Question

Use the given data to find the minimum sample size required to estimate a population proportion or percentage. Margin of error: seven percentage points, confidence level 95%, from a prior study,^p is estimated by the decimal equivalent of 42%

n=_____ (round to the nearest integer.)

Answer 1

`n = 191`

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence interval
`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`.

In this problem, we have that:

`M = 0.07, \pi = 0.42`

We have to find n

`0.07 = 1.96*\sqrt{(0.42*0.58)/(n)}`

`0.07√(n) = 1.96*√(0.42*0.58)`

`0.07√(n) = 0.9674`

`√(n) = (0.9674)/(0.07)`

`√(n) = 13.82`

`√(n)^(2) = (13.82)^(2)`

`n = 190.9`

So, rounded to the nearest integer

`n = 191`