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	27
Midsize Car	Conventional	23
Midsize Car	Hybrid	32
Midsize Car	Conventional	25
Small SUV	Hybrid	27
Small SUV	Conventional	21
Small SUV	Hybrid	28
Small SUV	Conventional	22
Midsize SUV	Hybrid	25
Midsize SUV	Conventional	17
Midsize SUV	Hybrid	26
Midsize SUV	Conventional	16

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

Answer 1

Excel > Data > Anova : Two-factor with replication > Select data > OK

Excel Output:

Anova: Two-Factor With Replication
SUMMARY	Small car	Midsize car	Small SUV	Midsize SUV	Total
Hybrid
Count	2	2	2	2	8
Sum	77	59	55	51	242
Average	38.5	29.5	27.5	25.5	30.25
Variance	24.5	12.5	0.5	0.5	33.64285714
conventional
Count	2	2	2	2	8
Sum	64	48	43	33	188
Average	32	24	21.5	16.5	23.5
Variance	8	2	0.5	0.5	37.42857143
Total
Count	4	4	4	4
Sum	141	107	98	84
Average	35.25	26.75	24.5	21
Variance	24.91666667	14.91666667	12.33333333	27.33333333
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Type	182.25	1	182.25	29.75510204	0.000605233	5.317655072
Class	441.25	3	147.0833333	24.01360544	0.00023556	4.066180551
Interaction	7.25	3	2.416666667	0.394557823	0.760517326	4.066180551
Within	49	8	6.125
Total	679.75	15

	test statistic	p-value	conclusion
class	24.01360544	0.00023556	the p-value ≤ α = 0.05, class is significant.
type	29.75510204	0.000605233	the p-value ≤ α = 0.05, type is significant
interaction	0.394557823	0.760517326	p-value > α = 0.05, interaction between class and type is not significant.