Question

A researcher wishes to estimate, with 95% confidence, the proportion who own a laptop. A previous study shows that 70% of those interviewed had a laptop computer. How large a sample should the researcher select so that the estimate will be within 3% of the true proportion?

Answer 1

The sample needs to be at least n = 897 in size.

Explanation:

Given

Confidence Level = 95%

Let p = those who had laptop

p = 70% = 0.7

Margin of Error = 3% = 0.03

Let n = required sample.

To estimate the proportion of laptop owners with 95% confidence, first the z-value is calculated.
`z_((a/2))`

Given that confidence level = 95%

α = 1 − 95 %

α = 1 − 0.95

α = 0.05

So;

`z_((a/2))` =
`z_((0.05/2))`

`z_((a/2))` =
`z_((0.025))`

From the z table

`z_((0.025)) = 1.96`

Using formula for margin of error

`M.E = z_((0.025)) * \sqrt{(pq)/(n) }`

where p + q =1

q = 1 - p

q = 1 - 0.7

q = 0.3

So,

`M.E = z_((0.025)) * \sqrt{(pq)/(n) }`

`0.03 = 1.96 * \sqrt{(0.7 * 0.3)/(n) }` --- Make n the subject of formula

First, square both sides

`0.03^(2) = 1.96^(2) * {(0.7 * 0.3)/(n) }` ------ Multiply both sides by
`{(n)/(0.03^(2))}`

`0.03^(2) * {(n)/(0.03^(2))} = 1.96^(2) * {(0.7 * 0.3)/(n) } * {(n)/(0.03^(2))}`

`n = 1.96^(2) * {(0.7 * 0.3)/(0.03^(2)) }`

`n = {(0.806736)/(0.0009) }`

`n = 896.473`

Hence, the sample needs to be at least n = 897 in size.