A consumer group claims that the average annual consumption of high fructose corn syrup by a person in the U.S. is 48.8 pounds. You believe it is higher. You take a simple random sample of 31 people in the U.S. and find an average of 53.4 pounds with a standard deviation of 5.2 pounds. Test at $10 \%$ significance. $H_{0}:$ Select an answerv Select an answerv $H_{A}:$ Select an answerv Select an answerv Test Statistic: P-value: Did something significant happen? Select an answer Select the Decision Rule: Select an answer There Select an answerv enough evidence to conclude Select an answer Submit Question

Step 1 ：State the hypotheses. The null hypothesis is that the average annual consumption of high fructose corn syrup by a person in the U.S. is 48.8 pounds. The alternative hypothesis is that the average annual consumption is higher than 48.8 pounds. So, $H_{0}: \mu = 48.8$ and $H_{A}: \mu > 48.8$

Step 2 ：Calculate the test statistic. The test statistic is calculated as follows: $t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $s$ is the standard deviation, and $n$ is the sample size. Substituting the given values, we get $t \approx 4.93$

Step 3 ：Find the p-value. The p-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. In this case, the p-value is approximately 0.0000144

Step 4 ：Make a decision. Since the p-value is less than the significance level of 0.10, we reject the null hypothesis. There is enough evidence 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 5 ：Final Answer: \(\boxed{Reject \: H_{0}}\)