Please help me and give me a right answer.. Thank you!! Instructions: If you are

Question

Please help me and give me a right answer.. Thank you!!

Instructions:

If you are performing a hypothesis test, make sure you state the hypotheses, the level of significance, the decision rule in terms of the critical value, the test statistic or p-value, your decision (whether to reject or not to reject the null hypothesis), and your conclusion in managerial terms. These steps must be completed in addition to your Minitab output.

2. The Human Resources (HR) Director would like to implement a three-week training program for the manufacturing department employees, which she believes would improve their productivity. In order to have supporting data to support her belief, with the approval of the VP Operations, the three-week training program is given to 15 employees chosen randomly from the manufacturing department. We observed the number of items produced by these 15 employees before and after their training. The observation period before and after the training was 20 working days long.
You have been tasked to conduct the study to validate the HR Director`s belief. Use a level of significance of 1%.

a. Using Minitab, do the appropriate hypotheses test for this study.
i. Report your finding to the HR Director by writing a one-paragraph report.
ii. Are the two data sets independent or not? Explain your answer.
b. Now show the manual calculations for the test in (a). You can use the Minitab descriptive statistics report to obtain the summaries to allow you to do the calculation.
i. Do you come to the same conclusion that you did in (a)?
c. Calculate manually the 99% confidence interval. What are your findings? Do you come to the same conclusion that you did in (a)?

excel file below..

Employee Before After 1 15 17 2 13 16 3 8 10 4 9 9 5 7 9 6 12 13 7 11 14 8 12 15 9 11 14 10 9 11 11 10 14 12 12 11 13 11 13 14 7 10 15 12 13

Explanation / Answer

a. Using Minitab, do the appropriate hypotheses test for this study.

Solution:

Here, we have to use paired t test. The null and alternative hypothesis for this test is given as below:

Null hypothesis: H0: A three-week training program does not improve their productivity for the manufacturing department employees.

Alternative hypothesis: Ha: A three-week training program improves their productivity for the manufacturing department employees.

H0: µBefore = µAfter Versus Ha: µBefore < µAfter

This is a one tailed test. This is a lower tailed or left tailed test.

We are given a level of significance as 1% or = 0.01.

Minitab output for this test is given as below:

Paired T-Test and CI: Before , After

Paired T for Before - After

                  N      Mean     StDev   SE Mean

Before           15    10.600     2.261     0.584

After            15    12.600     2.501     0.646

Difference       15    -2.000     1.309     0.338

99% upper bound for mean difference: -1.113

T-Test of mean difference = 0 (vs < 0): T-Value = -5.92 P-Value = 0.000

i. Report your finding to the HR Director by writing a one-paragraph report.

We reject null hypothesis if P-value < , otherwise we do not reject the null hypothesis. Here, P-value = 0.00 < = 0.01. So, we reject the null hypothesis that a three-week training program does not improve their productivity for the manufacturing department employees. There is sufficient evidence to conclude that a three-week training program improves their productivity for the manufacturing department employees.

ii. Are the two data sets independent or not? Explain your answer.

Data sets are not independent because it is observed that the productivity is improved after launching the training program.

b. Now show the manual calculations for the test in (a). You can use the Minitab descriptive statistics report to obtain the summaries to allow you to do the calculation.
i. Do you come to the same conclusion that you did in (a)?

Solution:

Here, we have to use paired t test. The null and alternative hypothesis for this test is given as below:

Null hypothesis: H0: A three-week training program does not improve their productivity for the manufacturing department employees.

Alternative hypothesis: Ha: A three-week training program improves their productivity for the manufacturing department employees.

H0: µBefore = µAfter Versus Ha: µBefore < µAfter

This is a one tailed test. This is a lower tailed or left tailed test.

We are given a level of significance as 1% or = 0.01.

Test statistic formula is given as below:

t = Dbar / [ Sd/sqrt(n)]

We have

Dbar = -2

Sd = 1.3093

n = 15

t = -2.00/[1.3093/sqrt(15)]

t = -5.9161

df = n – 1 = 15 – 1 = 14

Critical value = -2.6245 (by using t-table)

P-value = 0.00 (by using t-table)               

P-value = 0.00 < = 0.01

So, we reject the null hypothesis that a three-week training program does not improve their productivity for the manufacturing department employees. There is sufficient evidence to conclude that a three-week training program improves their productivity for the manufacturing department employees.

Results are same as part a.

c. Calculate manually the 99% confidence interval. What are your findings? Do you come to the same conclusion that you did in (a)?

Solution:

Confidence interval = Dbar -/+ t*Sd/sqrt(n)

df = n – 1 = 15 – 1 = 14

Confidence level = 99%

Critical t value = 2.9768

Confidence interval = -2 -/+ 2.9768* 1.3093/sqrt(15)

Confidence interval = -2 -/+ 1.0064

Lower limit = -2 - 1.0064 = -3.0064

Upper limit = -2 + 1.0064 = -0.9936

Confidence interval = (-3.0064, -0.9936)

The mean difference ‘0’ is not included in the above confidence interval, so we reject the null hypothesis. We come with same conclusion as we did in part a.

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