Question

Suppose that we wish to assess whether more than 60 percent of all U.S. households in a particular income class bought life insurance last year. That is, we wish to assess whether p, the proportion of all U.S. households in the income class that bought life insurance last year, exceeds 0.60. Assume that an insurance survey is based on 1,000 randomly selected U.S. households in the income class and that 640 of these households bought life insurance last year.

a) Assuming that p equals 0.60 and the sample size is 1,000, what is the probability of observing a sample proportion that is at least 0.64?
b) Based on your answer in part (a), do you think more than 60 percent of all U.S. households in the income class bought life insurance last year? Explain.

Answer 1

A. 0.0049

B. Yes

Explanation:

Sample proportion = 0.64

N = 1000

Population proportion = 0.60

We solve for standard deviation

= √p(1-p)/n

= √0.60(1-0.60)/1000

= √0.60x0.40/1000

= √0.00024

= 0.0155

A.

The probability of sample >=0.64

Z>=0.64-0.60/0.0155

Z >= 0.04/0.0155

So z >= 2.5806

Using excel this equal to 0.0049

0.0049 is probability of sample proportion being 0.64 at least.

B.

This answer in a shows that than 60% of households in the united states income class purchased life insurance last year.