Question

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 4167 patients treated with the​ drug, 148 developed the adverse reaction of nausea. Construct a 95​% confidence interval for the proportion of adverse reactions. ​a) Find the best point estimate of the population proportion p. nothing ​(Round to three decimal places as​ needed.)

Answer 1

Answer:

a) $\hat p=\frac{148}{4167}=0.036$

b) $0.0355 - 1.96\sqrt{\frac{0.036(1-0.036)}{4167}}=0.030$

$0.0355 + 1.96\sqrt{\frac{0.036(1-0.036)}{4167}}=0.041$

Step-by-step explanation:

The sample proportion have the following distribution

$\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Part a

The estimated proportion for this case can be calculated like this:

$\hat p=\frac{148}{4167}=0.036$

Part b

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Replacing we got:

$0.0355 - 1.96\sqrt{\frac{0.036(1-0.036)}{4167}}=0.030$

$0.0355 + 1.96\sqrt{\frac{0.036(1-0.036)}{4167}}=0.041$