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 .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)   Based on the p value you calculate with a sample size of 1,000, what is the confidence level of the sample indicating that 60 percent of households bought life insurance?

b)  Based on your answer in part a, how does the statistical test indicate that more than 60 percent of all U.S. households in the income class bought life insurance last year? Explain.

Answer 1

Null Hypothesis, H₀ : 60 Percent of the US households bought life insurance, i.e P = 0.6

Alternate Hypothesis, H_a: More than 60 percent of the US households bought life insurance i.e P > 0.6

In this case, Sample size, n = 1000

Households that bought life insurance = 640

Therefore, sample proportion, = 640 / 1000 = 0.64

Now, in this case, Z = = = 2.582

Here , Probability that proportion will be greater than 0.6 is given by,  P(Z < 2.582) = 0.9941 & p value is 0.0058.

Therefore, we can say with 99.41% confidence that proportion of households with insurance is greater than 0.6

b. Now, in this case, since p value is 0.0058, which is less than 0.05, we can conclude that the test results are significant .

Also, Alternate Hypothesis is accepted & Null Hypothesis is rejected,  which means that More than 60 percent of the US households bought life insurance.