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.

Answer 1

Answer:

a) The probability is P=0.3982.

b) No. There is no enough evidence to say that the proportion of the population is not 0.6.

Step-by-step explanation:

a) In this question, we have a sample of the population, of size n=1000. To know what is the probability of observing a sample proportion that is at least 0.64 we have to know the parameters of the sampling distribution.

The parameters of the sampling distribution of the proportion will be:

$\mu=\pi=0.6\\\\ \sigma=\sqrt{\frac{\pi(1-\pi)}{N} } =\sqrt{\frac{0.6(1-0.6)}{1000} }= 0.0155$

With these parameters, we can calculate the z-value for p=0.64 as:

$z=\frac{p-\pi}{\sigma}=\frac{0.64-0.6}{0.0155}= 0.258$

Then, we can calculate the probability of having a sample with p≥0.64:

$P(p\geq0.64)=P(z\geq0.258)=0.3982$

b) To know if the proportion of the population is greater than 0.6 the rigth thing to do is perform a hypothesis test, in which we test the following hypothesis:

$H_0: \pi\leq0.6\\\\H_1:\pi>0.6$

If the null hypothesis is rejected, we can conclude that there is evidence that the proportion of the population is greater than 0.6.

First, we assume a significance level of 0.05.

Second, we calculate the z value:

$z=\frac{p-\pi-0.5/N}{\sigma} =\frac{0.64-0.6-0.5/1000}{0.0155} =\frac{0.0395}{0.0155} =2.54$

The P-value for this z is P=0.4. The P-value is greater than the significance level, what means that there is no evidence to reject the null hypothesis.

In other words, a sample mean of 0.64 is a quite probable value even if the proportion of the population is 0.6.