A recent Gallup Poll interviewed a random sample of 1523 adults. Of these, 868 bought a lottery ticket in the past year.
Suppose that in fact (unknown to Gallup) exactly 60% of all adults bought a lottery ticket in the past year. If Gallup took many simple random samples of 1523 people, the sample proportion who bought a ticket would vary from sample to sample. The sampling distribution would be close to normal with
A) mean 0.6 and standard deviation 0.00016.
B) mean 0.6 and standard deviation 0.0126.
C) mean 0.6 and standard deviation 0.4899.
D) mean 0.6 and standard deviation 0.0251.

Question

Quizplus · Accepted Answer

The standard deviation of the sampling distribution of a proportion is calculated using the formula $\sqrt{\frac{p(1-p)}{n}}$, where $p$ is the population proportion and $n$ is the sample size. Using $p = 0.6$ and $n = 1523$, the calculation is $\sqrt{\frac{0.6(1-0.6)}{1523}} = \sqrt{\frac{0.6(0.4)}{1523}} = \sqrt{\frac{0.24}{1523}} \approx 0.0126$.

A Recent Gallup Poll Interviewed a Random Sample of 1523