An automobile dealer conducted a test to determine if the time in minutes needed to complete...

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

Analyzer
ComputerizedElectronic
CarCompact5042
Intermediate5745
Full-sized6145

Use α = 0.05 to test for any significantdifferences.

State the null and alternative hypotheses.

H0: μCompact ≠μIntermediate ≠μFull-sized
Ha: μCompact =μIntermediate =μFull-sizedH0:μComputerized ≠μElectronic
Ha: μComputerized =μElectronic    H0:μCompact = μIntermediate =μFull-sized
Ha: μCompact ≠μIntermediate ≠μFull-sizedH0:μComputerized =μElectronic
Ha: μComputerized ≠μElectronicH0:μComputerized = μElectronic= μCompact = μIntermediate= μFull-sized
Ha: Not all the population means are equal.

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

Find the p-value. (Round your answer to three decimalplaces.)

p-value =

State your conclusion.

Do not reject H0. There is sufficientevidence to conclude that the mean tune-up times are not the samefor both analyzers.Do not reject H0. There isnot sufficient evidence to conclude that the mean tune-up times arenot the same for both analyzers.    RejectH0. There is sufficient evidence to concludethat the mean tune-up times are not the same for bothanalyzers.Reject H0. There is not sufficientevidence to conclude that the mean tune-up times are not the samefor both analyzers.

Answer & Explanation Solved by verified expert
3.6 Ratings (609 Votes)

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: μComputerized = μElectronic
Ha: μComputerized ≠ μElectronic

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


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