Using the z-distribution, it is found that since the test statistic is less than the critical value for the right-tailed test, it is found that this does not provide convincing evidence that the proportion of pennies in her containers that are pre-1982 copper pennies is greater than 0.132.

At the null hypothesis, it is tested if the proportion of pennies in her containers that are pre-1982 copper pennies not greater than 0.132, that is:

$H_0: p \leq 0.132$

At the alternative hypothesis, it is tested if it is greater, that is:

$H_1: p > 0.132$

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:

$p = 0.132, n = 50, \overline{p} = \frac{11}{50} = 0.22$

Hence, the value of the test statistic is given by:

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

$z = \frac{0.22 - 0.132}{\sqrt{\frac{0.5(0.5)}{50}}}$

$z = 1.24$

The critical value for a right-tailed test, as we are testing if the proportion is greater than a value, using a 0.05 significance level, is of $z^{\ast} = 1.645$ .

Since the test statistic is less than the critical value for the right-tailed test, it is found that this does not provide convincing evidence that the proportion of pennies in her containers that are pre-1982 copper pennies is greater than 0.132.

You can learn more about the use of the z-distribution to test an hypothesis at brainly.com/question/16313918

6 0

2 years ago