# 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 th

Question

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

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1 year 2021-08-21T22:12:28+00:00 1 Answers 0 views 0

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.