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).

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What is negative -6 in simplest form

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it takes 45 pounds of seeds to completely plant a 5 acre field how many acres can be planted per pound of seed

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