Question

A company wants to estimate, at a 95% confidence level, the proportion of all families who own its product. A preliminary sample showed that 30.0% of the families in this sample own this company's product. The sample size that would limit the margin of error to be within 0.047 of the population proportion is:

Answer 1

The sample size is of 366.

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 z-score that has a p-value of
`1 - (\alpha)/(2)`.

The margin of error is of:

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

A preliminary sample showed that 30.0% of the families in this sample own this company's product.

This means that
`\pi = 0.3`

95% confidence level

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

The sample size that would limit the margin of error to be within 0.047 of the population proportion is:

This is n for which
`M = 0.047`. So

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

`0.047 = 1.96\sqrt{(0.3*0.7)/(n)}`

`0.047√(n) = 1.96√(0.3*0.7)`

`√(n) = (1.96√(0.3*0.7))/(0.047)`

`(√(n))^2 = ((1.96√(0.3*0.7))/(0.047))^2`

`n = 365.2`

Rounding up:

The sample size is of 366.