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

Answer:

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

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

Sample of 421 new car buyers, 75 preferred foreign cars. So $n = 421, \pi = \frac{75}{421} = 0.178$

85% confidence level

So $\alpha = 0.15$, z is the value of Z that has a pvalue of $1 - \frac{0.15}{2} = 0.925$, so $Z = 1.44$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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