Question

An automobile dealer conducted a test to determine if the time in minutes needed to complete a minor engine tune-up depends on whether a computerized engine analyzer or an electronic analyzer is used. Because tune-up time varies among compact, intermediate, and full-sized cars, the three types of cars were used as blocks in the experiment. The data obtained follow.

		Analyzer
		Computerized	Electronic
Car	Compact	50	41
	Intermediate	57	45
	Full-sized	61	46

Use α = 0.05 to test for any significant differences.

*State the null and alternative hypotheses.

H₀: μ_Compact = μ_Intermediate = μ_Full-sized
H_a: μ_Compact ≠ μ_Intermediate ≠ μ_Full-sized

H₀: μ_Compact ≠ μ_Intermediate ≠ μ_Full-sized
H_a: μ_Compact = μ_Intermediate = μ_Full-sized    

H₀: μ_Computerized ≠ μ_Electronic
H_a: μ_Computerized = μ_Electronic

H₀: μ_Computerized = μ_Electronic = μ_Compact = μ_Intermediate = μ_Full-sized
H_a: Not all the population means are equal.

H₀: μ_Computerized = μ_Electronic
H_a: μ_Computerized ≠ μ_Electronic

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

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

p-value = ____

*State your conclusion.

Do not reject H₀. There is not sufficient evidence to conclude that the mean tune-up times are not the same for both analyzers.

Reject H₀. There is sufficient evidence to conclude that the mean tune-up times are not the same for both analyzers.    

Reject H₀. There is not sufficient evidence to conclude that the mean tune-up times are not the same for both analyzers.

Do not reject H₀. There is sufficient evidence to conclude that the mean tune-up times are not the same for both analyzers.

Answer 1

H₀: μ_Computerized = μ_Electronic
H_a: μ_Computerized ≠ μ_Electronic

Anova: Two-Factor Without Replication
SUMMARY	Count	Sum	Average	Variance
subject 1	2	91	45.5	40.500
subject 2	2	102	51	72.000
subject 3	2	107	53.5	112.500
Computerized	3	168	56	31.000
Electronic	3	132	44	7.000
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Rows	67.00	2	34	7.44	0.1184	19.00
Columns	216.00	1	216	48.00	0.0202	18.51
Error	9.00	2	5
Total	292.00	5

test stat = 48.00

p value=0.020

Reject H₀. There is sufficient evidence to conclude that the mean tune-up times are not the same for both analyzers.