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Construct a 95% confidence interval to estimate the population proportion with a sample proportion equal to 025 and a sample size equal to 120 Click the icon to view a portion of the Cumulative Probabilities for the Standard Normal Distribution table. A 95% confidence interval estimates that the population proportion is between a lower limit of | Round to three decimal places as needed.) | and an upper limit of

Explanation / Answer

#1.

CI for 95%

n = 120

p = 0.25

z-value of 95% CI = 1.9600

SE = sqrt(p*(1-p)/n) = sqrt(0.25*0.75/120) = 0.03953

ME = z*SE = 0.07747

Lower Limit = p - ME = 0.17253

Upper Limit = p + ME = 0.32747

95% CI (0.1725 , 0.3275 )

#2.

CI for 90%

n = 450

p = 0.6

z-value of 90% CI = 1.6449

SE = sqrt(p*(1-p)/n) = 0.02309

ME = z*SE = 0.03799

Lower Limit = p - ME = 0.56201

Upper Limit = p + ME = 0.63799

90% CI (0.562 , 0.638 )

#3.

a)

n = 120

p = 0.45

z-value of 90% CI = 1.6449

SE = sqrt(p*(1-p)/n) = sqrt(0.45*0.55/120) = 0.04541

ME = z*SE = 1.6449*0.04541 = 0.07470

#4.

a)

CI for 99%

n = 320

p = 212/320 = 0.6625

z-value of 99% CI = 2.5758

SE = sqrt(p*(1-p)/n) = sqrt(0.6625((1-0.6625)/320) = 0.02643

ME = z*SE = 0.06809

Lower Limit = p - ME = 0.59441

Upper Limit = p + ME = 0.73059

99% CI (0.5944 , 0.7306 )

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