Question

A company manufactures printers and fax machines at plants located in Atlanta, Dallas, and Seattle. To measure how much employees at these plants know about quality management, a random sample of 6 employees was selected from each plant and the employees selected were given a quality awareness examination. The examination scores for these 18 employees are shown in the following table. The sample means, sample variances, and sample standard deviations for each group are also provided. Managers want to use these data to test the hypothesis that the mean examination score is the same for all three plants.

	Plant 1 Atlanta	Plant 2 Dallas	Plant 3 Seattle
	86	70	58
	74	74	65
	83	73	63
	76	75	69
	70	69	74
	79	89	67
Sample mean	78	75	66
Sample variance	34.8	52.4	29.6
Sample standard deviation	5.90	7.24	5.44

Set up the ANOVA table for these data. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.)

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p-value
Treatments
Error
Total

Answer 1

Ans:

df(treatments)=3-1=2

df(error)=18-3=15

Source	Sum	Degrees	Mean	F	p-value
of Variation	of Squares	of Freedom	Square
Treatments	468	2	234	6.01	0.0121
Error	584	15	38.93
Total	1052	17

Test statistic:

F=234/38.93=6.01

p-value=FDIST(6.01,2,15)=0.0121