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)   Assuming that p equals .60 and the sample size is 1,000, what is the probability of observing a sample proportion that is at least .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.

at least 0.64 given that p equals 0.60 (with a sample size of 1,000) is 0.0049. This is based on the assumption that the sample proportion has a normal distribution that enables us to compute the value of Z given by

(p1 - p)/SQRT[p*(1-p)/n] = 2.58 where p1 is the sample proportion and p is 0.60.

Using a Z table, Probability (Z >2.58) = 0.0049.

For, part b),

Null Hypothesis, H₀ : p ≤ 0.6

Alternative Hypothesis, H_a = p > 0.6

QUESTION::::The above testing of the hypothesis was based on a one tailed hypothesis where we are only testing that the sample proportion is higher than 0.60. If instead we conduct a two tailed test where we are testing whether the sample proportion is different from 0.60 (higher or lower), how will the hypothesis statement change in such case and what will be our finding from the test conducted?

It is a common practice to assume a significance level of 0.05 for testing the hypothesis, although we can also pick other significance levels as well. Since the probability of 0.0049 is less than the chosen significance level of 0.05, it allows us to reject the Null hypothesis and accept the alternative that more than 60 percent of all U.S. households in the income class bought life insurance last year.

Answer 1

The null and alternative hypothesis for two-tailed test are:

; i.e., the true proportion of U.S. households in the income class bought life insurance last year is not different than 0.60.

; i.e., the true proportion of U.S. households in the income class bought life insurance last year is different than 0.60.

At significance level we need to test this two-tailed hypothesis.

Given:

sample size, ,

sample proportion, ,

null-hypothesized value,

Test-statistic:

So, the test-statistic is calculated as

P-value: For the test-statistic the p-value is calculated to be as-

So, the p-value for the test-statistic is calculated as

Conclusion: and

Since,

So, at the sample data provides sufficient evidence to reject null hypothesis H₀ . Hence we conclude that, "The proportion of U.S. households in the income class bought life insurance last year is different than 0.60"

In two-tailed hypothesis, we can only conclude that the true proportion of U.S. households in the income class bought life insurance last is different than 0.60