Question

According to data from the 2010 United States Census, 71.6% of the total United States population was over the age of 20. Suppose Maria, a researcher, takes a random sample of 70 Americans and finds that 42 are over the age of 20. Let p represent the sample proportion of Americans that were over the age of 20.
What are the mean and standard deviation of the sampling distribution of p? Provide your answer rounded to three decimal places.

Answer 1

Answer:

Mean is 0.6.

Standard deviation is 0.059

Step-by-step explanation:

The first step is finding the sample proportion p.

Suppose Maria, a researcher, takes a random sample of 70 Americans and finds that 42 are over the age of 20.

This means that $p = \frac{42}{70} = 0.6$

What are the mean and standard deviation of the sampling distribution of p?

Mean:

The mean of a proportion is the same as the proportion. So the mean is 0.6

Standard deviation:

Proportion p in a sample of size n. The standard deviation is:

$s = \sqrt{\frac{p(1-p)}{n}}$

Sample size 70, so:

$s = \sqrt{\frac{0.6*0.4}{70}} = 0.059$