Question

An automotive manufacturer wants to know the proportion of new car buyers who prefer foreign cars over domestic. Step 2 of 2 : Suppose a sample of 421 new car buyers is drawn. Of those sampled, 75 preferred foreign over domestic cars. Using the data, construct the 85% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars. Round your answers to three decimal places.

Answer 1

The 85% onfidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.151, 0.205).

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)`.

For this problem, we have that:

Sample of 421 new car buyers, 75 preferred foreign cars. So
`n = 421, \pi = (75)/(421) = 0.178`

85% confidence level

So
`\alpha = 0.15`, z is the value of Z that has a pvalue of
`1 - (0.15)/(2) = 0.925`, so
`Z = 1.44`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.178 - 1.44\sqrt{(0.178*0.822)/(421)} = 0.151`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.178 + 1.44\sqrt{(0.178*0.822)/(421)} = 0.205`

The 85% onfidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.151, 0.205).