2. Turnover rates in the US and Japan. High job turnover rates are often associated with high product defect rates. In a recent study 5 Japanese and 5 US plants that manufacture air conditioners were randomly sampled; their turnover rates are listed in the table (10 pts)
US Plants (%) |
Japanese Plants (%) |
t-Test: Two-Sample Assuming Equal Variances |
US Plants |
Japanese Plants |
|
7.11 |
3.52 |
||||
6.06 |
2.02 |
Mean |
6.562 |
3.118 |
|
8 |
4.91 |
Variance |
1.482 |
1.506 |
|
6.87 |
3.22 |
Observations |
5 |
5 |
|
4.77 |
1.92 |
Pooled Variance(sp2) |
1.494 |
? | |
Hypothesized Mean Difference |
0 |
||||
Df |
? | ? | |||
t Stat |
? | ? | |||
P(T<=t) one-tail |
0.001 |
? | |||
t Critical one-tail |
? | ? | |||
P(T<=t) two-tail |
0.002 |
? | |||
t Critical two-tail |
? | ? |
A. Use the Excel output above to describe the two samples involved in the comparison:
Mean turnover rate in the US = 1 =
Japan = 2=
Pooled Variance (sp2)=
Combined degrees of freedom (df): (show calculation)
B. Set up the null and alternate hypothesis in terms of the Mean difference (m1-m2)to test the hypothesis that turnover rates in the US plants exceedsthe turnover rates in Japanese plants.
C. Refer to the Excel output and calculate the test-statistic value (using the pooled variance provided)
D. The correct criticalt-valueata= 0.05 based on your hypothesis above is: (show the critical value(s) on a graph)
Get Answers For Free
Most questions answered within 1 hours.