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:

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

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

Answer 2

According to the data given:

• The best point estimate of the population proportion p is 0.036.
• The 95​% confidence interval for the proportion of adverse reactions is (0.03, 0.042).

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$.

Out of 4167 patients treated with the​ drug, 148 developed the adverse reaction of nausea, hence:

$n = 4167, \pi = \frac{148}{4167} = 0.036$

The best point estimate of the population proportion p is 0.036.

95% confidence level

So $\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so $z = 1.96$.  

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.036 - 1.96\sqrt{\frac{0.036(0.964)}{4167}} = 0.03$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.036 + 1.96\sqrt{\frac{0.036(0.964)}{4167}} = 0.042$

The 95​% confidence interval for the proportion of adverse reactions is (0.03, 0.042).

A similar problem is given at https://brainly.com/question/16807970