Statistics Study Guide
Chapter 9
Hypothesis Testing
Case Problem 1: Quality Associates, Inc.
1. The hypothesis testing results are shown below:
Sample 1 Sample 2 Sample 3 Sample 4
Sample Size 30 30 30 30
Mean 11.959 12.029 11.889 12.081
Standard Deviation 0.220 0.220 0.207 0.206 Level of Significance (alpha) 0.010 0.010 0.010 0.010
Critical Value (lower tail) -2.576 -2.576 -2.576 -2.576
Critical Value (upper tail) 2.576 2.576 2.576 2.576 Hypothesized value 12 12 12 12
Standard Error 0.040 0.040 0.038 0.038
Test Statistic -1.027 0.713 -2.935 2.161 p-value 0.304 0.476 0.003 0.031
Only sample 3 leads to the rejection of the hypothesis H0: µ = 12. Thus, corrective action is warranted for sample 3. The other samples indicate H0 cannot be rejected and thus from all we can tell, the process is operating satisfactorily. Sample 3 with = 11.89 shows the process is operating below the desired mean. Sample 4 with = 12.08 is on the high side, but the p-value of .03 is not sufficient to reject H0.
2. The sample standard deviations for all four samples are in the .20 to .22 range. It appears that the process population standard deviation assumption of .21 is good.
3. With α = .01, z.005 = 2.576. Using the standard error of the mean =0.0383, the upper and lower control limits are computed as follows:
Upper Control Limit = 12 + 2.576 (0.0383) = 12.0987
Lower Control Limit = 12 - 2.576 (0.0383) = 11.9013
As long as a sample mean is between these two limits, the process is in control and no corrective action is required. Note that sample 3 with a mean of 11.89 shows corrective action is necessary because the sample mean is outside the control limits.
4. Increasing the level of significance will cause the null hypothesis to be rejected more often. While this may mean quicker corrective action when the process is out of control, it also means that there will be a higher error probability of stopping the process and attempting
Hypothesis Testing
Case Problem 1: Quality Associates, Inc.
1. The hypothesis testing results are shown below:
Sample 1 Sample 2 Sample 3 Sample 4
Sample Size 30 30 30 30
Mean 11.959 12.029 11.889 12.081
Standard Deviation 0.220 0.220 0.207 0.206 Level of Significance (alpha) 0.010 0.010 0.010 0.010
Critical Value (lower tail) -2.576 -2.576 -2.576 -2.576
Critical Value (upper tail) 2.576 2.576 2.576 2.576 Hypothesized value 12 12 12 12
Standard Error 0.040 0.040 0.038 0.038
Test Statistic -1.027 0.713 -2.935 2.161 p-value 0.304 0.476 0.003 0.031
Only sample 3 leads to the rejection of the hypothesis H0: µ = 12. Thus, corrective action is warranted for sample 3. The other samples indicate H0 cannot be rejected and thus from all we can tell, the process is operating satisfactorily. Sample 3 with = 11.89 shows the process is operating below the desired mean. Sample 4 with = 12.08 is on the high side, but the p-value of .03 is not sufficient to reject H0.
2. The sample standard deviations for all four samples are in the .20 to .22 range. It appears that the process population standard deviation assumption of .21 is good.
3. With α = .01, z.005 = 2.576. Using the standard error of the mean =0.0383, the upper and lower control limits are computed as follows:
Upper Control Limit = 12 + 2.576 (0.0383) = 12.0987
Lower Control Limit = 12 - 2.576 (0.0383) = 11.9013
As long as a sample mean is between these two limits, the process is in control and no corrective action is required. Note that sample 3 with a mean of 11.89 shows corrective action is necessary because the sample mean is outside the control limits.
4. Increasing the level of significance will cause the null hypothesis to be rejected more often. While this may mean quicker corrective action when the process is out of control, it also means that there will be a higher error probability of stopping the process and attempting