Testing Claims About Variation. In Exercises 5–16, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Assume that a simple random sample is selected from a normally distributed population.

Mint Specs Listed below are weights (grams) from a simple random sample of “wheat” pennies (from Data Set 29 “Coin Weights” in Appendix B). U.S. Mint specifications now require a standard deviation of 0.0230 g for weights of pennies. Use a 0.01 significance level to test the claim that wheat pennies are manufactured so that their weights have a standard deviation equal to 0.0230 g. Does the Mint specification appear to be met?

2.5024 2.5298 2.4998 2.4823 2.5163 2.5222 2.4900 2.4907 2.5017

The weights of the sampled pennies are recorded.

The standard deviation of the pennies is required to be 0.0230 g.

The level of significance is 0.01.

To test the claim that the standard deviation of the pennies’ weight is equal to the requirement.

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For applying the hypothesis test, first set up a null and an alternative hypothesis.

The null hypothesis is the statement about the value of a population parameter, which is equal to the claimed value. It is denoted by\({H_0}\).

The alternate hypothesis is a statement that the parameter has a value that is opposite to the null hypothesis. It is denoted by\({H_1}\).

Let\(\sigma \)be the actual standard deviation of the wheat pennies.

From the claim, the null and alternative hypotheses are as follows.

\(\begin{array}{l}{H_0}:\sigma = 0.0230\,\\{H_1}:\sigma \ne 0.0230\end{array}\)

LetX be the simple random sample ofweights of wheat pennies.

2.5024 2.5298 2.4998 2.4823 2.5163 2.5222 2.4900 2.4907 2.5017

The sample mean of weights is as follows.

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{{\rm{n}}}\\ = \frac{{2.5024 + 2.5298 + ... + 2.5017}}{9}\\ = 2.5039\end{array}\)

The sample standard deviation is calculated as follows.

\(\begin{array}{c}s = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{{\rm{n}} - 1}}} \\ = \sqrt {\frac{{{{\left( {2.5024 - 2.5039} \right)}^2} + {{\left( {2.5298 - 2.5039} \right)}^2} + ... + {{\left( {2.5017 - 2.5039} \right)}^2}}}{{9 - 1}}} \\ = 0.0159\end{array}\)

Thus, the sample standard deviation \(s\) is 0.0159.

To conduct a hypothesis test of a claim about a population standard deviation\(\sigma \) or population variance\({\sigma ^2}\),the test statistics is computed as follows.

\(\begin{array}{c}{\chi ^2} = \frac{{\left( {{\rm{n}} - 1} \right) \times {s^2}}}{{{\sigma ^2}}}\\ = \frac{{\left( {9 - 1} \right) \times {{0.0159}^2}}}{{{{0.0230}^2}}}\\ = 3.8135\end{array}\).

Thus, the value of the test statistic is 3.813.

The degree of freedom is as follows.

\(\begin{array}{c}df = n - 1\\ = 9 - 1\\ = 8\end{array}\)

For a two-sided hypothesis, divide the\(\alpha \)equally between the two tails as\(\frac{\alpha }{2}\)and\(1 - \frac{\alpha }{2}\).

Mathematically,

\(\begin{array}{l}P\left( {{\chi ^2} < \chi _{\frac{\alpha }{2}}^2} \right) = 1 - \frac{\alpha }{2}\\P\left( {{\chi ^2} > \chi _{\frac{{0.01}}{2}}^2} \right) = 1 - \frac{{0.01}}{2}\\P\left( {{\chi ^2} > \chi _{\frac{{0.01}}{2}}^2} \right) = 0.995\end{array}\)

Also,

\(\begin{array}{l}P\left( {{\chi ^2} > \chi _{\frac{\alpha }{2}}^2} \right) = \frac{\alpha }{2}\\P\left( {{\chi ^2} > \chi _{\frac{{0.01}}{2}}^2} \right) = \frac{{0.01}}{2}\\P\left( {{\chi ^2} > \chi _{\frac{{0.01}}{2}}^2} \right) = 0.005\end{array}\)

Referring to the chi-square table, the critical value with 8 degrees of freedom corresponding to the right-tailed area of 0.005 is observed as 21.955.

Referring to the chi-square table, the critical value with 8 degrees of freedom corresponding to the right-tailed area of 0.995 is observed as 1.344.

The decision rule for the test is as follows.

If the test statistic lies between the critical values, reject the null hypothesis at the given level of significance. Otherwise, fail to reject the null hypothesis.

As it is observed that \({\chi ^2} = 3.813\,\) lies between the critical values1.344 and 21.955,the null hypothesis is failed to berejected.

Thus, there is enough evidence to supportthe claim that the wheat pennies are manufactured so that their weights have a standard deviation\(\sigma \) equal to 0.0230g, and the requirement for the mint specification appears to have been met.