Statistics Discussion Sample Essay.

**Category: Others**

Quality Associates, Inc., a consulting firm, advises its clients about sampling and statisti- cal 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 while that client’s process was operating satisfactorily. The sample standard deviation for these data was 0.21; hence, with so much data, the population standard deviation was assumed to be 0.21. Quality Associates then suggested that random samples of size 30 be taken periodi- cally 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 that the mean for the process should be 12. The hypothesis test suggested by Quality Associates is as follows:

Ho: Mu = 12

Ha: Mu not equal 12

Corrective action will be taken any time H0 is rejected.

The samples listed in the following table were collected at hourly intervals during the

first day of operation of the new statistical process control procedure. These data are available in the file Quality.

Use Excel and Word to upload the managerial report that contains the following. Make sure to do all the testing in excel as outlined in this chapter.

Managerial Report

- Conduct a hypothesis test for each sample at the 0.01 level of significance and determine what action, if any, should be taken. Provide the test statistic and p value for each test.
- Compute the standard deviation for each of the four samples. Does the conjecture of 0.21 for the population standard deviation appear reasonable?
- Compute limits for the sample mean x around m 5 12 such 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 x 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.
- 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?

Requirements: COMPLETE THE ASSIGNMENT

**Statistics**

Student’s Name

Institution

Course

Date

**Statistics**

**P-Value.**

The null hypothesis shows that our experiment has no observed impact. There is usually an equal sign in the mathematical formulation of the null hypothesis. H0 refers to the hypothesis. We are trying to uncover evidence in the hypothesis test against the null hypothesis. If the p-value is modest enough to be below our alpha threshold, we are justified in rejecting the null hypothesis. One cannot discard the null hypothesis if the p-value is more significant compared to alpha. An alternative hypothesis is that experiment will have an experiential effect. Typically there will be inequality or not equal to a symbol in a mathematical representation of the alternative hypothesis. Either Ha or H1 denotes this theory. The alternative hypothesis, through our hypothesis test, is what we want to prove indirectly. We cannot accept the alternative hypothesis when the null hypothesis is not refuted. In all samples, the Null, and alternative hypothesis, and population are constant.

H0: μ=12

Ha: μ≠ 12

Population, n=30

- Sample 1

X

X=11.9587, s=0.22

Calculation

p-value is 0.31

- Sample 2

X=

X=12.0287, s=0.22

Calculation

p-value is 0.48

- sample 3

X=

X=11.8890, s=0.21

Calculation

p-value is 0.07

- sample 4

x=

X=12.08, s=0.2061

Calculation

p-value is 0.39

**2. **** ****Standard deviation**** and ****assumption of .21 for the population **

Population in all samples is n=30

- Sample 1.

s= ∑= 0.22

- Sample 2.

s= ∑= 0.22

- Sample 3.

s= ∑= 0.23

- Sample 4.

s= ∑= 0.21

**3. Limits for the Sample Mean ´x around μ=12**

Sample 1

11.8764<μ<12.04

Answer = (11.88, 12.04)

Sample 2

11.95<μ<12.11

The answer is (11.95, 12.11)

Sample 3

11.81<μ<11.97

Answer = (11.811, 11.9664)

Sample 4

12.00<μ<12.16

The answer is (12.00, 12.16)

**4. Level Of Significance to a Larger Value.**

From the calculations, the margin of error rises, and the size of the interval of confidence is reduced as the level of significance rises.