Question

Sample 1|Sample 2 Sample 3 Sample 4 11.91 11.55 11.62 11.62 11.69 11.52 11.59 11.75 11.82 1.90 11.97 11.64 11.80 11.87 12.03 11.94 12.01 11.92 11.99 12.13 12.09 12.16 11.93 12.21 12.32 12.39 11.93 11.85 11.92 11.7611.83 11.7012.38 11.77 11 12.00 12.07 12.04 11.98 12.05 12.30 12.37 12.18 12.25 11.97 12.17 12.24 11.85 11.92 12.30 12.37 12.15 12.22 12.02 11.36 12.02 11.75 12.05 11.95 12.18 12.14 11.72 12.07 12.05 11.85 11.64 12.16 12.39 11.65 12.11 11.90 12.22 11.56 11.88 11.95 12.03 12.35 12.06 12.09 11.7611.77 11.82 12.29 .84 11.89 11.79 1.60 12.30 11.95 12.27 1.96 12.29 12.47 11.75 12.03 12.04 11.96 12.17 11.95 11.94 11.89 11.97 88 12.23 12.25 11.61 12.10 11.91 12.12 11.61 12.21 12.20 12.00 12.28 12.00 12.01 12.11 12.22 11. 11.93 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 a client gave Quality Associates 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. 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. The hypothesis test suggested by Quality Associates follows Ho: μ-12 H1 : μ * 12 Corrective action will be taken any time Ho is rejected. Managerial Report . Conduct a hypothesis test for each sample at the 01 level of significance and determine what action, if any, should be taken. Provide the test statistic and p-value for each test. 2. Compute the standard deviation for each of the four samples. Does the assumption of .21 for the population standard deviation appear reasonable? 3, Compute limits for the sample mean x around μ-12 such that, as long as a new sample mean is within those limits, the process will be considered to be operating satisfactorily. If K exceeds the upper limit or if is below the lower limit, corrective action will be taken. These limits are referred to as upper and lower control limits for quality control purposes. 4. Discuss the implications of changing the level of significance to a larger value. What mistake or error could increase if the level of significance is increased? te meca ione ldchndiean Sample 4 Sample 1 Sample 2 Sample 3 Standard Dev Test Statistic p-value Corrective action taken? Does the assumption of .21 for the population standard deviation appear reasonable? hypothesis to be Increasing the level of significance will cause the action when the process is out of control. It also means that there will bea when the process is operating often. This may mean corrective error probability of stopping the process and attempting corrective action This would be in the probability of making a error.

Answer 1

sample 1

standard deviation = 0.2219

Test statistic

olVn

11.94- 12 0.21/V/30

= -1.48

P value = 0.1389 >0.01 (not significant )

Fail to reject H0

Upper control limit =

μ +3o/Vn 12 + 2.58 * 0.0383-12.10

Lower control limit =

μ-3c/Vn 12-2.58 * 0.0383 11.90

Since $\bar{X}$ is within the limits , NO corrective action taken .

sample 2

standard deviation = 0.2267

Test statistic

olVn

12.03 - 12 0.21/V30

= 0.88

P value = 0.3789  >0.01 (not significant )

Fail to reject H0.

Upper control limit =

μ +3o/Vn 12 + 2.58 * 0.0383-12.10

Lower control limit =

μ-3c/Vn 12-2.58 * 0.0383 11.90

Since $\bar{X}$ is within the limits , NO corrective action taken .

Note : we compute control limit using alpha =0.01 , that is zc =2.58

sample 3

standard deviation = 0.2025

Test statistic

olVn

11.88 12 - 0.21/V/30

= -3,10

P value = 0.0019 < 0.01 (significant)

Reject H0.

Upper control limit =

μ +3o/Vn 12 + 2.58 * 0.0383-12.10

Lower control limit =

μ-3c/Vn 12-2.58 * 0.0383 11.90

Since $\bar{X}$ is not within the limits , YES  corrective action taken .

sample 4

standard deviation = 0.2085

Test statistic

olVn

12.08 12 - 021/V30

= 2.20

P value = 0.0278 >0.01 (not significant)

Fail to reject H0

NOTE: If significance level were 0.05 , p value < 0.05 , we reject H0.

Upper control limit =

μ +3o/Vn 12 + 2.58 * 0.0383-12.10

Lower control limit =

μ-3c/Vn 12-2.58 * 0.0383 11.90

Since $\bar{X}$ is within the limits , NO corrective action taken .

YES assumption of population standard deviation seem reasonable

Increasing the level of significance cause the null hypothesis to be rejected (in sample 4) .This may mean taking corrective action when the process is out of control.

It also means there will be a increase in  error probability of stopping the process and making corrective action when the process is operating satisfactorily .This would be increase in the probability of making type 1 error .