Question

These reaction times (in tenths of a second) were recorded for group of subjects after each had been given a drug pain.



Drug A
4
7
6
3
4
3

Drug B
9
11
12
8
10
11

Drug C
8
6
7
6
5
7

Formulate the H0, and H1:
C.V. at alpha .01 level of significance
Compute the test value: ANOVA – F ratio
Decision
Interpretation

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c} & Drug\ A & Drug\ B & Drug\ C \\ \hline & 4 & 9 & 8 \\ & 7 & 11 & 6 \\ & 6 & 12 & 7 \\ & 3 & 8 & 6 \\ & 4 & 10 & 5 \\ & 3 & 11 & 7 \\ & & & \\ Sum= & 27 & 61 & 39 \\ Average= & 4.5 & 10.167 & 6.5 \\ \sum_ix_{ij}^2= & 135 & 631 & 259 \\ St.Dev= & 1.643 & 1.472 & 1.049 \\ SS= & 13.5 & 10.833 & 5.5 \\ n= & 6 & 6 & 6 \end{array}$$

$$\bar{x}={1\over n}\sum_ix_i$$ $$s^2={1\over n-1}\sum_i(x_i-\bar{x})^2$$ $$\bar{x}_A=4.5, s_A^2=2.7,n_A=6$$

$$\bar{x}_B={61\over 6}, s_B^2={13\over 6},n_B=6$$

$$\bar{x}_C=6.5, s_C^2=1.1,n_C=6$$

The total sample size is $N=18.$ Therefore, the total degrees of freedom are:

$$df_{total}=18-1=17$$

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

$$df_{within}=df_{total}-df_{between}=17-2=15$$

Compute the total sum of values and the grand mean. The following is obtained

$$\displaystyle\sum_{i,j}x_{ij}=27+61+39=127$$

The sum of squared values is

$$\displaystyle\sum_{i,j}x_{ij}^2 =135+631+259=1025$$

The total sum of squares is computed as follows

$$SS_{total}=\displaystyle\sum_{i,j}x_{ij}^2-{1\over N}(\displaystyle\sum_{i,j}x_{ij})^2=1025-{127^2\over 18}=128.944$$

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

$$SS_{within}=\displaystyle\sum SS_{within\ groups}=$$ $$=13.5+10.833+5.5=29.833$$

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:

$$MS_{between}={SS_{between}\over df_{between}}={128.944-29.833\over 2}=49.556$$

$$MS_{within}={SS_{within}\over df_{within}}={29.833\over 15}=1.989$$

The F-statistic is computed as follows:

$$F=\dfrac{MS_{between}}{MS_{within}}=\dfrac{49.556}{1.989}=24.916$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2=\mu_3$

$H_1:$ 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 $\alpha=0.01,$ and the degrees of freedom are $df_1=2, df_2=2,$ therefore, the rejection region for this F-test is $R=\{F: F>F_C=6.359\}$

Since it is observed that $F=24.916>6.359=F_C,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that not all 3 population means are equal, at the $\alpha=0.01$  significance level.

Using the P-value approach: The p-value is $p=0,$ and since $p=0<0.01,$ it is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that not all 3 population means are equal, at the $\alpha=0.01$ significance level.