Question

In a clinical test with 2161 subjects, 1214 showed improvement from the treatment. Find the margin of error for the 95% confidence interval used to estimate the population proportion.

Answer 1

Answer:

The margin of error for the 95% confidence interval used to estimate the population proportion is of 0.0209.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, 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 $1 - \frac{\alpha}{2}$.

The margin of error is of:

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

In a clinical test with 2161 subjects, 1214 showed improvement from the treatment.

This means that $n = 2161, \pi = \frac{1214}{2161} = 0.5618$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

Margin of error:

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

$M = 1.96\sqrt{\frac{0.5618*0.4382}{2161}}$

$M = 0.0209$

The margin of error for the 95% confidence interval used to estimate the population proportion is of 0.0209.