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 768 new car buyers is drawn. Of those sampled, 192 preferred foreign over domestic cars. Using the data, construct the 80% 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 80% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.23, 0.27).

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)`.

Suppose a sample of 768 new car buyers is drawn. Of those sampled, 192 preferred foreign over domestic cars.

This means that
`n = 768, \pi = (192)/(768) = 0.25`

80% confidence level

So
`\alpha = 0.2`, z is the value of Z that has a pvalue of
`1 - (0.2)/(2) = 0.95`, so
`Z = 1.28`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.25 - 1.28\sqrt{(0.25*0.75)/(768)} = 0.23`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} =  0.25 + 1.28\sqrt{(0.25*0.75)/(768)} = 0.27`

The 80% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.23, 0.27).