In a poll 592 of 1002 adults surveyed said they wanted to lo

In a poll, 592 of 1002 adults surveyed said they wanted to lose weight.

Find a 95% confidence interval for the proportion of peope in the population who wanted to lose weight? Round to three decimal places as needed.

Would a 99% confidence interval be wider or narrower? Is value z smaller or greater?

Solution

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

p=592/1002 =0.5908184

So the lower bound is

p - Z*sqrt(p*(1-p)/n ) =0.5908184 - 1.96*sqrt(0.5908184*(1-0.5908184)/1002) =0.560

So the upper bound is

p + Z*sqrt(p*(1-p)/n ) =0.5908184 + 1.96*sqrt(0.5908184*(1-0.5908184)/1002) =0.621

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a 99% confidence interval would be wider because Z is greater.