Use the Scheffé test to test each pair of means in Example 12–1 to see if a significant difference exists between each pair of means. Use 𝛼 = 0.05.

Question

Accepted Answer

The [latex]F[/latex] critical value for Example 12-1 is [latex]4.26[/latex]. Then the critical value for the individual tests with d.f.N. [latex]=2[/latex] and d.f.D. [latex]=9[/latex] is

[latex]F^{\prime}=(k-1)(\mathrm{C} . \mathrm{V} .)=(3-1)(4.26)=8.52[/latex]

a. For [latex]\bar{X}_{1}[/latex] versus [latex]\bar{X}_{2}[/latex],

[latex]F_{S}=\frac{\left(\bar{X}_{1}-\bar{X}_{2}\right)^{2}}{s_{W}^{2}\left[\left(1 / n_{1}\right)+\left(1 / n_{2}\right)\right]}=\frac{(37.25-35.4)^{2}}{25.106\left(\frac{1}{4}+\frac{1}{5}\right)}=0.30[/latex]

Since [latex]0.30<8.52[/latex], the decision is that [latex]\mu_{1}[/latex] is not significantly different from [latex]\mu_{2}[/latex].

b. For [latex]\bar{X}_{1}[/latex] versus [latex]\bar{X}_{3}[/latex],

[latex]F_{S}=\frac{\left(\bar{X}_{1}-\bar{X}_{3}\right)^{2}}{s_{W}^{2}\left[\left(1 / n_{1}\right)+\left(1 / n_{3}\right)\right]}=\frac{(37.25-26)^{2}}{25.106\left(\frac{1}{4}+\frac{1}{3}\right)}=8.64[/latex]

Since [latex]8.64>8.52[/latex], the decision is that [latex]\mu_{1}[/latex] is significantly different from [latex]\mu_{3}[/latex].

c. For [latex]\bar{X}_{2}[/latex] versus [latex]\bar{X}_{3}[/latex],

[latex]F_{S}=\frac{\left(\bar{X}_{2}-\bar{X}_{3}\right)^{2}}{s_{W}^{2}\left[\left(1 / n_{2}\right)+\left(1 / n_{3}\right)\right]}=\frac{(35.4-26)^{2}}{25.106\left(\frac{1}{5}+\frac{1}{3}\right)}=6.60[/latex]

Since [latex]6.60<8.64[/latex], the decision is that [latex]\mu_{2}[/latex] is not significantly different from [latex]\mu_{3}[/latex]. Hence, only the mean of the small cars is not equal to the mean of luxury cars.

Question 12.3: Use the Scheffé test to test each pair of means in Example 1......

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