Question

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

Drug A Drug B Drug C

4 9 8

7 11 6

6 12 7

3 8 6

4 10 5

3 11 7

1. Formulate the H0, and H1:

2. C.V. at alpha .01 level of significance

3. Compute the test value: ANOVA - F ratio

4. Decision

5. Interpretation

Answer 1

The total sample size is N=18.

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

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

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

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

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

$MS_{between}={SS_{between}\over df_{between}}={128.944-29.833\over 2}=49.556; MS_{within}={SS_{within}\over df_{within}}={29.833\over 15}=1.989$

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

The following null and alternative hypotheses need to be tested:

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

H₁: 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 df₁=2, df₂=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   α =0.01

  significance level.

Using the P-value approach: The p-value is  p=0, using table 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  α =0.01significance level.