Answer:
(1) The critical value is 2.797
(2) The pooled standard deviation is 7
(3) There is no difference in the mean productivity of employees in Detroit and Bochum
Step-by-step explanation:
(1) Total number of observation = n1 + n2 = 13+13 = 26, degree of freedom = total number of observation - 2 = 26 - 2 = 24
Using the t-distribution table, the critical value corresponding to 24 degrees of freedom and 0.01 significance level is 2.797
(2) Pooled variance = [variance of Detroit (number of observation in Detroit - 1) + variance of Bochum (number of observation in Bochum - 1)] ÷ (number of observation in Detroit + number of observation in Bochum - 2) = 42(13 - 1) + 56(13 - 1) ÷ (13 + 13 - 2) = (42×12) + (56×12) ÷ (26 - 2) = (504+672) ÷ 24 = 1176 ÷ 24 = 49
Pooled standard deviation = √(pooled variance) = √49 = 7
(3) Null hypothesis: There is no difference in the mean productivity of employees in Detroit and Bochum
Alternate hypothesis: There is a difference in the mean productivity of employees in Detroit and Bochum
Decision rule:
Reject the null hypothesis if p-value is less than or equal to the significance level
Fail to reject the null hypothesis if p-value is greater than the significance level
The p-value (0.08) is greater than the significance level (0.01), fail to reject the null hypothesis
Conclusion: There is no difference in the mean productivity of employees in Detroit and Bochum
