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	60
	74	76	65
	83	74	63
	75	73	68
	71	69	75
	86	88	59
Sample mean	79	75	65
Sample variance	41.2	47.2	34.8
Sample standard deviation	6.42	6.87	5.90

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.

State the null and alternative hypotheses.

H₀: μ₁ = μ₂ = μ₃
H_a: Not all the population means are equal.

Find the value of the test statistic. (Round your answer to two decimal places.)

Find the p-value. (Round your answer to four decimal places.)

p-value =

State your conclusion.

Do not reject H₀. There is not sufficient evidence to conclude that the means for the three plants are not equal. Reject H₀. There is not sufficient evidence to conclude that the means for the three plants are not equal.     Do not reject H₀. There is sufficient evidence to conclude that the means for the three plants are not equal. Reject H₀. There is sufficient evidence to conclude that the means for the three plants are not equal.

Answer 1

The statistic software output for this problem is:

Analysis of Variance results:
Data stored in separate columns.

Column statistics

Column	n	Mean	Std. Dev.	Std. Error
1	6	79	6.4187226	2.6204325
2	6	75	6.8702256	2.8047579
3	6	65	5.8991525	2.4083189

ANOVA table

Source	DF	SS	MS	F-Stat	P-value
Columns	2	624	312	7.60	0.0053
Error	15	616	41.07
Total	17	1240

The null and alternative hypothesis are :

H₀: μ₁ = μ₂ = μ₃
H_a: Not all the population means are equal.

Test statistics = 7.60

P-value = 0.0053

Reject H₀. There is sufficient evidence to conclude that the means for the three plants are not equal.