In a survey of 4059 adults, 705 say they have seen a ghost. Construct a 99% conf

Question

In a survey of 4059 adults, 705 say they have seen a ghost. Construct a 99% confidence interval for the population proportion. Interpret the results. A 99% confidence interval for the population proportion is (Round to three decimal places as needed.) Interpret your results. Choose the correct answer below. With 99% confidence, it can be said that the sample proportion of adults who say they have seen a ghost is between the endpoints of the given confidence interval. With 99% confidence, it can be said that the population proportion of adults who say they have seen a ghost is between the endpoints of the given confidence interval. With 99% probability, the population proportion of adults who say they have not seen a ghost is between the endpoints of the given confidence interval. The endpoints of the given confidence interval show that 99% of adults have seen a ghost.

Explanation / Answer

p = 706/4059 = .174

A 99% CI for pop. proportion = .174 +/- 2.575*sqrt(.174 *.826/4059) = .16 to .19

Mean = 50
Stdev = 1.2

a. P(X>=51) = P(Z> (51.5-50) / (1.2/sqrt(11)) = P(Z>4.15) = .00017
b. P(X>=41) = P(Z> (51.5-50) / (1.2/sqrt(43)) = The P-Value is < 0.00001

99% Confidence interval is a confidence level not a probability .So, C and D is false.

Also, CI is not about sample proportion of adults. A is also false therefore.

B is the answer

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