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Question 12.5: Gasoline Consumption A researcher wishes to see whether the ......

Gasoline Consumption

A researcher wishes to see whether the type of gasoline used and the type of automobile driven have any effect on gasoline consumption. Two types of gasoline, regular and highoctane, will be used, and two types of automobiles, two-wheel- and all-wheel-drive, will be used in each group. There will be two automobiles in each group, for a total of eight automobiles used. Using a two-way analysis of variance, determine if there is an interactive effect, an effect due to the type of gasoline used, and an effect due to the type of vehicle driven.

The data (in miles per gallon) are shown here, and the summary table is given in Table 12-6.

 

Gas

Type of automobile
Two-wheel-drive All-wheel-drive
Regular 26.7 28.6
25.2 29.3
High-octane 32.3 26.1
32.8 24.2
TABLE 12–6 ANOVA Summary Table for Example 12–5

Source

SS d.f. MS

F

Gasoline A

3.920

Automobile B

9.680

Interaction (A × B )

54.080

Within (error)

  3.300

Total

70.980

Step-by-Step
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Step 1 State the hypotheses. The hypotheses for the interaction are

H_{0} : There is no interaction effect between type of gasoline used and type of automobile a person drives on gasoline consumption.

H_{1} : There is an interaction effect between type of gasoline used and type of automobile a person drives on gasoline consumption.

The hypotheses for the gasoline types are

H_{0} : There is no difference between the means of gasoline consumption for two types of gasoline.

H_{1} : There is a difference between the means of gasoline consumption for two types of gasoline.

The hypotheses for the types of automobile driven are

H_{0} : There is no difference between the means of gasoline consumption for two-wheel-drive and all-wheel-drive automobiles.

H_{1} : There is a difference between the means of gasoline consumption for two-wheel-drive and all-wheel-drive automobiles.

Step 2 Find the critical values for each F test. In this case, each independent variable, or factor, has two levels. Hence, a 2 \times 2 ANOVA table is used. Factor A is designated as the gasoline type. It has two levels, regular and high-octane; therefore, a=2. Factor B is designated as the automobile type. It also has two levels; therefore, b=2. The degrees of freedom for each factor are as follows:

Factor A: \quad d.f.N. =a-1=2-1=1

Factor B: \quad d.f.N. =b-1=2-1=1

Interaction (A \times B): \quad d.f.N. =(a-1)(b-1)=(2-1)(2-1)=1 \cdot 1=1

Within (error): \quad d.f.D. =a b(n-1)=2 \cdot 2(2-1)=4

where n is the number of data values in each group. In this case, n=2.

The critical value for the F_{A} test is found by using \alpha=0.05, d.f.N. =1, and d.f.D. =4. In this case, F_{A}=7.71. The critical value for the F_{B} test is found by using \alpha=0.05, d.f.N. =1, and d.f.D. =4; also F_{B} is 7.71.

Finally, the critical value for the F_{A \times B} test is found by using d.f.N. =1 and d.f.D. =4; it is also 7.71.

Note: If there are different levels of the factors, the critical values will not all be the same. For example, if factor A has three levels and factor b has four levels, and if there are two subjects in each group, then the degrees of freedom are as follows:

\begin{aligned}\text { d.f.N. } & =a-1=3-1=2 & & \text { factor } A \\\text { d.f.N. } & =b-1=4-1=3 & & \text { factor } B \\\text { d.f.N. } & =(a-1)(b-1)=(3-1)(4-1) & & \\& =2 \cdot 3=6 & & \text { factor } A \times B \\\text { d.f.D. } & =a b(n-1)=3 \cdot 4(2-1)=12 & & \text { within (error) factor }\end{aligned}

Step 3 Complete the ANOVA summary table to get the test values. The mean squares are computed first.

\begin{aligned}\mathrm{MS}_{A} & =\frac{\mathrm{SS}_{A}}{a-1}=\frac{3.920}{2-1}=3.920 \\\mathrm{MS}_{B} & =\frac{\mathrm{SS}_{B}}{b-1}=\frac{9.680}{2-1}=9.680 \\\mathrm{MS}_{A \times B} & =\frac{\mathrm{SS}_{A \times B}}{(a-1)(b-1)}=\frac{54.080}{(2-1)(2-1)}=54.080 \\\mathrm{MS}_{W} & =\frac{\mathrm{SS}_{W}}{a b(n-1)}=\frac{3.300}{4}=0.825\end{aligned}

The F values are computed next.

\begin{array}{rlr}F_{A}=\frac{\mathrm{MS}_{A}}{\mathrm{MS}_{W}}=\frac{3.920}{0.825}=4.752 & \text { d.f.N. }=a-1=1 & \text { d.f.D. }=a b(n-1)=4 \\F_{B}=\frac{\mathrm{MS}_{B}}{\mathrm{MS}_{W}}=\frac{9.680}{0.825}=11.733 & \text { d.f.N. }=b-1=1 & \text { d.f.D. }=a b(n-1)=4 \\F_{A \times B}=\frac{\mathrm{MS}_{A \times B}}{\mathrm{MS}_{W}}=\frac{54.080}{0.825}=65.552 & \text { d.f.N. }=(a-1)(b-1)=1 & \text { d.f.D. }=a b(n-1)=4\end{array}

The completed ANOVA table is shown in Table 12-7.

Step 4 Make the decision. Since F_{B}=11.733 and F_{A \times B}=65.552 are greater than the critical value 7.71, the null hypotheses concerning the type of automobile driven and the interaction effect should be rejected. Since the interaction effect is statistically significant, no decision should be made about the automobile type without further investigation.

Step 5 Summarize the results. Since the null hypothesis for the interaction effect was rejected, it can be concluded that the combination of type of gasoline and type of automobile does affect gasoline consumption.

TABLE 12–7 Completed ANOVA Summary Table for Example 12–5
Source SS d.f. MS F
Gasoline A 3.920 1 3.920 4.752
Automobile B 9.680 1 9.680 11.733
Interaction (A × B ) 54.080 1 54.080 65.552
Within (error)  3.300  4  0.825
Total 70.980 7

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