Question

An automotive engineer wanted to determine whether the octane of gasoline used in a car increases gas mileage. He selected six different brands of car and assigned a driver to each car. Then 5 gallons of the gasoline selected was placed in the car. The car was driven around a closed track at 40 miles per hour until it ran out of gas. The number of miles driven was recorded and then divided by 5 to obtain the miles per allon, He obtained the following results Car Brand Chev. Impala 89 Octane 28.4 26.9 26.1 26.4 28.9 25.1 87 Octane 28.3 27.1 26.4 26.1 28.4 25.3 92 Octane 28.7 27.2 26.8 27.3 29.1 25.8 300M F. Taurus Linc. LS Toy. Ca Volvo S60 a.Normal probability plots for each treatment indicate that the requirement of normality has been satisfied. Verify that the requirement of equal population variances for each treatment has been satisfied. Include MINITAB plots b. State the population characteristic to be tested. State Ho: Ha: d.Is there sufficient evidence that the mean miles per gallon are different among the three octane levels at the 5% significance level? Copy ANOVA results. e.If the null hypothesis from part (c) was rejected, use Tukey's test to determine which pairwise means differ. Include MINITAB outputs. f.Based on your results in part (e), make a mean bar graph, and state your conclusion.

One-Way ANOVA and Randomized Black Experiment (Two-Way ANOVA)

Answer 1

Solution:

We will use MINITAB software to answer this question,

a)

The normal probability plot of response Y is

So the data is normally distributed.

b) here we will test for Octane

so we will test for means of Octane at 3 different levels

c)

H0: to test all 3 levels of Octane are equal.

H1 : at least one level of Octane is different from others.

d) the MINITAB output of one way ANOVA is

here the p-value for the test is 0.721>0.05. Hence we can accept H0 and claim that there is no difference in means of 3 groups.

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