Green Valley Assembly Company
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.
Table 1: Mean and standard deviation of output per hour
| Group |
| Old | Job-enriched |
Mean | 11 | 9.7 |
Standard deviation | 1.2 | 0.9 |
sample size | 50 | 50 |
The above table 1 shows the mean and standard deviation of output per hour for old and job-enriched employees.
In the first part of the analysis, whether there was any change in the mean output per hour between the two groups is evaluated with the help of z test for two samples. Let μ1 be population mean output of old group and μ2 be population mean output of job-enriched group.
The null hypothesis tested is Ho: μ1=μ2
Against alternative hypothesis H1: μ1≠μ2
Let level of significance α=0.05
The decision rule is Reject Ho if calculated value of z statistic is greater than 1.96 or z value is less than -1.96 at 5% level of significance
The test statistic is
z= (mean1-mean2)/√(s12/n1+s22/n2)
=(11-9.7)/√((1.2)2/50+(0.9)2/50)
z= 6.128259
p value =0.0000
Since the calculated value is greater than 1.96, Hence Ho is rejected.
Conclusion: The evidence shows that there is a significant change in the mean output per hour between the two groups.
Regarding the second part of the analysis, whether the employees consistency is affected by the new programme, this part is evaluated using data of proportion of defectively assembled products by the two groups. This part is answered with the help of z test for proportion of two samples.
Table 2: proportion of defectives assembled by the two groups
| Group |
| Old | Job-enriched |
Proportion (p) | 0.12 | 0.09 |
q =1-p | 0.88 | 0.91 |
sample...
