Question

According to data from the 2010 United States Census, 11.4% of all housing units in the United States were vacant. Suppose Maria, a researcher, takes a random sample of 200 housing units in the United States and finds that 15 are vacant. Let ^ p represent the sample proportion of housing units that were vacant. What are the mean and standard deviation of the sampling distribution of ^ p

Answer 1

The mean of the sampling distribution of ^ p is 0.075 = 7.5% and the standard deviation is 0.0186 = 1.86%

Explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For the distribution of the sampling proportions of a proportion p in a sample of size n, we have that the mean is
`p` and the standard deviation is
`s = \sqrt{(p(1-p))/(n)}`

In this problem, we have that:

`n = 200, p = (15)/(200) = 0.075`

So

Mean 0.075

Standard deviation
`s = \sqrt{(0.075*0.925)/(200)} = 0.0186`

The mean of the sampling distribution of ^ p is 0.075 = 7.5% and the standard deviation is 0.0186 = 1.86%