Question

According to a poll, 76% of California adults (385 out of 506 surveyed) feel that education is one of the top issues facing California. We wish to construct a 90% confidence interval for the true proportion of California adults who feel that education is one of the top issues facing California. Find the error bound. (Round your answer to three decimal places.)

Answer 1

Answer:

The error bound is 3.125%.

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 interval $1-\alpha$, we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

A sample of 506 California adults.. This means that $n = 506$.

76% of California adults (385 out of 506 surveyed) feel that education is one of the top issues facing California. This means that $\pi = 0.76$

We wish to construct a 90% confidence interval

So $\alpha = 0.10$, z is the value of Z that has a pvalue of $1 - \frac{0.10}{2} = 0.95$, so $z = 1.645$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.76 - 1.645\sqrt{\frac{0.76*0.24}{506}} = 0.7288$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.76 + 1.645\sqrt{\frac{0.76*0.24}{506}} = 0.7913$

The error bound of the confidence interval is the division by 2 of the subtraction of the upper limit by the lower limit. So:

$EBM = \frac{0.7913 - 0.7288}{2} = 0.03125$

The error bound is 3.125%.