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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)

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1 vote

Answer:

c. (0.248, 0.392)

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} = √(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)

User Yelliver
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Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

In a particular large Midwest city, a simple random sample of 400 people reveals that-example-1
User Venera
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