Final answer:
Using a Z-test, the hypothesis test for a sample mean of 12.03 oz and standard deviation of 0.14 oz with 36 cans did not provide enough evidence to reject the null hypothesis. Therefore, it cannot be claimed that the true mean content of the soda cans differs from the labeled 12.00 oz at a 0.05 significance level.
Step-by-step explanation:
To test the claim that the contents of soda cans have a mean different from 12.00 oz, we need to conduct a hypothesis test using the sample data provided. We are given that the sample mean (μ) is 12.03 oz, the standard deviation (s) is 0.14 oz, and the sample size (n) is 36 cans. Since n is greater than 30, we can use the Z-test for this hypothesis test.
The null hypothesis (H0) is that the true mean is equal to the claimed amount, which is 12.00 oz (H0: μ = 12.00). The alternative hypothesis (H1) is that the true mean is not equal to the claimed amount (H1: μ ≠ 12.00), indicating a two-tailed test.
First, we calculate the standard error (SE) of the mean using the formula SE = s / √n. For our data, SE = 0.14 / √36 = 0.14 / 6 = 0.0233 oz. Then, we find the Z-score for our sample mean. The Z-score is calculated as Z = (Sample Mean - Population Mean) / SE. This gives us Z = (12.03 - 12.00) / 0.0233 = 1.29.
Now we compare the calculated Z-score with the critical Z-score for a significance level of 0.05 in a two-tailed test. The critical Z-score for a two-tailed test at the 0.05 level is roughly ±1.96. Since ±1.29 does not exceed ±1.96, we fail to reject the null hypothesis. Therefore, we do not have sufficient evidence to say that the mean amount of soda in the cans is different from 12.00 oz.