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 60% plan to vote for Smith for govener.

Margin of Error (as a percentage):

Confidence interval _____ % to ______ %

b) a survey of 1500 people finds that 39% support stricter penalties for child abuse.

Margin of Error ( as a percentage):

Confidence interval ______% to ______%

Answer 1

Solution:

Part a

We are given n = 500, P = 0.60

Confidence level = 95%

Z = 1.96

(by using z-table)

Margin of error = E = Z*sqrt(P*(1 – P)/n)

Margin of error = E = 1.96*sqrt(0.60*(1 – 0.60)/500)

Margin of error = E = 0.0429

Confidence interval = P -/+ E

Lower limit = P – E = 0.60 – 0.0429 = 0.5571

Upper limit = P + E = 0.60 + 0.0429 = 0.6429

Confidence interval = 55.71% to 64.29%

Part b

We are given n = 1500, P = 0.39

Confidence level = 95%

Z = 1.96

(by using z-table)

Margin of error = E = Z*sqrt(P*(1 – P)/n)

Margin of error = E = 1.96*sqrt(0.39*(1 – 0.39)/1500)

Margin of error = E = 0.0247

Confidence interval = P -/+ E

Lower limit = P – E = 0.39 – 0.0247 = 0.3653

Upper limit = P + E = 0.39 + 0.0247 = 0.4147

Confidence interval = 36.53% to 41.47%