Question

An operations manager with General Motors corporate offices wishes to determine if the true mean productivity of employees at its Detroit, Michigan plant is different from its Opel/Vauxhall Plant in Bochum, Germany. For this analysis, productivity will be measured by the number of cars that employees assemble per day. Random samples of employees in Detroit and Bochum are selected, and the productivity scores at each assembly plant are recorded. The following results are obtained.

t-Test: Two-Sample Assuming Equal Variances

Detroit

Bochum

Mean

58.5

54.3

Variance

42

56

Observations

13

13

Pooled Variance

XX

Hypothesized Mean Difference

0

df

XX

t Stat

XXXXX

P(T<=t) one-tail

XXXXX

t Critical one-tail

XXXXX

P(T<=t) two-tail

XXXXX

t Critical two-tail

XXXXX

1. Using a significance level of 0.01, the critical value of the appropriate test statistic to conduct the test of hypothesis for the belief that there is a difference in the mean productivity is:

2. What is the pooled standard deviation?

3. Assume that the p-value associated with the test statistic is 0.08. What is the conclusion of the hypothesis test? Use the significance level from question above.

Answer 1

(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

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