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 897 new car buyers is drawn. Of those sampled, 304 preferred foreign over domestic cars. Using the data, construct the 95% 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 95% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.308, 0.37)

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:

$n = 897, \pi = \frac{304}{897} = 0.339$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.339 - 1.96\sqrt{\frac{0.339*0.661}{897}} = 0.308$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.339 + 1.96\sqrt{\frac{0.339*0.661}{897}} = 0.37$

The 95% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.308, 0.37)