Question

It is believed that as many as 23% of adults over 50 never graduated from high school. We wish to see if this percentage is the same among the 25 to 30 age group. Question 1. How many of this younger age group must we survey in order to estimate the proportion of non-grads to within .10 with 90% confidence? Use the value of p from the over-50 age group. (Round up to the nearest integer.)

Answer 1

We must survey at least 48 people from this younger age group.

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)}`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

How many of this younger age group must we survey in order to estimate the proportion of non-grads to within .10 with 90% confidence? Use the value of p from the over-50 age group.

This is n when
`M = 0.1, \pi = 0.23`

So

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

`0.1 = 1.645\sqrt{(0.23*0.77)/(n)}`

`0.1√(n) = 1.645√(0.23*0.77)`

`√(n) = (1.645√(0.23*0.77))/(0.1)`

`(√(n))^(2) = ((1.645√(0.23*0.77))/(0.1))^(2)`

`n = 47.92`

Rouding up,

We must survey at least 48 people from this younger age group.