Question

Data from the U.S. Department of Education indicates that 46% of business graduate students from private universities had student loans. Suppose you randomly survey a sample of graduate business students from private universities. Consider the sampling distribution (sample size n = 215) for the proportion of these students who have loans.What is the mean of this distribution?What is the standard deviation of this sampling distribution (i.e., the standard error)?

Answer 1

The mean of this distribution is 0.46 and the standard deviation is 0.034.

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 sampling distributions of samples of size n of a proportion p, the mean is
`\mu = p` and the standard deviation is
`s = \sqrt{(p(1-p))/(n)}`

In this question:

`n = 215, p = 0.46`

So

`\mu = 0.46, s = \sqrt{(0.46*0.54)/(215)} = 0.0340`

The mean of this distribution is 0.46 and the standard deviation is 0.034.

Answer 2

For this case the mean is given by:

`\mu = p =0.46`

And the standard deviation would be:

`\sigma = \sqrt{(0.46*(1-0.46))/(215)}= 0.0340`

Explanation:

For this case we have the following info given :

`n = 215` represent the sample size

`p = 0.46` represent the proportion of business graduate students from private universities had student loans

For this case we want to find the distribution for the sample proportion and we know that this distribution is given by:

`\hat p \sim N (p , \sqrt{(p(1-p))/(n)})`

And for this case the mean is given by:

`\mu = p =0.46`

And the standard deviation would be:

`\sigma = \sqrt{(0.46*(1-0.46))/(215)}= 0.0340`