Question

NCP manufactures printers and fax machines at plants located in City 1, City 2, and City 3. 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 table given below. 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	Plant 2	Plant 3
	City 1	City 2	City 3
	74	61	82
	76	71	82
	74	63	84
	86	74	81
	81	68	82
	71	65	93
Sample mean	77	67	84
Sample variance	30.4	24.4	20.4
Sample standard deviation	5.51	4.94	4.52

Set up the ANOVA table and test for any significant difference in the mean examination score for the three plants. Use a=0.5.

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square (to 2 decimals)	(to 2 decimals)	-value (to 4 decimals)
Treatments
Error
Total

The p-value is - Select your answer -less than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 11 .

What is your conclusion?

Answer 1

excel data analysis tool for one factor anova,steps are:

write data>menu>data>data analysis>anova :one factor>enter required labels>ok, and following o/p Is obtained,

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
City 1	6	462	77	30.4
City 2	6	402	67	24.4
City 3	6	504	84.00	20.40
ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	876.000	2	438.00	17.47	0.000
Within Groups	376.00	15	25.07
Total	1252.00	17

the p value is less than 0.01

since,p-value <α=0.05, reject Ho

hence, there is enough confidence that there is a significant difference in the mean examination score for the three plants at α=0.05