The total sample size is N=500. Therefore, the total degrees of freedom are:
dftotal=500−1=499The between-groups degrees of freedom are dfbetween=5−1=4, and the within-groups degrees of freedom are:
dfwithin=dftotal−dfbetween=499−4=495i,j∑Xij=499712i,j∑Xij2=499691630SStotal=i,j∑Xij2−N1(i,j∑Xij)2=267464.112SSwithin=i∑(ni−1)si2=266084.42SSbetween=i∑ni(xˉ−xˉˉ)2=1379.692MSbetween=dfbetweenSSbetween=41379.692=344.923MSwithin=dfwithinSSwithin=495266084.42=537.544F=MSwithinMSbetween=537.544344.923=0.642The following null and alternative hypotheses need to be tested:
H0:μ1=μ2=μ3=μ4=μ5
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 α=0.05, and the degrees of freedom are df1=4 and df2=4, therefore, the rejection region for this F-test is R={F:F>Fc=2.39}.
Test Statistics
F=MSwithinMSbetween=537.544344.923=0.642Since it is observed that F=0.642<2.39=Fc, 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 α=0.05 significance level.
Using the P-value approach: The p-value is p=0.633, and since p=0.633≥0.05,
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 α=0.05 significance level.
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