# Answer in Statistics and Probability for Gag #249439

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.

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

“df_{total}=500-1=499”

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

“df_{within}=df_{total}-df_{between}=499-4=495”“displaystylesum_{i,j}X_{ij}=499712”“displaystylesum_{i,j}X_{ij}^2=499691630”“SS_{total}=displaystylesum_{i,j}X_{ij}^2-dfrac{1}{N}(displaystylesum_{i,j}X_{ij})^2=267464.112”“SS_{within}=266084.42”“SS_{between}=1379.692nu200b”“MS_{between}=dfrac{SS_{between}}{df_{between}}=dfrac{1379.692}{4}=344.923”“MS_{within}=dfrac{SS_{within}}{df_{within}}=dfrac{266084.42}{495}=537.544”“F=dfrac{MS_{between}}{MS_{within}}=dfrac{344.923}{537.544}=0.642”

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

“F=dfrac{MS_{between}}{MS_{within}}=dfrac{344.923}{537.544}=0.642”

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=0.633geq0.05,”

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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