The CEO of a company wants to estimate the percent of employees that use company computers to go on Facebook during work hours with 95% confidence. He selects a random sample of 150 of the employees and finds that 53 of them logged onto Facebook that day. Compute the 95% confidence interval for the population proportion.
A) $0.35 \pm 1.96 \sqrt { \frac { 0.35 ( 1 - 0.35 ) } { 150 } }$
B) $0.35 \pm 1.645 \sqrt { \frac { 0.35 ( 1 - 0.35 ) } { 150 } }$
C) $0.53 \pm 1.645 \sqrt { \frac { 0.53 ( 1 - 0.53 ) } { 150 } }$
D) $0.53 \pm 1.96 \sqrt { \frac { 0.53 ( 1 - 0.53 ) } { 150 } }$

Question

Quizplus · Accepted Answer

The correct answer is A because the formula for the confidence interval for a proportion is:sample proportion +/- margin of errorwhere the margin of error is:z * sqrt((sample proportion * (1 - sample proportion)) / sample size)where z is the z-score corresponding to the desired confidence level. For a 95% confidence level, z = 1.96.Plugging in the values from the question, we get:0.35 +/- 1.96 * sqrt((0.35 * (1 - 0.35)) / 150)= 0.35 +/- 1.96 * sqrt((0.35 * 0.65) / 150)= 0.35 +/- 1.96 * sqrt(0.00143)= 0.35 +/- 1.96 * 0.038= 0.35 +/- 0.074= (0.276, 0.424)

The CEO of a Company Wants to Estimate the Percent 0.35±1.960.35(1−0.35)1500.35 \pm 1.96 \sqrt { \frac { 0.35 ( 1 - 0.35 ) } { 150 } }0.35±1.961500.35(1−0.35)​​

The CEO of a Company Wants to Estimate the Percent $0.35 \pm 1.96 \sqrt { \frac { 0.35 ( 1 - 0.35 ) } { 150 } }$