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. …show more content…
Based on the standard deviations calculated for each sample, the assumption of .21 for the population standard deviation appears reasonable. An average of the 4 individual sample standard deviations is equal to .2135, which can be rounded down to .21.
The third question asked to compute limits for the sample mean equal to 12. Condition was that as long as a new sample mean is within those limits, the process will be considered to be operating satisfactorily. If X exceeds the upper limit or if the X is below the lower limit, corrective actions will be taken.
The formula for calculating upper and lower control limits is
Using x bar equal to 12, z alpha/2 equal to 2.576 and the standard error equal to .0383, the upper limit was equal to 12.0987 and the lower limit was equal to 11.9013.
Based on the upper (12.0987) and lower (11.9013) control limits calculated for a mean equal to 12, sample 3 falls outside the control limit with a mean of 11.8890. Because the mean exceeds the lower limit, it indicates that corrective action needs to be taken.
Increasing the level of significance to a larger value will lead to rejecting the null hypothesis more often. If the level of significance is increased to .05, both samples 3 and 4 will provide