Question

Find the margin of error and 95% confidence interval for the following surveys. Round all answers to 2 decimal places.

a) A survey of 500 people finds that 67% plan to vote for Smith for governor. Margin of Error (as a percentage): answer  _____ Confidence Interval: answer _____ % to answer ____ %

b) A survey of 1500 people finds that 44% support stricter penalties for child abuse.

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

Answer 1

Solution :

Given that,

a) n = 500

Point estimate = sample proportion = = 0.67

1 - = 1 - 0.67 = 0.33

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)

E = 1.96 (((0.67 * 0.33) / 500)

E = 0.0412

E = 4.12%

A 95% confidence interval for population proportion p is ,

± E

= 0.67  ± 0.0412

= ( 0.6288, 0.7112 )

( 62.88%, 71.12% )

b) n = 1500

Point estimate = sample proportion = = 0.44

1 - = 1 - 0.44 = 0.56

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)

E = 1.96 (((0.44 * 0.56) / 1500)

E = 0.0251

E = 2.51%

A 95% confidence interval for population proportion p is ,

± E

= 0.44 ± 0.0251

= ( 0.4149, 0.4651)

( 41.49%, 46.51% )