Question

Students at a certain school were​ surveyed, and it was estimated that 20​% of college students abstain from drinking alcohol. To estimate this proportion in your​ school, how large a random sample would you need to estimate it to within 0.08 with probability 0.95​, if before conducting the study​ (a) you are unwilling to predict the proportion value at your school and​ (b) you use the results from the surveyed school as a guideline.

Answer 1

a) A sample size of at least 251 students is needed.

b) A sample of at least 97 students is needed.

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:

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

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

(a) you are unwilling to predict the proportion value at your school

We need a sample size of at least n.

n is found when M = 0.08.

We wont predict a proportion value for the school, so we use
`\pi = 0.5`

So

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

`0.08 = 1.96\sqrt{(0.5*0.5)/(n)}`

`0.08√(n) = 1.96√(0.5*0.5)`

`√(n) = (1.96√(0.5*0.5))/(0.08)`

`(√(n))^(2) = ((1.96√(0.5*0.5))/(0.08))^(2)`

`n = 150.1`

Rounding up

A sample size of at least 251 is needed.

(b) you use the results from the surveyed school as a guideline.

Now we have that
`\pi = 0.2`

So

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

`0.08 = 1.96\sqrt{(0.2*0.8)/(n)}`

`0.08√(n) = 1.96√(0.2*0.8)`

`√(n) = (1.96√(0.2*0.8))/(0.08)`

`(√(n))^(2) = ((1.96√(0.2*0.8))/(0.08))^(2)`

`n = 96.04`

Rounding up

A sample of at least 97 students is needed.