Step 1 :This is a one-sample t-test problem. We are testing the null hypothesis that the average annual consumption of high fructose corn syrup by a person in the U.S. is 48.8 pounds against the alternative hypothesis that it is higher. The sample size is 31, the sample mean is 52.9 pounds, and the sample standard deviation is 4.2 pounds. The significance level is 1%.
Step 2 :We can calculate the test statistic using the formula: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \] where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size.
Step 3 :Substituting the given values into the formula, we get: \[ t = \frac{52.9 - 48.8}{4.2 / \sqrt{31}} \]
Step 4 :The calculated test statistic is approximately 5.435.
Step 5 :We can find the p-value by looking up the test statistic in the t-distribution table with 30 degrees of freedom. The calculated p-value is approximately 0.00000341.
Step 6 :Since the p-value is less than the significance level of 0.01, we reject the null hypothesis.
Step 7 :There is enough evidence at the 1% significance level to conclude that the average annual consumption of high fructose corn syrup by a person in the U.S. is higher than 48.8 pounds.
Step 8 :The final answer is: The test statistic is approximately \(\boxed{5.435}\) and the p-value is approximately \(\boxed{0.00000341}\). We reject the null hypothesis. There is enough evidence at the 1% significance level to conclude that the average annual consumption of high fructose corn syrup by a person in the U.S. is higher than 48.8 pounds.