Question

Ford provides an option of a sunroof and side air bag package for its Windstar model. This package costs$1400. Ford wants todetermine the percentage of Windstar buyers who would pay $1400 extra for the sunroof and side air bags. How many Windstar buyers must be surveyed if we want to be 95% confident that the sample percentage is within four percentage points of the true percentage for all Windstar buyers

Answer 1

601 buyers must be surveyed.

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`.

How many Windstar buyers must be surveyed if we want to be 95% confident that the sample percentage is within four percentage points of the true percentage for all Windstar buyers?

We need to survey n buyers. n is found when
`M = 0.04`. We dont have an estimate for the proportion, so we use
`\pi = 0.5`, which is when the largest sample will be needed.

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

`0.04 = 1.96\sqrt{(0.5*0.5)/(n)}`

`0.04√(n) = 1.96*0.5`

`√(n) = (1.96*0.5)/(0.04)`

`(√(n))^2 = ((1.96*0.5)/(0.04))^2`

`n = 600.3`

Rounding up

601 buyers must be surveyed.