Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Cans of Coke Data Set 26 “Cola Weights and Volumes” in Appendix B includes volumes (ounces) of a sample of cans of regular Coke. The summary statistics are n = 36, x = 12.19 oz, s = 0.11 oz. Use a 0.05 significance level to test the claim that cans of Coke have a mean volume of 12.00 ounces. Does it appear that consumers are being cheated?

The summary statistics for the volumes of sample cans of regular Coke are as follows.

Sample size\(n = 36\).

Sample mean\(\bar x = 12.19\;oz\).

Sample standard deviation\(s = 0.11\;oz\).

Level of significance\(\alpha = 0.05\).

The claim states that the Coke volume has a mean volume of 12.00 ounces.

Null hypothesis\({H_0}\): The cans of Coke have a mean volume of 12 ounces.

Alternative hypothesis\({H_1}\): The cans of Coke do not have a mean volume of 12 ounces.

The hypotheses are formulated as follows.

\(\begin{array}{l}{H_0}:\mu = 12\;{\rm{oz}}\\{H_1}:\mu \ne 12\;{\rm{oz}}\end{array}\).

The simple random sample is collected from a normally distributed population. For an unknown population standard deviation, use student t-distribution.

The test statistic is given as follows.

\(\begin{array}{c}t = \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\\ = \frac{{12.19 - 12}}{{\frac{{0.11}}{{\sqrt {36} }}}}\\ = 10.3636\end{array}\).

The test statistic is 10.3636.

The level of significance is\(\alpha = 0.05\).

The degree of freedom is computed as follows.

\(\begin{array}{c}df = n - 1\\ = 36 - 1\\ = 35\end{array}\).

Refer to the t-table for the critical value corresponding to 35 degrees of freedom and a level of significance of 0.05 for the two-tailed test, which is\({t_{0.05}} = 2.030\).

Thus, the critical values are \( \pm 2.030\), and the non-rejection area is \(\left( {t: - 2.030 < t < 2.030} \right)\).

The decision rule states the following:

If the test statistic is greater than the critical value, the null hypothesis will be rejected.

If the test statistic is not greater than the critical value, the null hypothesis will fail to be rejected.

Here, the test statistic is 10.364, which is greater than 2.030. Thus, the null hypothesis is rejected at a 0.05 level of significance.

As the null hypothesis is rejected, it can be concluded that there is insufficient evidence to support the claim that the mean volume of the Coke cans is 12.00 ounces.

As the sample mean is 12.19 ounces, which is greater than the claimed volume of 12.00 ounces, the consumers will get slightly more than the claimed amount. Thus, the consumers will not be cheated.