Question

# A survey was run by a high school student in order to determine what proportion of...

A survey was run by a high school student in order to determine what proportion of mortgage-holders in his town expect to own their house within 10 years. He surveyed 41 mortgage holders and found that the proportion of these that did expect to own their house within 10 years is 0.53. The student decides to construct a 95% confidence interval for the population proportion. a)Calculate the margin of error that the high school student will have. Give your answer as a decimal to 2 decimal places. Margin of error =

A university student finds the survey results of the high school student and believes he should have had a larger sample. The university student surveys 82 mortgage holders in her town and finds that the proportion of these that do expect to own their house within 10 years is again 0.53. This student also constructs a 95% confidence interval for the population proportion. b)Calculate the margin of error that the university student will have. Give your answer as a decimal to 2 decimal places. Margin of error =

a)

sample proportion, = 0.53
sample size, n = 41
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.53 * (1 - 0.53)/41) = 0.0779

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0779
ME = 0.15

b)

sample proportion, = 0.53
sample size, n = 82
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.53 * (1 - 0.53)/82) = 0.0551

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0551
ME = 0.11

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