Question

Beth wants to determine a 80 percent confidence interval for the true proportion of high school students in the area who attend their home basketball games. how large of a sample must she have to get a margin of error less than 0.04? [if no estimate is known for p, let p^ = 0.5]

Answer 1

We need to sample at least 256 high school students.

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 for the interval is:

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

80% confidence level

So
`\alpha = 0.2`, z is the value of Z that has a pvalue of
`1 - (0.2)/(2) = 0.9`, so
`Z = 1.28`.

How large of a sample must she have to get a margin of error less than 0.04?

We need a sample of at least n high school students, in which n is found when M = 0.04.

We use
`\pi = 0.5`, since no estimate is know. So

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

`0.04 = 1.28\sqrt{(0.5*0.5)/(n)}`

`0.04√(n) = 1.28*0.5`

`√(n) = (1.28*0.5)/(0.04)`

`(√(n))^(2) = ((1.28*0.5)/(0.04))^(2)`

`n = 256`

We need to sample at least 256 high school students.

Answer 2

TO solve this question on "how large of a sample must Beth have to get a margin of error less than 0.04", we will use the margin of error formula:

`E=Z\sqrt{(\hat p(1-\hat p))/(n)}`

Here, for a 80% confidence level,
`Z=1.28`

`\hat p =0.5` (given)

Thus, margin of error, E=0.04

rearranging we get:

`n=((Z)/(E))^2\hat p(1-\hat p)`

Plugging in gives us:

`n=((1.28)/(0.04))^2* 0.5(1-0.5)=1024*0.25=256`

Thus, Beth's sample for the true proportion of high school students in the area who attend their home basketball games must have 256 students.