Question

A news article estimated that only 5% of those age 65 and older who prefer to watch the news, rather than to read or listen, watch the news online. This estimate was based on a survey of a large sample of adult Americans. Consider the population consisting of all adult Americans age 65 and older who prefer to watch the news, and suppose that for this population the actual proportion who prefer to watch online is 0.05. A random sample of n = 100 people will be selected from this population and p, the proportion of people who prefer to watch online, will be calculated.

(a) What are the mean and standard deviation of the sampling distribution of p? (Round your standard deviation to four decimal places.)
(b) Is the sampling distribution of p approximately normal for random samples of size n 100? Explain.
i. The sampling distribution of p is approximately normal because np is less than 10.
ii. The sampling distribution of p is approximately normal because np is at least 10.
iii. The sampling distribution of p is not approximately normal because np is less than 10
iv. The sampling distribution of p is not approximately normal because np is at least 10
v. The sampling distribution of p is not approximately normal because n(1 - p) is less than 10.

Answer 1

a) The mean is 0.05 and the standard deviation is 0.0218.

b) v. The sampling distribution of p is not approximately normal because n(1 - p) is less than 10.

Explanation:

Central Limit Theorem

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)}`.

If
`np \geq 10` and
`n(1-p) \geq 10`, the sampling distribution of the sample proportion p will be approximately normal.

Consider the population consisting of all adult Americans age 65 and older who prefer to watch the news, and suppose that for this population the actual proportion who prefer to watch online is 0.05.

This means that
`p = 0.05`

A random sample of n = 100 people

So
`n = 100`

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

Mean
`\mu = p = 0.05`

Standard deviation
`s = \sqrt{(p(1-p))/(n)} = \sqrt{(0.05*0.95)/(100)} = 0.0218`

The mean is 0.05 and the standard deviation is 0.0218.

(b) Is the sampling distribution of p approximately normal for random samples of size n 100? Explain.

`np = 100*0.95 = 95 > 10`

`n(1-p) = 100*0.05 = 5 < 10`

As n(1-p) < 10, it is not approximately normal, option v.