Question

What type of experimental design is employed in this problem? 1. 3. Use a multiple boxplot in order to compare responses B,C, D) are to bo oompared for read wear aner 24,000 mles of demng Anexpennent 4. State clearly the hypothesis that we are testing in this probiem 5. Run ANOVA on the data and ganerate the output with plots f there is eny significent difference fferent mean? (Hint use the boxplots from part ia) of residuals to comment on the normality of eeror tem (residuab) in this ANOVA model 2. After running the model in Miritab: L. Interpret the estimate of 50 in the words of the probilem Do the same for 81. 00-1.49 in the original pulp( check the test, you must compare th* shape to ฮ normal distribution pat. 4. Generate the fitted line plot for this regression s. Generate prediction interval for y" and confidence interval for (v) when( average of to run the regression model again

can you answer step by step please

Answer 1

A2

1) Anova one way
2) dependent variable - tread wear
treatments - different brands

3)

4 3 2 寸 cd

there is difference in more than 2 levels

4)
Ho : muA = muB= muC= muD
Ha: at least one mean is different

Group 1 Group 2 15 14 14 13 Group 3 Group 4 12 13 12 10 12 56 14 786 0.816 2 4 48 12 578 0.816 36 Sum Average 10.25 326 0.816 2 427 St. Dev. - 6.75 4 4

The total sample size is N 16. Therefore, the total degrees of freedom are: d ftotal 16-1 15 degrees of freedom are dAtween = 4-1 = 3, and the within-groups degrees of freedom are dfwithindtotal dfbtween 15- 3 12 First, we need to compute the total sum of values and the grand mean. The following is obtained Xij 48 +5636+ 41 181 Also, the sum of squared values is Σ x2 5784 7864 3264 427-2117 Based on the above calculations, the total sum of squares is computed as follows Xy | = 2117 X total =

The within sum of squares is computed as shown in the calculation below: S Swithin- S Swithingroups 2+2+2+ 6.75 - 12.75 The between sum of squares is computed directly as shown in the calculation below: The between sum of squares is computed directly as shown in the calculation below Now that sum of squares are computed, we can proceed with computing the mean sum of squares M Setueen56.688 dfbetween S Suithin 12.75 M Sd fwithin - 1.063 Now that sum of squares are computed, we can proceed with computing the mean sum of squares 56.68818.896 M Sbet tueend fbetueen M SvithinSSuithin12.75 dfuthin12 - 1.063 within - Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows MS between 17.784 M Swithin 1.063

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ha: Not all means are equal The above hypotheses will be tested using an F-ratio for a One-Way ANOVA. 2) Rejection Region Based on the information provided, the significance level is α-0.05, and the degrees of freedom are dfi 3 and df 3, therefore, the rejection region for this F-test is R- F:FFe 3.493. (3) Test Statistics een 18.896 MSuithin1.063 = 17.784 (4) Decision about the null hypothesis Since it is observed that F = 17.784 > Fe = 3.49, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0001, and since p = 0.0001 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 4 population means are equal, at the α-_ 0.05 significance level.