[ ]

Name: B) identify the value of the margin of error (E)C)Construct the confid
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Description: B) identify the value of the margin of error (E)C)Construct the confidence interval ### Understanding Confidence Intervals and Proportions in Clinical Trials#### Problem Statement:Use the sample data and confidence level given below to complete parts

Answered: Use the sample data and confidence…

MATLAB: An Introduction with Applications

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Author:Amos Gilat

Publisher:Amos Gilat

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B) identify the value of the margin of error (E)

C)Construct the confidence interval

### Understanding Confidence Intervals and Proportions in Clinical Trials

#### Problem Statement:
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 4939 patients treated with the drug, 166 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.)*

[ ]

Expert Solution

Step 1

Given
Total number of patients treated with drug = n = 4939
The total number of patients developed the adverse reaction of nausea = 166
The proportion of patients developed the adverse reaction of nausea = p = 166/4939 = 0.0336

The value of n*p and n(1-p) are 166 and 4773 both are more than 10, so we assume normal approximation as the sample size is large.

The 99% critical Z value is 2.575

The point estimate = p = 0.0336

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