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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:
[tex]H_0: p \leq 0.132[/tex]
At the alternative hypothesis, it is tested if it is greater, that is:
[tex]H_1: p > 0.132[/tex]
The test statistic is given by:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
In which:
- [tex]\overline{p}[/tex] is the sample proportion.
- p is the proportion tested at the null hypothesis.
- n is the sample size.
In this problem, the parameters are:
[tex]p = 0.132, n = 50, \overline{p} = \frac{11}{50} = 0.22[/tex]
Hence, the value of the test statistic is given by:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.22 - 0.132}{\sqrt{\frac{0.5(0.5)}{50}}}[/tex]
[tex]z = 1.24[/tex]
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 [tex]z^{\ast} = 1.645[/tex].
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 https://brainly.com/question/16313918
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