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 972 new car buyers is drawn. Of those sampled, 700 preferred domestic rather than foreign 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.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The estimated proportion for this case is [tex]\hat p =\frac{700}{972}=0.720[/tex]

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 85% of confidence, our significance level would be given by [tex]\alpha=1-0.85=0.15[/tex] and [tex]\alpha/2 =0.075[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.44, z_{1-\alpha/2}=1.44[/tex]

The confidence interval for the mean is given by the following formula:

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

If we replace the values obtained we got:

[tex]0.72 - 1.44\sqrt{\frac{0.72(1-0.72)}{700}}=0.696[/tex]

[tex]0.72 + 1.44\sqrt{\frac{0.72(1-0.72)}{700}}=0.744[/tex]

The 85% confidence interval would be given by (0.696;0.744)