Question

A company is evaluating the effectiveness of two different training programs to speed up the production process.   The employees were placed into three groups and their assembly times were recorded in minutes.

No Training	Training Method 1	Training Method 2
44	48	38
47	52	35
43	43	42
51	39	47
57	49	40
42	41	38
50	37	36
46	40	39
44	45	35
45		41

Test the hypothesis at the .01 level that there is no difference in the population mean times for the three groups.

What is the value of your test statistic from step 3?

Select one:

a. .0023.

b. 13.4333.

c. 2.575.

d. 7.7683.

What is your decision/summarize results (steps 4 and 5)?

Select one:

a. Fail to reject H0. There is a difference in the mean times.

b. Fail to reject H0. There is no difference in the mean times.

c. Reject H0. There is a difference in the mean times.

d. Reject H0. There is no difference in the mean times.

Test the hypothesis at the .01 level that there is no difference in the population mean times for the three groups.

What is your critical value for step 2?

Select one:

a. 4.64.

b. 4.82.

c. 5.53.

d. 7.72.

Answer 1

Here we need to use one way ANOVA. Hypotheses are:

H0: There is no difference in the population mean times for the three groups.

Ha: There is a difference in the population mean times for the three groups.

Following is the output of one way ANOVA:

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
No Training	10	469	46.9	20.98888889
Training Method 1	9	394	43.77777778	25.69444444
Training Method 2	10	391	39.1	13.43333333
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	307.954789	2	153.9773946	7.768252845	0.002264978	5.526334714
Within Groups	515.355556	26	19.82136752
Total	823.310345	28

The test statistics is:

F = 7.7683

Correct option is d.

The p-value is: 0.0023

Since p-value is less than 0.01 so we reject the null hypothesis.

c. Reject H0. There is a difference in the mean times.

-----------------

The critical value is: 5.53

Correct option is c. 5.53.