Question

Find the margin of error and 95% confidence interval for the survey result described. According to a poll of 1100 households, about 71% of adults over the age 24 have graduated from high school

margin of error= ___%

confidence interval between ___% and ___%

Answer 1

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

n = 1100

P = x/n = 0.71

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

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

Margin of error = 1.96* sqrt(0.71*(1 – 0.71)/1100)

Margin of error = 0.0268

Margin of error = 2.68%

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.71 ± 0.0268

Lower limit = 0.71 - 0.0268 = 0.6832

Upper limit = 0.71 + 0.0268 = 0.7368

Confidence interval = (0.6832, 0.7368)

Confidence interval between 68.32% and 73.68%