Question

A publisher reports that 40% of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 200 found that 31% of the readers owned a particular make of car. Determine the P-value of the test statistic. Round your answer to four decimal places.

Answer 1

`p-value = 0.9953`

Explanation:

Given

`P = 40\%` --- proportion of population

`n = 200` -- samples

`p = 31\%` -- proportion of samples

Required

Determine the p value

First, calculate the standard deviation (
`\sigma`)

`\sigma = \sqrt{(P*(1 - P))/(n)}`

`\sigma = \sqrt{(40\%*(1 - 40\%))/(200)}`

`\sigma = \sqrt{(0.40*(1 - 0.40))/(200)}`

`\sigma = \sqrt{(0.40*0.60)/(200)}`

`\sigma = \sqrt{(0.24)/(200)}`

`\sigma = √(0.0012)`

`\sigma = 0.0346`

Next, we calculate the z score

`z = (p - P)/(\sigma)`

`z = (40\% - 31\%)/(0.0346)`

`z = (9\%)/(0.0346)`

`z = (0.09)/(0.0346)`

`z = 2.6012`

`z \approx 2.60`

From the z table, the p value of
`z = 2.60` is: 0.9953

Hence:

`p-value = 0.9953`