Answer to Question #129509 in Statistics and Probability for danu

Question #129509
The life of a 60- watt light bulb in hours is known to be normally distributed with σ = 25
hours. Create 5 different random samples of 100 bulbs each which has a mean life of x_bar ~
1000 hours and perform one-way ANOVA with state it.
1
Expert's answer
2020-08-13T18:45:16-0400

The total sample size is N=500.N=500. Therefore, the total degrees of freedom are:


dftotal=5001=499df_{ total}=500−1=499

The between-groups degrees of freedom are dfbetween=51=4,df_{between}=5-1=4, and the within-groups degrees of freedom are:


dfwithin=dftotaldfbetween=4994=495df_{within}=df_{total}-df_{between}=499-4=495i,jXij=499712\displaystyle\sum_{i,j}X_{ij}=499712i,jXij2=499691630\displaystyle\sum_{i,j}X_{ij}^2=499691630SStotal=i,jXij21N(i,jXij)2=267464.112SS_{total}=\displaystyle\sum_{i,j}X_{ij}^2-\dfrac{1}{N}(\displaystyle\sum_{i,j}X_{ij})^2=267464.112SSwithin=266084.42SS_{within}=266084.42SSbetween=1379.692SS_{between}=1379.692 ​MSbetween=SSbetweendfbetween=1379.6924=344.923MS_{between}=\dfrac{SS_{between}}{df_{between}}=\dfrac{1379.692}{4}=344.923MSwithin=SSwithindfwithin=266084.42495=537.544MS_{within}=\dfrac{SS_{within}}{df_{within}}=\dfrac{266084.42}{495}=537.544F=MSbetweenMSwithin=344.923537.544=0.642F=\dfrac{MS_{between}}{MS_{within}}=\dfrac{344.923}{537.544}=0.642


The following null and alternative hypotheses need to be tested:

H0:μ1=μ2=μ3=μ4=μ5H_0: \mu_1=\mu_2=\mu_3=\mu_4=\mu_5

H1: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 α=0.05,\alpha=0.05, and the degrees of freedom are df1=4df_1=4 and df2=4,df_2=4, therefore, the rejection region for this F-test is R={F:F>Fc=2.39}.R=\{F:F>F_c=2.39\}.

Test Statistics


F=MSbetweenMSwithin=344.923537.544=0.642F=\dfrac{MS_{between}}{MS_{within}}=\dfrac{344.923}{537.544}=0.642

Since it is observed that F=0.642<2.39=Fc,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 α=0.05\alpha=0.05 significance level.


Using the P-value approach: The p-value is p=0.633,p=0.633, and since p=0.6330.05,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 α=0.05\alpha=0.05 significance level.



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