Question

It has been found that 50.3% of U.S. households own stocks and mutual funds. A random sample of 300 heads of households indicated that 171 owned some type of stock. At what level of significance in a two- tailed test would you conclude that this was a significant difference

Answer 1

Finding the p-value of the test using the z-distribution, as we are working with a proportion, it is found that this was a significant difference for levels of significance of 0.02 and lower.

What are the hypothesis test?

At the null hypothesis, we test if the proportion is of 50.3%, hence:

$H_0: p = 0.503$

We have a two-tailed test, hence at the alternative hypothesis we test if the proportion is different, that is:

$H_1: p \neq 0.503$

What is the test statistic?

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

• $\overline{p}$ is the sample proportion.

• p is the proportion tested at the null hypothesis.

• n is the sample size.

In this problem, the parameters are:

$n = 300, \overline{p} = \frac{171}{300} = 0.57, p = 0.503$

Hence:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.57 - 0.503}{\sqrt{\frac{0.503(0.497)}{300}}}$

$z = 2.32$

What is the p-value?

Considering a two-tailed test, using a z-distribution calculator, the p-value is of 0.0203. Hence this was a significant difference for levels of significance of 0.02 and lower.

More can be learned about the z-distribution at brainly.com/question/26454209