Question

As part of a study designed to compare hybrid and similarly equipped conventional vehicles, a group tested a variety of classes of hybrid and all-gas model cars and sport utility vehicles (SUVs). Suppose the following data show the miles-per-gallon rating obtained for two hybrid small cars, two hybrid midsize cars, two hybrid small SUVs, and two hybrid midsize SUVs; also shown are the miles per gallon obtained for eight similarly equipped conventional models.

Class	Type	MPG
Small Car	Hybrid	35
Small Car	Conventional	30
Small Car	Hybrid	42
Small Car	Conventional	34
Midsize Car	Hybrid	29
Midsize Car	Conventional	21
Midsize Car	Hybrid	34
Midsize Car	Conventional	23
Small SUV	Hybrid	27
Small SUV	Conventional	21
Small SUV	Hybrid	28
Small SUV	Conventional	22
Midsize SUV	Hybrid	23
Midsize SUV	Conventional	19
Midsize SUV	Hybrid	24
Midsize SUV	Conventional	18

At the α = 0.05 level of significance, test for significant effects due to class, type, and interaction.

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

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

p-value =

State your conclusion about class.

Because the p-value > α = 0.05, class is not significant.Because the p-value > α = 0.05, class is significant.    Because the p-value ≤ α = 0.05, class is significant.Because the p-value ≤ α = 0.05, class is not significant.

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

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

p-value =

State your conclusion about type.

Because the p-value > α = 0.05, type is significant.Because the p-value > α = 0.05, type is not significant.    Because the p-value ≤ α = 0.05, type is not significant.Because the p-value ≤ α = 0.05, type is significant.

Find the value of the test statistic for interaction between class and type. (Round your answer to two decimal places.)

Find the p-value for interaction between class and type. (Round your answer to three decimal places.)

p-value =

State your conclusion about interaction between class and type.

Because the p-value > α = 0.05, interaction between class and type is significant.Because the p-value > α = 0.05, interaction between class and type is not significant.    Because the p-value ≤ α = 0.05, interaction between class and type is not significant.Because the p-value ≤ α = 0.05, interaction between class and type is significant.

Answer 1

Anova: Two-Factor With Replication
SUMMARY	small car	midsize car	small SUV	midsize SUV	Total
Hybrid
Count	2	2	2	2	8
Sum	77	63	55	47	242
Average	38.5	31.5	27.5	23.5	30.25
Variance	24.5	12.5	0.5	0.5	40.5
Conventional
Count	2	2	2	2	8
Sum	64	44	43	37	188
Average	32	22	21.5	18.5	23.5
Variance	8	2	0.5	0.5	31.14286
Total
Count	4	4	4	4
Sum	141	107	98	84
Average	35.25	26.75	24.5	21
Variance	24.91667	34.91667	12.33333	8.666667
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Sample (Type)	182.25	1	182.25	29.7551	0.000605	5.317655
Columns (class)	441.25	3	147.0833	24.01361	0.000236	4.066181
Interaction	11.25	3	3.75	0.612245	0.625881	4.066181
Within	49	8	6.125
Total	683.75	15

.

the value of the test statistic for class= 24.01

p-value = 0.000

Because the p-value ≤ α = 0.05, class is significant.

=====================

value of the test statistic for type = 29.76

p-value = 0.001

Because the p-value ≤ α = 0.05, type is significant.

==================

value of the test statistic for interaction between class and type= 0.61

p-value = 0.626

Because the p-value > α = 0.05, interaction between class and type is not significant.