Answer to Question #141487 in Statistics and Probability for ASM

Question

It is suspected that higher–priced automobiles are assembled with greater car than lower-priced automobiles. To investigate whether there is any basis for this feeling, large luxury model A, medium-sized Sedan B, and a subcompact hatchback C were compared for defects when they arrived at the dealers’ showroom. All cars were manufactured by the same company to test that the average number of defects is the same for the 3 model.

Answer 1

$$\begin{matrix} & Model \ A & Model \ B & Model\ C \\ & 4 & 5 & 8 \\ & 7 & 1 & 6 \\ & 6 & & 8 \\ Sum= & 17 & 6 & 22 \\ Average=& 5.6667 & 3 & 7.3333 \\ \sum_iX_{ij}^2= & 101 & 26 & 164 \\ St.Dev.= & 1.528 & 2.828 & 1.155 \\ SS= & 4.666667 & 8 & 2.666667 \\ n= & 3 & 2 & 3 \end{matrix}$$

The total sample size is $N=8.$ Therefore, the total degrees of freedom are:

$$df_{total}=8-1=7$$

Also, the between-groups degrees of freedom are $df_{between}=3-1=2,$ and the within-groups degrees of freedom are:

$$df_{within}=df_{total}-df_{between}=7-2=5$$

First, we need to compute the total sum of values and the grand mean. The following is obtained

$$\sum\limits_{i,j}X_{i,j}=17+6+22=45$$

Also, the sum of squared values is

$$\sum\limits_{i,j}X_{i,j}^2 =101+26+164=291$$

Based on the above calculations, the total sum of squares is computed as follows

$$SS_{total}=\sum\limits_{i,j}X_{i,j}^2-\dfrac{1}{N}\big(\sum\limits_{i,j}X_{i,j}\big)^2=$$

$$=291-\dfrac{45^2}{8}=37.875$$

The within sum of squares is computed as shown in the calculation below:

$$SS_{within}=\sum SS_{witingroups}=$$

$$=4.666667+8+2.666667=15.333$$

$$SS_{between}=SS_{total}-SS_{within}=$$

$$=37.875-15.333=22.542$$

$$MS_{between}=\dfrac{SS_{between}}{df_{between}}=\dfrac{22.542}{2}=11.271$$

$$MS_{within}=\dfrac{SS_{within}}{df_{within}}=\dfrac{15.333}{5}=3.067$$

The F-statistic is computed as follows:

$$F=\dfrac{MS_{between}}{MS_{within}}=\dfrac{11.271}{3.067}=3.675$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2=\mu_2$

$H_1:$ Not all means are equal

The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.

Based on the information provided, the significance level is $\alpha=0.05,$ and the degrees of freedom are $df_1=2, df_2=5,$ therefore, the rejection region for this

F-test is $R=\{F:F>F_c=5.786\}.$  

Test Statistics

$$F=\dfrac{MS_{between}}{MS_{within}}=\dfrac{11.271}{3.067}=3.675$$

Since it is observed that $F=3.675<5.786=F_c,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that not all 3 population means are equal, at the $\alpha=0.05$ significance level.

Using the P-value approach:

The p-value is $p=0.104294,$ and since $p=0.104294.0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that not all 3 population means are equal, at the $\alpha=0.05$ significance level.