Question

The manager of a customer support center takes an SRS of 40 reported issues and finds that 22% of the sampled issues required more than one call to resolve. The manager may take an SRS like this each month. Suppose that it is really an average of 25% of the approximately 1000 issues reported per month that require more than one call.

Let represent the proportion of a sample of 40 reported issues that require more than one call to resolve. What are the mean and standard deviation of the sampling distribution of p?

Answer 1

The mean of the sampling distribution of p is 0.25 and the standard deviation is 0.0685.

Explanation:

Central Limit Theorem

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 a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`

25% of the approximately 1000 issues reported per month that require more than one call.

This means that
`p = 0.25`

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

Sample of 40 means that
`n = 40`.

By the Central Limit Theorem,

The mean is
`\mu = p = 0.25`

The standard deviation is
`s = \sqrt{(p(1-p))/(n)} = \sqrt{(0.25*0.75)/(40)} = 0.0685`

The mean of the sampling distribution of p is 0.25 and the standard deviation is 0.0685.