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vichka [17]
3 years ago
6

we believe that 42% 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
Mathematics
1 answer:
Maksim231197 [3]3 years ago
3 0

Answer:

A sample of 1077 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}}

42% of freshmen do not visit their counselors regularly.

This means that \pi = 0.42

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.

How large of a sample size is required?

A sample size of n is required, and n is found when M = 0.035. So

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

0.035 = 2.327\sqrt{\frac{0.42*0.58}{n}}

0.035\sqrt{n} = 2.327\sqrt{0.42*0.58}

\sqrt{n} = \frac{2.327\sqrt{0.42*0.58}}{0.035}

(\sqrt{n})^2 = (\frac{2.327\sqrt{0.42*0.58}}{0.035})^2

n = 1076.8

Rounding up:

A sample of 1077 is required.

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