Green Valley Assembly Company

Topics: Statistics, Statistical hypothesis testing, Sample size Pages: 2 (287 words) Published: October 11, 2012
Analysis of Green Valley Assembly Company

Case 10.3
The Green valley company has introduced a job enrichment program, where one group of employee will be receiving job enriched program in one wing and another group of employee will be continued with the current method in other wing. At the end six month of program, the company evaluated whether there was any change in mean output of the two groups | Old | Job-enriched |

Mean | 11 | 9.7 |
Standard deviation | 1.2 | 0.9 |
sample size | 50 | 50 |
State the Null and Alternative Hypothesis
Was there any change in the mean output per hour between the two groups?

Ho: μ1=μ2
HA: μ1≠μ2

Level of significance α=0.05

Decision Rule and Test Statistic
Reject Ho if z< -1.96 or z> 1.96 at significance level of α=0.05

Test Statistic
z= (mean1-mean2)/√(s12/n1+s22/n2)
=(11-9.7)/√((1.2)2/50+(0.9)2/50)
z= 6.1283

Part 2 of the Analysis
Was employee consistency affected by the new job-enrichment program?

| Old| Job-enriched|
Proportion (p)| 0.12| 0.09|
q =1-p | 0.88| 0.91|
Sample Size| 50| 50|

Standard Error
√p1q1/n1 + p2q2/n2| 0.0527|

Let π1 be population proportion of defective assembled products by the old group and π2 be population proportion of defectives assembled products by the job-enriched group. Part 2 continued…
Ho: π 1= π 2
H1: π1 ≠ π 2
α=0.05

Reject Ho if z< -1.96 or z> 1.96 at significance level of α=0.05

Test Statistic
z= (p1-p2/)√p1q1/n1 + p2q2/n2
=(0.12-0.09)/0.0527
=0.5692

P-Value = 0.28
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