A magazine reported the results of a random telephone poll commissioned by a television network. Of...

A magazine reported the results of a random telephone poll commissioned by a television network. Of the 1144 men who responded, only 32 said that their most important measure of success was their work. Complete parts a through c. a right parenthesis Estimate the percentage of all males who measure success primarily from their work. Use a 95% confidence interval. Don't forget to check the conditions first.

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Expert Solution

sample proportion, = 0.028
sample size, n = 1144
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.028 * (1 - 0.028)/1144) = 0.0049

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 * 0.0049
ME = 0.01

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.028 - 1.96 * 0.0049 , 0.028 + 1.96 * 0.0049)
CI = (0.0184 , 0.0376)
Therefore, based on the data provided, the 95% confidence interval for the population proportion is 0.0184 < p < 0.0376 , which indicates that we are 95% confident that the true population proportion p is contained by the interval (0.0184 , 0.0376)

orchestra answered 1 year ago

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