Question

If a manufacturer conducted a survey among randomly selected target market households and wanted to be 95​% confident that the difference between the sample estimate and the actual market share for its new product was no more than 9​%, what sample size would be​ needed?

Answer 1

We need a sample size of least 119

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)}`

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`.

Sample size needed

At least n, in which n is found when
`M = 0.09`

We don't know the proportion, so we use
`\pi = 0.5`, which is when we would need the largest sample size.

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

`0.09 = 1.96\sqrt{(0.5*0.5)/(n)}`

`0.09√(n) = 1.96*0.5`

`√(n) = (1.96*0.5)/(0.09)`

`(√(n))^(2) = ((1.96*0.5)/(0.09))^(2)`

`n = 118.6`

Rounding up

We need a sample size of least 119