Hi there, Thanks a lot! The human resources department of a very large organisat

Question

Hi there,

Thanks a lot!

The human resources department of a very large organisation is trying to determine the proportion of all employees that are satisfied with their current position. They randomly select 100 employees and ask them: "Are you satisfied with your current position?" 68 replied yes they were. Construct a 95% confidence interval to estimate the true proportion of all employees at this workplace who were satisfied with their position A 95% confidence interval for the true proportion of all employees at this workplace who were satisfied with their position is between (round your answers to 2 dp) and

Explanation / Answer

Ans:

Sample proportion=68/100=0.68

confidence level,c=0.95

alpha=1-0.95=0.05

alpha/2=0.025

critical z value=normsinv(0.025) or normsinv(0.975)=+/-1.96

95% confidence interval for true proportiom

=0.68+/-1.96*sqrt(0.68*(1-0.68)/100)

=0.68+/-0.09

=(0.59, 0.77)

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