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

A sample size of 1031 is required.

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is of:

`M = z\sqrt{(\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 - (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{(\pi(1-\pi))/(n)}`

`0.035 = 2.327\sqrt{(0.37*0.63)/(n)}`

`0.035√(n) = 2.327√(0.37*0.63)`

`√(n) = (2.327√(0.37*0.63))/(0.035)`

`(√(n))^2 = ((2.327√(0.37*0.63))/(0.035))^2`

`n = 1030.4`

Rounding up:

A sample size of 1031 is required.