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	42
	Intermediate	57	45
	Full-sized	61	45

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-sizedH₀: μ_Computerized ≠ μ_Electronic
H_a: μ_Computerized = μ_Electronic    H₀: μ_Compact = μ_Intermediate = μ_Full-sized
H_a: μ_Compact ≠ μ_Intermediate ≠ μ_Full-sizedH₀: μ_Computerized = μ_Electronic
H_a: μ_Computerized ≠ μ_ElectronicH₀: μ_Computerized = μ_Electronic = μ_Compact = μ_Intermediate = μ_Full-sized
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 three decimal places.)

p-value =

State your conclusion.

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

Answer 1

using excel data analysis tool for two factor anova, following o/p Is obtained : write data>menu>data>data analysis>anova :two factor without replication>enter required labels>ok, and following o/p Is obtained,

Anova: Two-Factor Without Replication
SUMMARY	Count	Sum	Average	Variance
compact	2	92	46	32.000
intermediate	2	102	51	72.000
full sized	2	106	53	128.000
computerized	3	168	56	31.000
electronic	3	132	142	3.000
			134
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Rows	52.00	2	26.00	3.25	0.24	19.00
Columns	216.00	1	216.00	27.00	0.04	18.51
Error	16.00	2	8.00
Total	284.00	5

a)

H₀: μ_Computerized = μ_Electronic
H_a: μ_Computerized ≠ μ_Electronic

value of the test statistic.=3.25

p value=0.035

p value <α=0.05, reject Ho

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