7/29/2019 Harshal MR Assignment

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Marketing Research: One way ANOVA

Submitted to: Dr. M. Sivagnanasundaram Submitted by: Harshal Gandhi (12A1HP027)

Problem 1: Car Loan interest rates and city (sig. level = 0.05)

H0: The new car loan interest rates do not depend on the city which you apply.H1: The new car loan interest rates depend on the city which you apply.

ANOVA

rate

Sum of Squares df Mean Square F Sig.

Between Groups 3.918 4 .979 1.986 .115

Within Groups 19.723 40 .493

Total 23.641 44

Here the sig. > 0.05 so the null hypothesis is accepted and we can say that the new car loan interest

rates do not depend on the city which you apply.

Problem 2: Auto insurance premiums & company (sig. level = 0.01)

H0: Auto insurance premiums paid per month by all drivers insured by each of these four companies are

the same

H1: Auto insurance premiums paid per month by all drivers insured by each of these four companies are

different

ANOVA

Premium

Sum of Squares df Mean Square F Sig.

Between Groups 818.938 3 272.979 1.338 .308

Within Groups 2448.000 12 204.000

Total 3266.938 15

Here the sig. > 0.01 so the null hypothesis is accepted and we can say that the auto insurance

premiums paid per month by all drivers insured by each of these four companies are the same

Problem 3: Mean pressure applied to the drivers head & the type of the car

(sig. level = 0.05)

H0: The mean pressure applied to the drivers head during a crash test is equal for each type of car

H1: The mean pressure applied to the drivers head during a crash test is not equal for each type of car

Test of Homogeneity of Variances

mean pressure

Levene Statistic df1 df2 Sig.

.224 2 6 .806For Levene test, sig. > 0.05 so equal variance is assumed.

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ANOVA

mean pressure

Sum of Squares df Mean Square F Sig.

Between Groups 86049.556 2 43024.778 25.175 .001

Within Groups 10254.000 6 1709.000

Total 96303.556 8

Here the sig. < 0.05 so null hypothesis is rejected and we can say that the mean pressure applied to thedrivers head during a crash test is not equal for each type of car. Now, from the Contrast test, we can

check for which car the mean is different.

Contrast Coefficients

Contrast

type of car

compact mid size full size

1 -2 1 1

2 0 -1 1

Contrast Tests

Contrast

Value of

Contrast Std. Error t df Sig. (2-tailed)

mean pressure Assume equal

variances

1 -412.3333 58.46366 -7.053 6 .000

2 -26.3333 33.75401 -.780 6 .465

Does not assume equal

variances

1 -412.3333 51.78159 -7.963 5.395 .000

2 -26.3333 37.21410 -.708 3.896 .519

For contrast 1, the sig. < 0.05, & for contrast 2, the sig. > 0.05 so we can say that compact car

differs in the mean, when the mid size and full size car does not differ.