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.

Cola Cans A random sample of 20 aluminum cola cans with thickness 0.0109 in. is selected and the axial loads are measured and the standard deviation is 18.6 lb. The axial load is the pressure applied to the top that causes the can to crush. Use a 0.05 significance level to test the claim that cans with thickness 0.0109 in. have axial loads with the same standard deviation as the axial loads of cans that are 0.0111 in. thick. The thicker cans have axial loads with a standard deviation of 27.8 lb (based on Data Set 30 “Aluminum Cans” in Appendix B). Does the thickness of the cans appear to affect the variation of the axial loads?

A random sample of 20 aluminum cola cans with athickness of 0.0109 in measured for the axial loads has astandard deviation of 18.6 lb.

The significance level is 0.05.

The claim to be tested is that the axial load standard deviation for cans with a thickness of 0.0109 in is the same as the cans with a thickness of 0.0111 in.

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 opposite to the null hypothesis. It is denoted by\({H_1}\).

The claim is that cans with a thickness of 0.0109 in have axial loads with the same standard deviation\(\left( \sigma \right)\)of 27.8 lb.

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

\(\begin{array}{l}{{\rm{H}}_0}:\sigma = 27.8\,\,\,\,\,\,\,\\{{\rm{H}}_1}:\sigma \ne 27.8\end{array}\)

Here, \(\sigma \) is the measure of standard deviation for axial loads with a thickness of 0.0109 in.

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

\(\begin{array}{c}{\chi ^2} = \frac{{\left( {{\rm{n}} - 1} \right) \times {s^2}}}{{{\sigma ^2}}}\\ = \frac{{\left( {20 - 1} \right) \times {{18.6}^2}}}{{{{27.8}^2}}}\\ = 8.5053\end{array}\).

Thus, the value of the test statistic is 8.5053.

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

Mathematically,

\(P\left( {{\chi ^2} < \chi _{\frac{\alpha }{2}}^2} \right) = 1 - \frac{\alpha }{2}\)and

\(P\left( {{\chi ^2} > \chi _{\frac{\alpha }{2}}^2} \right) = \frac{\alpha }{2}\).

Referring to chi-square table, the critical value of\({\chi ^2}\)is obtained corresponding to row 19 and significance level 0.025 as 32.852.

Referring to Table A-4, chi-square table, the critical value of\({\chi ^2}\)is obtained corresponding to row 19 and significance level 0.975 as as 8.907 from the area to the right of 0.975

Thus, the critical values are 8.907 and 32.852, and the non-rejection region is \(\left( {{\chi ^2}:8.907 < {\chi ^2} < 32.852} \right)\).

The decision rule for the test is as follows.

If the value of the test statistic lies beyond the non-region, reject the null hypothesis at the given level of significance; otherwise, fail to reject the null hypothesis.

Thus, it is observed that\({\chi ^2} = 8.505\)lies beyond the non-rejection region.

Therefore, it is concluded that the null hypothesis is rejected, and it can be stated that there is not sufficient evidence to support the claim that cans with a thickness of 0.0109 in have the same load as the ones with 0.0111 in thickness.

Hence, it can be concluded that the thickness of the cans appears to affect the variation of the axial loads.