SOLUTION: 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 55% plan to vote for

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Question 1206714: 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 55% plan to vote for Smith for governor.
Margin of Error (as a percentage):
Confidence Interval:
% to
%
b) A survey of 1500 people finds that 57% support stricter penalties for child abuse.
Margin of Error (as a percentage):
Confidence Interval:
% to
%
Answer by MathLover1(20850) About Me  (Show Source):
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a) A survey of 500+people finds that 55% plan to vote for Smith for governor.

margin of error = z+%2A+sqrt%28p+%2A+%281+-+p%29+%2F+n%29
where z is the z-score for the 95% confidence level (z=1.96), p+is the sample proportion (p=0.55), and n is the sample size (n=500).
margin of error = 1.96+%2A+sqrt%280.55+%2A+%281+-+0.55%29+%2F+500%29
margin of error = 1.96+%2A+sqrt%280.000495%29
margin of error =+1.96+%2A+0.022248595461286987
margin of error =+0.0436+

confidence interval = p +- margin of error
confidence interval = 0.55+%2B-+0.0436
confidence interval is:
0.55+-+0.0436=0.5064+ or 50.64%
0.55+%2B+0.0436=0.5936 or 59.36%

We can be 95% confident that the true proportion of people who plan to vote for Smith for governor is between 50.64% and 59.36%.


b) A survey of 1500 people finds that 57% support stricter penalties for child abuse.

margin of error = 1.96+%2A+sqrt%280.57+%2A+%281+-+0.57%29+%2F+1500%29
margin of error = 1.96+%2A+0.01278280094502
margin of error = 0.02505

confidence interval =+0.57+%2B-0.02505

confidence interval is:
0.57+-0.02505=0.5450
0.57+%2B0.02505=0.5951

We can be 95% confident that the true proportion of people who support stricter penalties for child abuse is between 54.50% and 59.51%.