Case Study Corrected

1) 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 the p-value for each test.
At 0.01 level of significance │Zᾳ/2 │ = 2.58
Reject Ho when │Z│> │Zᾳ/2 │

Sample 1
Sample 2
Sample 3
Sample 4
SAMPLE MEAN
11.958
12.028
11.889
12.081
POPULATION MEAN
12
12
12
12
SAMPLE STANDARD DEVIATION
0.22
0.22
0.20
0.20
SAMPLE SIZE
30
30
30
30
Z
-1.028
0.712
-2.935
2.161
RESULT
ACCEPT
ACCEPT
REJECT
ACCEPT

Sample 3 is the only one that rejects the null hypothesis of u = 12.
2) Compute standard deviation for each of the four samples. Does the assumption of .21 for the population standard deviation appear reasonable?

Sample 1
Sample 2
Sample 3
Sample 4
SAMPLE MEAN
11.958
12.028
11.889
12.081
SAMPLE SIZE
30
30
30
30
SAMPLE STANDARD DEVIATION
0.22
0.22
0.20
0.20

From the table and observing the values the standard sample deviations of each sample fell in the range of {0.20, 0.22}. 0.21 fell in the observed range and therefore the assumption of population’s standard deviation as 0.21 is reasonable/good.
3) Compute limits for the sample mean around u = 12 such that, as long as a new sample mean is within those limits, the process will be considered to be operating satisfactorily. If sample mean exceeds the upper limit or if it 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.
Class Interval = {u – [Z*Error], u + [Z*Error]} = {12 – [2.58*0.038], 12 + [2.58*0.038]} = {11.901, 12.098}

Sample 1
Sample 2
Sample 3
Sample 4
SAMPLE MEAN
11.958
12.028
11.889
12.081

From the data it was found that the mean of sample 3 does not fall with class interval and therefore corrective action must be taken on it.
4) Discuss the implications of changing the level of significance to a larger value. What mistakes or error could increase if the level of significance is increased?
For 0.01