Question

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.

Answer 1

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:

$$df_{within}=df_{total}-df_{between}=499-4=495$$

$$\displaystyle\sum_{i,j}X_{ij}=499712$$

$$\displaystyle\sum_{i,j}X_{ij}^2=499691630$$

$$SS_{total}=\displaystyle\sum_{i,j}X_{ij}^2-\dfrac{1}{N}(\displaystyle\sum_{i,j}X_{ij})^2=267464.112$$

$$SS_{within}=266084.42$$

$$SS_{between}=1379.692 ​$$

$$MS_{between}=\dfrac{SS_{between}}{df_{between}}=\dfrac{1379.692}{4}=344.923$$

$$MS_{within}=\dfrac{SS_{within}}{df_{within}}=\dfrac{266084.42}{495}=537.544$$

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

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 $p=0.633\geq0.05,$

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.