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 66% plan to vote for Smith for governor.

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

(b) A survey of 1500 people finds that 61% support stricter penalties for child abuse.

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

Answer 1

Here' the answer to the question. Please write back in case you've doubts and give a thumbs up in case you're satisfied with the answer.

Margin of error formula, E = Z*sqrt(p*p'/n)

Margin of error for confidence interval of 95% will have a Z of 1.96

a.

n = 500,
Margon of Error = E = 1.96*sqrt(.66*.34/500) = 0.0415 or 4.15%
Confidence interval = Mean +/- E = 66% + /- 4.15% = 61.85% to 70.15%

b.

Similarly,
for n = 1500, p = 61%
E = 1.96*sqrt(.61*.39/1500) = 0.0247 or 2.47%
Confidence interval = Mean +/- E = 61% + /- 4.15% = 56.85% to 65.15%