A recent Gallup Poll interviewed a random sample of 1523 adults. Of these, 868 bought a lottery ticket in the past year.
A 95% confidence interval for the proportion of all adults who bought a lottery ticket in the past year is (assume Gallup used a simple random sample)
A) 0.57 ± 0.00016.
B) 0.57 ± 0.00032.
C) 0.57 ± 0.013.
D) 0.57 ± 0.025.
E) 0.57 ± 0.03.

Question

Quizplus · Accepted Answer

We can use the formula for a confidence interval for a proportion:

p̂ ± z*sqrt((p̂(1-p̂))/n)

where p̂ is the sample proportion, z is the z-score corresponding to the desired confidence level (in this case, 95% corresponds to a z-score of 1.96), and n is the sample size.

Plugging in the values given:

p̂ = 868/1523 = 0.570
z = 1.96
n = 1523

p̂ ± z*sqrt((p̂(1-p̂))/n) = 0.570 ± 1.96*sqrt((0.570(1-0.570))/1523)

Calculating this expression, we get:

0.570 ± 0.025

Therefore, the 95% confidence interval for the proportion of all adults who bought a lottery ticket in the past year is 0.570 ± 0.025 or (0.545, 0.595). Therefore, the best choice is D.

A Recent Gallup Poll Interviewed a Random Sample of 1523