Question

The manager of computer operation of a large company wants to study computer usage of two departments within the company-the accounting dept. and the research dept. A random sample of five jobs from the accounting dept. in the past week and six jobs from the research dept. in the past week are selected, and the processing time(in seconds) for each job is recorded. Use a level of significance of 0.05, 1. Please provide descriptive stat. for the data ( Excel) 2. Is there evidence that the mean processing time in the research dept. is greater 6 seconds? ( Hand calculated, but use mean and SD from Excel output) 3. Is there evidence of a difference between the mean processing times of these two depts.?(Please use Excel, then you will use hand calculation to double check your answer) For question 2 and 3, make sure that you write down H0 and H1, conclusion and summary.

Accounting 9 3 8 7 12

Research 4 13 10 9 9 6

Answer 1

1) We have to find the descriptive statistic using excel.

X1: Accounting

X2: Research

n1 = 5

n2 = 6

= Mean of accounting = 7.8 ( Use command =AVERAGE(select x1 values)

= Mean of research = 8.5 ( Use command =AVERAGE(select x2 values)

s1 = Standard deviation of accounting = 3.271085 ( Use command =STDEV(Select x1 values)

s2 = Standard deviation of accounting = 3.146427 ( Use command =STDEV(Select x2 values)

2) We have to check there is evidence that the mean processing time in the research dept. is greater than 6 or not.

The null and alternative hypothesis is

Level of significance = 0.05

Here population standard deviation is unknown so we have to use t-test statistic.
Test statistic is

Degrees of freedom = n - 1 = 5 - 1 = 4

Critical value = 2.132 ( Using t table)

Test statistic < critical vaue we fail to reject null hypothesis.

Conclusion: There is not sufficient evidence that the mean processing time in the research dept. is greater than 6 seconds.

3) Now we have to test there is evidence of a difference between the mean processing times of these two depts.

The null and alternative hypothesis is

Level of significance = 0.05

Test statistic is

Degrees of freedom = n1 + n2 - 2 = 5 + 6 - 2 = 9

Critical value = 2.262 ( Using t table)

| t | < critical value we fail to reject null hypothesis.

Conclusion: There is not sufficient evidence of a difference between the mean processing times of these two depts.

Now by using excel, we have to check our answers.

First, enter all values into excel.

Then click on Data --------> Data Analysis -------->t-Test: Two-Sample Assuming Equal Variance ----->ok

Variable 1 Range: Select x1 values.

Variable 2 Range: Select x2 values.

Hypothesized Mean difference: 0

Alpha: 0.05

Output range: Select any empty cell.

------------->ok

We get

t-Test: Two-Sample Assuming Equal Variances
	x1	x2
Mean	7.8	8.5
Variance	10.7	9.9
Observations	5	6
Pooled Variance	10.25556
Hypothesized Mean Difference	0
df	9
t Stat	-0.36098
P(T<=t) one-tail	0.363221
t Critical one-tail	1.833113
P(T<=t) two-tail	0.726443
t Critical two-tail	2.262157

So both answers are same.