chapter 9

Topics: Standard deviation, Statistics, Arithmetic mean Pages: 5 (638 words) Published: February 26, 2014
 Peter j. Dalley, The case study has given to us is regarding Quality Associates, Inc. It’s a consulting firm. Quality Associates, Inc. advises its clients about sampling and statistical procedures. In this particular case, Quality Associates, Inc. has provided samples to be analyzed, so it’s client can quickly learn whether the process is operating satisfactorily or corrective actions needs to be taken.

Summary of Statistics

 Sample 1
 Sample 2
 Sample 3
 Sample 4
Mean
11.96
12.10
12.14
12.15
Standard Error
0.04
0.04
0.04
0.02
Standard Deviation
0.22
0.23
0.23
0.16
Sample Variance
0.05
0.05
0.05
0.03
Sum
358.69
362.90
364.08
364.46

From the summary of statistics we can see that mean has an upward trend. Mean value differ from sample to sample. Here we can observe that mean value has a relation with the amount of sum. We can also notice that amount of sum is differing from sample to sample and follow the same trend as mean.

Introduction of the test:
By analyzing the new samples, the client would be able to know that, whether the process was operating satisfactorily or not. If the process was not operating satisfactorily, then corrective action would be taken to eliminate the problem. In this case the mean for the process is 12. The hypothesis test suggested by Quality Associates follows:

H0: μ=12
Ha: μ≠ 12

Corrective action will be taken if H0 is rejected. In this case α = .01 that means probability of making a Type I error when μ = 12 is .01.

Sample 1
Sample 2
Sample 3
Sample 4
Z statistics
-1.14
2.52
3.55
3.88
P value
0.26
0.01
0.00
0.00

Accept
Reject
Reject
Reject

Here we can see in Sample 1, P-value is higher than α. So sample 1 is operating satisfactorily, so in case of sample 1 we are going to accept null and no corrective action is necessary. On the other hand, Sample 2, 3 and 4 has equal or lower p-value than α. So we can assume that process was not going accordingly. So, corrective action is needed to take.

Use of Standard deviation:
Standard deviation shows the number of variations that can exist from an average. There are standard deviation of population and standard deviation of sample. Usually we try to find out the standard deviation of population. In this case, both sample and population standard deviations are 0.21. Now with the four new samples we can find out was that reasonable?

Sample
Standard Deviation
1
0.22
2
0.23
3
0.23
4
0.16
Average
0.21

From the table we can see that average standard deviation of the four new samples is 0.21. This is same as the standard deviation of the population. So we can come to a conclusion that assuming population standard deviations .21 was reasonable.

Control limit:
In this case the upper limit is 12.10 and the lower limit is. By considering the level of significance α =01, we can assume that we are 99% confidence that the mean distance for the population is between 11.90 and 12.10. If sample means stay within this limit manufacturing process is operating satisfactory.

Control limits Upper limit Lower limit...
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