12. A food manufacturer claims that at the time of purchase by a consumer the average age of its product is no more than 120 days. In an experiment to test this claim a random sample of 36 items are found to have ages at the time of purchase with a sample mean of x= 122.5 days and a sample standard deviation of s = 13.4 days. (a) Write down the null hypothesis and the alternative hypothesis that the experimenter should use for the test. (b) For what values of the t-statistic does the experimenter accept the null hypothesis with a level of significance a = 0.10? (c) Is the null hypothesis accepted or rejected with a = 0.10? (d) Write down an expression for the p-value and estimate it.

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**Hypothesis Testing Example**

12. A food manufacturer claims that at the time of purchase by a consumer, the average age of its product is no more than 120 days. In an experiment to test this claim, a random sample of 36 items is found to have ages at the time of purchase with a sample mean of \( \bar{x} = 122.5 \) days and a sample standard deviation of \( s = 13.4 \) days.

(a) **Null Hypothesis (\( H_0 \)) and Alternative Hypothesis (\( H_a \)):**
   - Null Hypothesis (\( H_0 \)): The average age of the product at the time of purchase is 120 days or less. (\( \mu \leq 120 \))
   - Alternative Hypothesis (\( H_a \)): The average age of the product at the time of purchase is greater than 120 days. (\( \mu > 120 \))

(b) **Values of the t-statistic at a significance level (\( \alpha = 0.10 \)):**
   - To accept the null hypothesis, we calculate the t-statistic and compare it to the critical t-value from the t-distribution with \( n-1 = 35 \) degrees of freedom.

(c) **Hypothesis Test Decision at \( \alpha = 0.10 \):**
   - Using the calculated t-statistic and comparing it against the critical value, decide whether to accept or reject the null hypothesis at the 10% significance level.

(d) **Expression for the p-value and Estimate:**
   - Calculate the p-value using the t-statistic. The p-value indicates the probability of observing a result as extreme as the sample mean under the null hypothesis.
   - Estimate the p-value using statistical software or a t-distribution table.

**Note:** Detailed calculations and the critical value lookup would typically be performed using statistical resources or software.
Transcribed Image Text:**Hypothesis Testing Example** 12. A food manufacturer claims that at the time of purchase by a consumer, the average age of its product is no more than 120 days. In an experiment to test this claim, a random sample of 36 items is found to have ages at the time of purchase with a sample mean of \( \bar{x} = 122.5 \) days and a sample standard deviation of \( s = 13.4 \) days. (a) **Null Hypothesis (\( H_0 \)) and Alternative Hypothesis (\( H_a \)):** - Null Hypothesis (\( H_0 \)): The average age of the product at the time of purchase is 120 days or less. (\( \mu \leq 120 \)) - Alternative Hypothesis (\( H_a \)): The average age of the product at the time of purchase is greater than 120 days. (\( \mu > 120 \)) (b) **Values of the t-statistic at a significance level (\( \alpha = 0.10 \)):** - To accept the null hypothesis, we calculate the t-statistic and compare it to the critical t-value from the t-distribution with \( n-1 = 35 \) degrees of freedom. (c) **Hypothesis Test Decision at \( \alpha = 0.10 \):** - Using the calculated t-statistic and comparing it against the critical value, decide whether to accept or reject the null hypothesis at the 10% significance level. (d) **Expression for the p-value and Estimate:** - Calculate the p-value using the t-statistic. The p-value indicates the probability of observing a result as extreme as the sample mean under the null hypothesis. - Estimate the p-value using statistical software or a t-distribution table. **Note:** Detailed calculations and the critical value lookup would typically be performed using statistical resources or software.
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