Please answer with details 250 out of 360 employees agree a

Please answer with details

250 out of 360 employees agree a company project. (1) Find 90% confidence interval of the proportion of agreed employees p normal distribution: (2) Find 95% confidence interval of p:

Solution

(1) phat=250/360 = 0.6944444

Given a=1-0.9=0.1, Z(0.05) = 1.645 (from standard normal table)

So the lower bound is

p - Z*sqrt(p*(1-p)/n) = 0.6944444 -1.645*sqrt(0.6944444*(1-0.6944444)/360) =0.6545071

So the upper bound is

p + Z*sqrt(p*(1-p)/n) = 0.6944444 +1.645*sqrt(0.6944444*(1-0.6944444)/360)=0.7343817

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(2) Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So the lower bound is

p - Z*sqrt(p*(1-p)/n) = 0.6944444 -1.96*sqrt(0.6944444*(1-0.6944444)/360) =0.6468596

So the upper bound is

p + Z*sqrt(p*(1-p)/n) = 0.6944444 +1.96*sqrt(0.6944444*(1-0.6944444)/360)=0.7420292