Question

A poll of 1769 U.S. adults found that 84% regularly used Facebook as a news source.

Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 90% level of confidence. Round all answers to 2 decimal places.

Margin of Error (as a percentage):
Confidence Interval: % to %

Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 95% level of confidence. Round all answers to 2 decimal places.

Margin of Error (as a percentage):
Confidence Interval: % to %

Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 99% level of confidence. Round all answers to 2 decimal places.

Margin of Error (as a percentage):
Confidence Interval: % to %

The more error we allow, the less precise our estimate. Therefore, as the confidence level increases, the precision of our estimate

• increases
• decreases
• stays roughly the same

Answer 1

Solution :

Given that,

n = 1769

Point estimate = sample proportion = = 0.84

1 - = 1 - 0.84 = 0.16

a) At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 (((0.84 * 0.16) / 1769 )

= 0.0143

A 90% confidence interval for population proportion p is ,

± E

= 0.84  ± 0.0143

( 0.8257, 0.8543 )

= ( 82.57%, 85.43% )

b) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.960}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 (((0.84 * 0.16) / 1769)

= 0.0171

A 95% confidence interval for population proportion p is ,

± E

= 0.84  ± 0.0171

( 0.8229, 0.8571 )

= ( 82.29%, 85.71% )

c) At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z_{0.005 = 2.576}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 (((0.84 * 0.16) / 1769)

= 0.0225

A 99% confidence interval for population proportion p is ,

± E

= 0.84  ± 0.0225

( 0.8175, 0.8625 )

= ( 81.75%, 86.25% )

The more error we allow, the less precise our estimate. Therefore, as the confidence level increases, the precision of our estimate increases