In a clinical trial of a certain drug, 16 subjects experience headaches among the 231 subjects treated with the drug. Construct a 95% (Wald) confidence interval estimate for the proportion of treated subjects who experience headaches, completing parts (a) through (d) below.

A. Find the best point estimate of the population proportion.

B. Identify the value of the margin of error E.

C. Construct the confidence interval.

D. write a statement that correctly interprets the confidence interval.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

Part a

The estimated proportion for this case is:

[tex]\hat p=\frac{16}{231}= 0.0693[/tex]

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 [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

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

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

The margin of error is given by:

[tex] ME= z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

And replacing we got:

[tex] ME= 1.96 *\sqrt{\frac{0.0693 (1-0.0693)}{231}}= 0.0328[/tex]

Part c

If we replace the values obtained we got:

[tex]0.0693 - 1.96\sqrt{\frac{0.0693(1-0.0693)}{231}}=0.0365[/tex]

[tex]0.0693 + 1.96\sqrt{\frac{0.0693(1-0.0693)}{231}}=0.1021[/tex]

The 95% confidence interval would be given by (0.0365;0.1021)

Part d

For this case we can say that with 95% of confidence the true proportion of subjects with headaches is between 0.0365 and 0.1021