I need to solve this case and I do not underestand: The Green Valley Assembly Co
ID: 3224314 • Letter: I
Question
I need to solve this case and I do not underestand:
The Green Valley Assembly Company assembles consumer electronics products for manufacturers that need temporary extra production capacity As such, it has periodic product Changes. Because the products Green Valley assembles are marketed under the label of well-known manufacturers, high quality is a must. Tom Bradley, of the Green Valley personnel department, has been very impressed by recent research Concerning job-enrichment programs. In particular, he has been impressed with the increases in quality that seem to be associated with these programs. However, some studies have shown no significant increase in quality, and they imply that the money spent on such programs has not been worthwhile Tom has talked to Sandra Hansen, the production manager, about instituting a job-enrichment program in the assembly operation at Green Valley. Sandra was somewhat pessimistic about the potential, but she agreed to introduce the program. The plan was to implement the program in one wing of the plant and Continue with the current method in the other wing. The procedure was to be in effect for six months. After that period, a test would be made to determine the effectiveness of the job-enrichment program After the six-month trial period, a random sample of employees from each wing produced the following output measures: Both Sandra and Tom Wonder whether the job-enrichment program has affected production output. They would like to use these sample results to determine whether the average output has changed and to determine whether the employees' Consistency has been affected by the new program. A second sample from each wing was selected. The measure was the quality of the products assembled. In the "old" wing, 79 products were tested and 12% were found to be defectively assembled. In the "job-enriched" wing, 123 products were examined and 9% were judged defectively assembled. With all these data, Sandra and Tom are beginning to get a little Confused. However, they realize that there must be some way to use the information to make a judgment about the effectiveness of the job-enrichment program.Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1> 2
Alternative hypothesis: 1 < 2
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 0.212
DF = 98
t = [ (x1 - x2) - d ] / SE
t = 6.13
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
The observed difference in sample means produced a t statistic of 6.13. We use the t Distribution Calculator to find P(t > 6.13) = less than 0.00001
Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.
From the above test we have sufficient evidence that production capacity has increased.
Test for defective assemblies.
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P1< P2
Alternative hypothesis: P1 > P2
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.
Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).
p = (p1 * n1 + p2 * n2) / (n1 + n2)
p = 0.1017
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.06045
z = (p1 - p2) / SE
z = 0.4963
where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.
Since we have a one-tailed test, the P-value is the probability that the z-score is more than 0.4963. We use the Normal Distribution Calculator to find P(z > 0.4963) = 0.3085
Interpret results. Since the P-value (0.3085) is greater than the significance level (0.01), we cannot reject the null hypothesis.