Question

In a particular large Midwest city, a simple random sample of 400 people reveals that 128 of them know how to ski. 1. A 99.8% confidence interval for the proportion of people in the city who know how to ski is

a. (0.272, 0.368)
b. (0.304, 0.336)
c. (0.248, 0.392)
d. (0.282, 0.358)

Answer 1

Answer

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Step-by-step explanation:

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Answer 2

Answer:

c. (0.248, 0.392)

Step-by-step explanation:

We have a large simple random sample of n = 400 people and 128 of them know how to ski. Let p be the true proportion of people in the city who know how to ski, then $\hat{p} = 128/400 = 0.32$ is an estimated of p. An approximation of the standard deviation of $\hat{p}$ is $\sqrt{\hat{p}(1-\hat{p})/n} = \sqrt{0.32(1-0.32)/400} = 0.0233$. Because we want a 99.8% confidence interval for the proportion of people in the city who know how to ski, we have that $\alpha = 0.002$, we need $z_{\alpha/2} = z_{0.002/2} = z_{0.001} = -3.0902$, i.e., the 0.1th quantile of the standard normal distribution. Therefore, the 99.8% confidence interval is given by $0.32\pm (3.0902)(0.0233)$, i.e., (0.2480, 0.3920)