An online site presented this question, \'Would the recent norovirus outbreak de
ID: 3150125 • Letter: A
Question
An online site presented this question, 'Would the recent norovirus outbreak deter you from taking a cruise?' Among the 34,889 people who responded, 61% answered 'yes'. Use the sample data to construct a 90% confidence interval estimate for the proportion of the population of all people who would respond 'yes' to that question. Does the confidence interval provide a good estimate of the population proportion (Round to three decimal places as needed.) Does the confidence interval provide a good estimate of the population proportion? Yes, the sample is large enough to provide a good estimate of the population proportion. N, the sample is a voluntary sample and might not be representative of the population. Yes, all the assumptions for a confidence interval are satisfied. No, the responses are not independent.Explanation / Answer
A)
Note that
p^ = point estimate of the population proportion = x / n = 0.61
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.002611277
Now, for the critical z,
alpha/2 = 0.05
Thus, z(alpha/2) = 1.644853627
Thus,
Margin of error = z(alpha/2)*sp = 0.004295169
lower bound = p^ - z(alpha/2) * sp = 0.605704831
upper bound = p^ + z(alpha/2) * sp = 0.614295169
Thus, the confidence interval is
( 0.605704831 , 0.614295169 ) [ANSWER]
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b)
This is an online poll, so only those who go to that website gets sampled. Hence,
OPTION B: No, the sample is a voluntary sample and might not be a representative of the population.