Case Synopsis
In this case, we have Quality Associates, Inc. a consulting firm advising its client about sampling and statistical procedures that can be used to control their manufacturing process. Their client has offered samples to be analyzed, so they can quickly learn whether the process is operating satisfactorily or corrective actions needs to be taken. The numbers given in the case were as follows: assumed population standard deviation is equal to .21, sample size is equal to 30 and the test value of the mean was 12. They also stated the two hypotheses to be tested: the null hypothesis that the population is equal to 12 and the alternative hypothesis that the mean is not equal to 12. This indicates a two tailed tests to determine whether or not to reject the null hypothesis. The 4 provided sample sizes each contained 30 observations, indicating a normal distribution and z test statistics.

Methodology
The first question required conducting a hypothesis test for each sample at the .01 level of significance. Based upon the test, determine if any corrective actions need to be taken. There are two approaches to hypothesis testing, the p-value approach and the critical value approach. The first step for the p-value approach was to calculate the mean for each sample. In order, they were: 11.9587| 12.0287| 11.8890| 12.0813|

Next, was to calculate standard error, by using the formula sigma divided by the square root of n. This came out to be .0383. To find the z test statistic subtract the test value of 12 from the sample mean and divide by the standard error.

The z test statistic for each sample were as follows:
-1.0966| 0.7493| -2.8982| 2.1227|

The 1 tail p-value could then be found by using the normsdist function in Excel. This function indicates probability to the left of the value, so positive numbers were subtracted from 1. Since this is a two tailed test, the values were...

...QualityAssociatesInc. |
To: | Peter Dalley |
From: | Harp Bhander |
Date: | 5/17/2013 |
Re: | Analysis Results |
QualityAssociates, a consulting firm has been approached by a client to analyze their manufacturing process. The client came to QualityAssociates with a sample of 800 observations that they had collected during times when the manufacturing process was running smoothly. The sample of 800 yielded a sample standard deviation of 0.21. QualityAssociates suggested that the company periodically take a sample of 30 to monitor their process on an ongoing basis to quickly learn if process was operating at a satisfactory level, and if not it could quickly be adjusted. The specifications of the design calls for the mean of the process to be 12.
Sample size=30 | Sample 1 | Sample 2 | Sample 3 | Sample 4 |
Mean | 12.54 | 12.04 | 11.04 | 12.05 |
Min | 12.01 | 11.62 | 10.58 | 11.64 |
Max | 12.82 | 12.48 | 11.46 | 12.54 |
SD | 0.19 | 0.18 | 0.20 | 0.25 |
Table 1
The Hypothesis test that will be conducted is as follows:
Ho: µ = 12
Ha: µ ≠12
α=0.01
Sample Number | Test Statistic | P Value | Do not Reject Ho | Reject Ho |
1 | 14.11 | 0.00 | | X |
2 | 1.08 | 0.22 | X | |
3 | -25.05 | 0.00 | | X |
4 | 1.33 | 0.16 | X | |
Table 2
The preceding table shows us that we should not reject Ho for processes 2...

...QualityAssociates, a consulting firm, advises its clients about sampling and statistical procedures that can be used to control their manufacturing processes. In particular application, a client gave QualityAssociates a sample of 800 observations taken during a time in which that client’s process was operating satisfactorily. The sample standard deviation for these data was 0.21; hence with so much data, the population standard deviation was assumed to be 0.21. QualityAssociates then suggested that random samples of size 30 be taken periodically to monitor the process on an ongoing basis. By analyzing news samples, the client could quickly learn whether the process was operation satisfactorily. When the process was not operation satisfactorily, corrective action could be taken to eliminate the problem. The design specification indicated the mean for the process should be 12. The hypothesis test suggested by QualityAssociates follows.
:
:
Corrective action will be taken any time is rejected.
The dataset “Quality.sav” contains each of four samples, each of size 30, collected at hourly intervals during the first day of operation of the new statistical control procedure.
Managerial report
1. Conduct a hypothesis test for each sample at the 0.01 level of significance and determine what action, if any, should be taken. Provide the test statistic and the...

...CASE 7 – QUALITYASSOCIATESINC.
BACKGROUND (CASE STUDY WRITE UP)
QualityAssociates, Inc. is a consulting firm who advises its clients about statistical and sampling methods that can be used to control their manufacturing procedures. In this particular case we consider a production line designed to fill bottles of a shampoo with a mean weight of 12 ounces of shampoo per bottle. QualityAssociates, Inc. made a quality control testing of the manufacturing machine of this client (Alibaba Machinery), to determine if the process is operating properly or if, perhaps, a machine malfunction has caused the process to begin underfilling or overfilling the bottles. The client picked a sample of 800 bottles taken during a time when the machine was operating satisfactorily. The sample standard deviation for these data was 0.21; and thus (with so much data) the population standard deviation was assumed to be 0.21. Then QualityAssociates recommended that random samples of size 30 should be taken periodically to monitor the machine operation on an ongoing basis. By analyzing the new samples, Alibaba Machinery could quickly learn whether the machine is operating satisfactorily. If the machine is not operating satisfactorily, corrective action can be taken to eliminate the problem. The design specification indicated that...

...QualityAssociates, Ins.
QualityAssociates, Ins., a consulting firm, advises its clients about sampling and statistical procedures that can be used to control their manufacturing processes. In one particular application, a client gave QualityAssociates a sample of 800 observations taken during a time in which that client’s process was operating satisfactorily. The sample standard deviation for these data was .21; hence, with so much data, the population standard deviation was assumed to be .21. QualityAssociates then suggested that random samples of size 30 be taken periodically to monitor the process on an ongoing basis. By analyzing the new samples, the client could quickly learn whether the process was operating satisfactorily. When the process was not operating satisfactorily, corrective action could be taken to eliminate the problem. The design specification indicated the mean for the process should be 12. The hypothesis test suggested by QualityAssociates follows.
Ho: = 12
Ha: 12
Corrective action will be taken any time Ho rejected.
The following samples were collected at hourly intervals during the first day of operation of the new statistical process control procedure. These data are available in the data set Quality.
Sample 1 | Sample 2 | Sample 3 | Sample 4 | Sample 1 | Sample 2 | Sample 3 |...

...QualityAssociates, Inc. Report
A company has approached QualityAssociates, Inc. to verify that their sampling and statistical procedures are working efficiently. A sample of 800 units has been taken for observations for the study while operations were known to be operating at satisfactory levels. .21 has been set as the sample standard deviation and likewise so has the population standard deviation assumed to be .21 because there are a substantial amount of observations. Random samples of 30 will allow for swift action if said processes are not producing at efficient levels or drop below an average of 12 units, the null hypothesis, which has been set for this particular operation. Corrective action shall be taken anytime the experimental hypothesis tested does not equal 12.
Taking 30 samples at continuous hourly intervals four times during the day and hypothesis testing them resulted in corrective action being taken one out of three times. The first sample had a test statistic of -1.027 which led to sample 1 being .303 above the given acceptable levels. The second hypothesis test statistic equaled .713 with a .169 above the first sample’s level of significance and was also not rejected. The third sample resulted in a test statistic reading of -2.935 with a level of significance .004 below the acceptance level of .01 and corrective action was taken. The last sample delivered a test...

...thinking on quality and subsequently coming up with the idea of Kaizen and making it their way of doing.
Quality circles
Quality circles is an important organ of Kaizen, they consist of an informal group of people that involves operators, supervisors, managers, etc. getting together to improve processes and procedures in the making of a product. These circles are embraced in a participative style of management whereby new ideas are generated and implemented
The concept is based on the observation that the operators are in a better position to contribute ideas that will lead to an improvement in the operation since they are closest to the operation. Thus improvement ideas come not only from the management but form all levels of the organization.
The informal nature of the quality circles is such that invisible barriers between the people from different levels or departments of the organization are overcome and everyone feels free and comfortable to share their ideas. Thus consultation and discussion are hastened throughout the organization and the employees attain an uplift of morale and motivation also improving productivity.
The group members are actively involved in the decision making process and thus feel as an important player within the organization where they can contribute towards creating a better quality product. The members feel more integrated in the organization and as a family work...

...Case Problem: Par, Inc.
Section I: Summary
Par, Inc., a major manufacturer of golf equipment believes that a cut-resistant, longer lasting golf ball could increase their market share. In addition to the requirement that the ball be longer lasting, they wanted to ensure that the new coating would not reduce driving distances, and would be comparable to the current product.
Section II: Relevant Statistical Results
Statistic
Current Model
New Model
Sample Mean
270.275
267.500
Standard Error
1.3840
1.5648
Standard Deviation
8.7530
9.8969
Sample Size
40
40
Confidence Level (95%)
2.62
3.14
Degrees of freedom
39
39
t = 1.33
Comparisons
1. A two tail hypothesis test was conducted based on the sample studies of 40 current and 40 new golf balls.
The testing was performed with a machine designed to hit the ball the same every time. Both samples were tested with the same machine to ensure consistent strikes.
2. The hypothesis for Par, Inc. to compare the driving distances of the current and new golf balls is:
H0: µ1 - µ2 ≥ 0 (mean distance of new greater than or equal to the old)
Ha: µ1 -µ2 < 0 (mean distance of new less than old)
3. Statistical basis
n1 = 40 n2 = 40
x1 = 270.28 x2 = 267.5
s1 = 8.75 s2 = 9.90
α = .05
df = 78
t = (270.28 – 267.5) – 0 = 1.33
P-value = .10
P-value > .05 (or α); do not reject H0
4. The 95% confidence interval for the sample mean driving distance for current golf balls is 267.66 to 272.90....

...Case #1
Par, Inc.
Date: October 8th 2011
Question 1
Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls
Ans:
The rational is that Par, Inc. is concerned with making sure that the market share or the new ball is comparable to the previous with the new cut-resistant, longer lasting ball. We need to compare the mean driving distance in order to be certain that the new ball efforts are worth it. Therefore we need to formulate a hypothesis test.
The text tells us that Par, Inc. would like to sell the new ball once it’s comparable to the previous. Therefore:
Ho: μ₁ -μ₂ ≥ 0 the mean driving distance of the new balls is greater than or equal to that of the old balls
Ha: μ₁ - μ₂ ‹ 0 the mean driving distance of the new balls is less than that of the old balls,
α = 0.05
CURRENT | NEW |
270.28 | 267.50 |
8.75 | 9.90 |
40 | 40 |
Mean
Sample Standard Deviation
N
t= (x̄₁ - x̄₂) – (μ₁ - μ₂)0
√Sp²[(1/n₁) + (1/n₂)]
Sp² = (n₁-1) s₁² + (n₂-1) s² = (40-1)(8.75) +(40-1)(9.90) ² = 6808.23 = 87.285
n₁ + n₂ - 2 40+40-2 78
Sp² = 87.285
Df= 78
t=(270.28 – 267.5) – 0 = 1.33
√87.28 (1/40 + 1/40)
t=1.33
Reject Ho: if p-Val ≤ α
P-val is...