Question

B-D

7.1.14-T Question Help 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 4487 patients treated with the drug, 110 developed the adverse reaction of nausea. Construct a 95% confidence interval for the proportion of adverse reactions. b) Identify the value of the margin of error E. (Round to three decimal places as needed.) Enter your answer in the answer box and then click Check Answer

Answer 1

Nimber of patients treated using drug to prevent blood clot = n = 4487

Number of patients showing adverse effect of nausea = X = 110

sample proportion is computed as follows

$p = \frac{X}{n} = \frac{110}{4487} = 0.025$

The critical value for α=0.05 is

zc​(upper) = 1.96 using excel formula =NORM.S.INV(0.975)

zc​(lower) = -1.96 using excel formula =NORM.S.INV(0.025)

So we use absolute value of z while calculating confidence interval and the corresponding confidence interval is computed as

$CI = \left ( p - z_{c}\sqrt{\frac{p(1-p)}{n}}, p + z_{c}\sqrt{\frac{p(1-p)}{n}} \right )$

$CI = \left ( 0.025 - 1.96\sqrt{\frac{0.025(1-0.025)}{4487}}, 0.025 + 1.96\sqrt{\frac{0.025(1-0.025)}{4487}} \right )$

$CI = \left ( 0.025 - 0.0046, 0.025 + 0.0046} \right )$

CI = (0.0204, 0.0296

The 95% confidence interval for the population proportion is  0.0204<p<0.0296, which indicates that we are 95% confident that the true population proportion of patients showing adverse effect of nausea is contained by the interval (0.0204, 0.0296).

Margin of error is

$E = z_{c}\sqrt{\frac{p(1-p)}{n}}$

$E = 1.96\sqrt{\frac{0.025(1-0.025)}{4487}}$

$E = 0.0046 \cong 0.005$

So margin of error is E = 0.005(Round to three decimal places)