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The accompanying data set includes volumes (ounces) of a sample of cans of regular Coke. The summary statistics are $n=36,\overline{x}=12.201\mathrm{oz},\mathrm{s}=0.106\mathrm{oz}$. Assume that a simple random sample has been selected. Use a 0.01 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.
$H_0:\mu =12.00$
$H_1:\mu \ne 12.00$
(Type integers or decimals. Do not rovind.)
Identify the test statistic.
(Round to two decimal places as needed.)
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Step 1 ：Identify the null and alternative hypotheses. The null hypothesis $H_0$ is that the mean volume of cans of Coke is 12.00 ounces, i.e., $H_0:\mu =12.00$. The alternative hypothesis $H_1$ is that the mean volume of cans of Coke is not 12.00 ounces, i.e., $H_1:\mu \ne 12.00$.

Step 2 ：Identify the test statistic. The test statistic can be calculated using the formula for a one-sample t-test, which is $(\text{sample mean}-\text{population mean})/(\text{sample standard deviation}/\sqrt{\text{sample size}})$. In this case, the sample mean is 12.201 ounces, the population mean is 12.00 ounces, the sample standard deviation is 0.106 ounces, and the sample size is 36.

Step 3 ：Calculate the test statistic. Using the values from the previous step, the test statistic is $(12.201-12.00)/(0.106/\sqrt{36})$.

Step 4 ：The test statistic is approximately 11.38. This value is quite high, which indicates that the sample mean is significantly different from the population mean. This suggests that the null hypothesis may be rejected, but we would need to compare this test statistic to a critical value or use it to calculate a p-value to make a final decision.

Step 5 ：Final Answer: The test statistic is approximately $\boxed{{\displaystyle 11.38}}$.