Question

Quality Associates, Inc., a consulting firm, advises its clients about sampling and statistical procedures that can be used to control their manufacturing processes. In one particular application, metal spheres are created for conducting scientific experiments. A client gave Quality Associates a sample of 800 observations taken during a time in which that client's process of creating metal spheres was operating satisfactorily. The sample standard deviation for the radii of these spheres was 0.21 cm; hence, with so much data, the population standard deviation was assumed to be 0.21 cm. Quality Associates 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 cm.

• A (brief) introduction that states the intent of your report.

•  A section in which you state the appropriate hypotheses for whether the production process is operating correctly. You should conduct a hypothesis test for each of the four samples using a 1% level of significance and determine what action, if any, should be taken.

•  A section in which you compute an interval around 12 cm such that, as long as a new sample mean is within those limits, the process will be considered to be operating satisfactorily using a 1% level of significance. Include in this section a discussion of the implications of changing the level of significance to 5%. What mistake or error could increase with this adjusted level of significance?

•  A section that contains conclusions and recommendations.

Sample 1	Sample 2	Sample 3	Sample 4
11.55	11.62	11.91	12.02
11.62	11.69	11.36	12.02
11.52	11.59	11.75	12.05
11.75	11.82	11.95	12.18
11.90	11.97	12.14	12.11
11.64	11.71	11.72	12.07
11.80	11.87	11.61	12.05
12.03	12.10	11.85	11.64
11.94	12.01	12.16	12.39
11.92	11.99	11.91	11.65
12.13	12.20	12.12	12.11
12.09	12.16	11.61	11.90
11.93	12.00	12.21	12.22
12.21	12.28	11.56	11.88
12.32	12.39	11.95	12.03
11.93	12.00	12.01	12.35
11.85	11.92	12.06	12.09
11.76	11.83	11.76	11.77
12.16	12.23	11.82	12.20
11.77	11.84	12.12	11.79
12.00	12.07	11.60	12.30
12.04	12.11	11.95	12.27
11.98	12.05	11.96	12.29
12.30	12.37	12.22	12.47
12.18	12.25	11.75	12.03
11.97	12.04	11.96	12.17
12.17	12.24	11.95	11.94
11.85	11.92	11.89	11.97
12.30	12.37	11.88	12.23
12.15	12.22	11.93	12.25

Answer 1

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Answer:

1)

The given data are based on the hourly intervals during the first day of operation of the new statistical process control procedure. The population standard deviation is 0.21.

The null and alternative hypotheses are given below:

Null hypothesis:

H0:μ=12H0:μ=12

Alternative hypothesis:

Ha:μ≠12Ha:μ≠12

Test statistic and p-value for sample 1:

Software Procedure:

A step-by-step procedure to obtain the test statistic and p-value using MINITAB software:

• Choose Stat > Basic Statistics > 1-Sample Z.
• In Samples in Column, enter the column of Sample 1.
• In Standard deviation, enter a value for s as 0.21.
• In Perform hypothesis test, enter the test mean as 12.
• Check Options; enter Confidence level as 99%.
• Choose not equal in alternative.
• Click OK.

Output using MINITAB software is given below:

From the output, the test statistic is −1.08, and the p-value is 0.281.

Test statistic and p-value for sample 2:

Software Procedure:

A step-by-step procedure to obtain the test statistic and p-value using MINITAB software:

• Choose Stat > Basic Statistics > 1-Sample Z.
• In Samples in Column, enter the column of Sample 2.
• In Standard deviation, enter a value for s as 0.21.
• In Perform hypothesis test, enter the test mean as 12.
• Check Options; enter Confidence level as 99%.
• Choose not equal in alternative.
• Click OK.

Output using MINITAB software is given below:

From the output, the test statistic is 0.75, and the p-value is 0.455.

Test statistic and p-value for sample 3:

Software Procedure:

A step-by-step procedure to obtain the test statistic and p-value using MINITAB software:

• Choose Stat > Basic Statistics > 1-Sample Z.
• In Samples in Column, enter the column of Sample 3.
• In Standard deviation, enter a value for s as 0.21.
• In Perform hypothesis test, enter the test mean as 12.
• Check Options; enter Confidence level as 99%.
• Choose not equal in alternative.
• Click OK.

Output using MINITAB software is given below:

From the output, the test statistic is –2.9, and the p-value is 0.004.

Test statistic and p-value for sample 4:

Software Procedure:

A step-by-step procedure to obtain the test statistic and p-value using MINITAB software:

• Choose Stat > Basic Statistics > 1-Sample Z.
• In Samples in Column, enter the column of Sample 4.
• In Standard deviation, enter a value for s as 0.21.
• In Perform hypothesis test, enter the test mean as 12.
• Check Options; enter Confidence level as 99%.
• Choose not equal in alternative.
• Click OK.

Output using MINITAB software is given below:

From the output, the test statistic is 2.12, and the p-value is 0.034.

Rejection rule:

If p-value≤αp-value≤α, reject the null hypothesis

If p-value>αp-value>α, do not reject the null hypothesis.

Conclusion for Sample 1:

Here, the p-value is greater than the level of significance.

That is, p-value(=0.281)>α(=0.01)p-value(=0.281)>α(=0.01).

By the rejection rule, the null hypothesis is not rejected.

Hence, the mean for the process should be 12.

Conclusion for Sample 2:

Here, the p-value is greater than the level of significance.

That is, p-value(=0.455)>α(=0.01)p-value(=0.455)>α(=0.01).

By the rejection rule, the null hypothesis is not rejected.

Hence, the mean for the process should be 12.

Conclusion for Sample 3:

Here, the p-value is less than the level of significance.

That is, p-value(=0.004)<α(=0.01)p-value(=0.004)<α(=0.01).

By the rejection rule, the null hypothesis is rejected.

Hence, the mean for the process should not be 12.

Here, the null hypothesis is rejected. Hence, the corrective action will be taken any time.

Conclusion for Sample 4:

Here, the p-value is greater than the level of significance.

That is, p-value(=0.034)>α(=0.01)p-value(=0.034)>α(=0.01).

By the rejection rule, the null hypothesis is not rejected.

Hence, the mean for the process should be 12.

2)

3)

If the level of significance is larger, the change for rejecting the null hypothesis will decrease. This does not give the most accurate result.