Question

Answer 1

Let the average tread lives be μ , μ3 and μ4 , respectively for four types of tires of trucks. Now the null and alternative statements of hypothesis are as below H At least one u, is different from others

The below set of R commands is used to read the data: > TL-=r e a d . table (file. choose ( ) ,header=T) attach (TL1) Further, to conduct the ANOVA Ftest, follow the below R commands > fit-aov (TL1svalueslas.factor (TL1sind); > anova (fit) anova (fit)

Thus, the ANOVA F test output obtained from the above R commands is shown below Analysis of Variance Table Response: TL1svalues as.factor (TL1Sind) 3 5.3178 1.77261 2.0515 0.1334 Df Sum Sq Mean Sq F value Pr (>F) Residuals 24 20.7374 0.86406 In the above output, the p-value (Pr) obtained is equal to 0.1334. The value of p is clearly greater than the given 0.1 level of significance. Therefore, the null hypothesis is not rejected

For making the procedure of test valid, some assumptions are required to be made. The assumptions are that the samples are independent and are taken from normal populations Another assumption is that the variances of population are equal

(1) According to the information given, the contrast here is given below Here, the average tread lives be μ.μ2., and μ4, respectively for four types of tires of trucks.

(ii) For making comparison in two brands, following are the null and alternative statements of hypothesis

The R commands are used further as written below: sm-by (values, ind, mean) var-by (values,ind,var): > t-(sm[1+m[21-sm[31-sm[1)/2; >st-sgrt (mean (svar) (1/4 2/7 2)) LL-t-qt (0.95,24) st; > UL-t+qt (0.95,24) st;

(iii) The outcome obtained from the specialized contrast test is clearly not in favor with the outcome that is obtained from the overall nuill hypothesis test in th is not equal to zero, then it means that the null hypothesis of part (a) will not be true. Also, the specialized contrast's t-test is way more powerful than the equality of all means' F test. e part (a). This is because if 6