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
	85	70	59
	75	74	63
	83	72	61
	75	73	69
	72	68	74
	90	87	64
Sample mean	80	74	65
Sample variance	49.6	45.2	30.8
Sample standard deviation	7.04	6.72	5.55

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

Test for any significant difference in the mean examination score for the three plants. Use

α = 0.05.

Answer 1

Given:

The total treatments or plants located in different locations (k) is 3.

The total number of employees or experimental units (N) is 18.

The provided data is,

	Plant 1	Plant 2	Plant 3
ni	6	6	6
x̄i	80	74	65
si	7.04	6.72	5.55

The overall mean (x̄) is,

The p-value obtained from the F-table at the degrees of freedom (2, 12) and the F-test statistic (8.17) is 0.0039.

The ANOVA table is,

Source of variation	Sum of squares (SS)	Degrees of freedom (df)	Mean square (MS)	F	p-value
Treatments	684	2	342	8.17	0.0039
Error	627.6125	15	41.84
Total	1311.6125	17

The statistical hypothesis is,

Ho: There is no significant difference in the mean examination score for the three plants.

Ha: There is a significant difference in the mean examination score for the three plants.

Since the p-value (0.0039) is less than the significance level (0.05), so the decision is to reject the null hypothesis. Therefore, it can be concluded that there is sufficient evidence to support that there is a significant difference in the mean examination score for the three plants at the significance level 0.05.