The total sample size is "N=500." Therefore, the total degrees of freedom are:
The between-groups degrees of freedom are "df_{between}=5-1=4," and the within-groups degrees of freedom are:
The following null and alternative hypotheses need to be tested:
"H_0: \\mu_1=\\mu_2=\\mu_3=\\mu_4=\\mu_5"
"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=4" and "df_2=4," therefore, the rejection region for this F-test is "R=\\{F:F>F_c=2.39\\}."
Test Statistics
Since it is observed that "F=0.642<2.39=F_c," it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that not all 5 population means are equal, at the "\\alpha=0.05" significance level.
Using the P-value approach: The p-value is "p=0.633," and since "p=0.633\\geq0.05," it is concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that not all 5 population means are equal, at the "\\alpha=0.05" significance level.
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