The accompanying data set includes volumes (ounces) of a sample of cans of regular Coke. The summary statistics are $n=36, \bar{x}=12.185$ oz, $s=0.129$ oz. Assume that a simple random sample has been selected. Use a 0.10 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?
Click the icon to view the data set of regular Coke can volumes.
Identify the null and alternative hypotheses.
(Type integers or decimals. Do not round.)
Identify the test statistic.
(Round to two decimal places as needed.)
Identify the P-value.

Step 1 ：First, we set up the null and alternative hypotheses. The null hypothesis is that the mean volume of Coke cans is 12.00 ounces, and the alternative hypothesis is that the mean volume is not 12.00 ounces. So, we have \(H_0: \mu = 12.00\) and \(H_a: \mu \neq 12.00\).

Step 2 ：Next, we calculate the test statistic using the formula \(t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu_0\) is the hypothesized population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. Substituting the given values, we get \(t = \frac{12.185 - 12.00}{0.129 / \sqrt{36}}\), which simplifies to \(t = 8.60\).

Step 3 ：Finally, we calculate the P-value, which is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. The P-value is approximately 0.00000000037.

Step 4 ：Since the P-value is less than the significance level of 0.10, we reject the null hypothesis. This suggests that the mean volume of Coke cans is not 12.00 ounces.

Step 5 ：The final answer is: The test statistic is approximately \(\boxed{8.60}\) and the P-value is approximately \(\boxed{0.00000000037}\).