Question

Based on historical data in Oxnard college, we believe that 37% of freshmen do not visit their counselors regularly. For this year, you would like to obtain a new sample to estimate the proportiton of freshmen who do not visit their counselors regularly. You would like to be 98% confident that your estimate is within 3.5% of the true population proportion. How large of a sample size is required

Answer 1

Answer:

A sample size of 1031 is required.

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

The margin of error is of:

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

37% of freshmen do not visit their counselors regularly.

This means that $\pi = 0.37$

98% confidence level

So $\alpha = 0.02$, z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$, so $Z = 2.327$.

You would like to be 98% confident that your estimate is within 3.5% of the true population proportion. How large of a sample size is required?

A sample size of n is required.

n is found when M = 0.035. So

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

$0.035 = 2.327\sqrt{\frac{0.37*0.63}{n}}$

$0.035\sqrt{n} = 2.327\sqrt{0.37*0.63}$

$\sqrt{n} = \frac{2.327\sqrt{0.37*0.63}}{0.035}$

$(\sqrt{n})^2 = (\frac{2.327\sqrt{0.37*0.63}}{0.035})^2$

$n = 1030.4$

Rounding up:

A sample size of 1031 is required.