Question 2 • Probability and Statistics with Applications Honors Sem-2-FL-1210300 (2024-2025)
A manufacturing company packages shipments in either large or small boxes. A random sample of 40 shipments that are packaged in large boxes is found to have a mean of 15 pounds and a standard deviation of 2.8 pounds. A separate random sample of 50 shipments that are packaged in small boxes is found to have a mean of 10 pounds and a standard deviation of 1.5 pounds. The manager would like to know if the data provide convincing evidence that the true mean weight of all shipments that are packaged in small boxes is less than the true mean weight of all shipments that are packaged in large boxes. The manager tests H0: μS – μL = 0, Ha: μS – μL < 0, where μL = the true mean weight of all shipments that are packaged in large boxes and μS = the true mean weight of all shipments that are packaged in small boxes. The conditions for inference have been met. What are the values of the test statistic and P-value for a t-test about a difference in means?Find the t-table here and the z-table here.
Answer
A
t = –10.19. The P-value is less than 0.0005.
B
z = –10.19. The P-value is less than 0.0002.
C
t = 10.19. The P-value is less than 0.0005.
D
z = 10.19. The P-value is less than 0.0002.
Explanation
The sample difference, small-box minus large-box, is 10 − 15 = −5 pounds. Its standard error is √(1.5²/50 + 2.8²/40) ≈ 0.491, so the test statistic is t = −5/0.491 ≈ −10.19. This is an extremely negative t-statistic for the alternative that the small-box mean is less, giving a one-sided P-value less than 0.0005; the answer is t = −10.19 with P-value less than 0.0005.
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