Can the manager conclude from the data that the belief is correct? The sample mean is 460.38 and the sample standard deviation is 38.89.

Steps to Solve

1. State the Hypotheses We need to formulate our null and alternative hypotheses.

• Null Hypothesis ($H_0$): The average number of packages processed by trainees is less than or equal to 450 packages per hour.
• Alternative Hypothesis ($H_a$): The average number of packages processed by trainees is greater than 450 packages per hour.

2. Determine the Sample Statistics From the provided data, we have:

• Sample mean ($\bar{x}$) = 460.38
• Sample standard deviation ($s$) = 38.89
• Sample size ($n$) = 50

3. Calculate the Test Statistic We will use the t-test formula to calculate the test statistic since the sample size is less than 30: $$ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} $$ Where:

• $\mu_0$ = 450 (the value under the null hypothesis)

Plugging in the values: $$ t = \frac{460.38 - 450}{38.89 / \sqrt{50}} $$

4. Calculate the Critical Value Using a t-distribution table, find the critical t-value for one-tailed test at a significance level ($\alpha$) of 0.05 with $n - 1 = 49$ degrees of freedom.

5. Decision Rule If the calculated test statistic is greater than the critical value from the t-table, we reject the null hypothesis.

6. Compare and Conclude Compare the calculated t-value with the critical t-value to make a conclusion about the average number of packages processed.