Question

A previous survey reported that 53% of respondents increased their portfolio value over the past 3 years. How large should a sample be if the margin of error is .03 for a 91% confidence interval

Answer 1

A sample of 796 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)}`

In this question, we have that:

`\pi = 0.53`

91% confidence level

So
`\alpha = 0.09`, z is the value of Z that has a pvalue of
`1 - (0.09)/(2) = 0.955`, so
`Z = 1.695`.

How large should a sample be if the margin of error is .03 for a 91% confidence interval

We need a sample of n, which is found when
`M = 0.03`. So

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

`0.03 = 1.695\sqrt{(0.53*0.47)/(n)}`

`0.03√(n) = 1.695√(0.53*0.47)`

`√(n) = (1.695√(0.53*0.47))/(0.03)`

`(√(n))^2 = ((1.695√(0.53*0.47))/(0.03))^(2)`

`n = 795.2`

Rounding up

A sample of 796 is needed.