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:

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

$\begin{matrix} Sum= & 99645 & 99684 & 99687 & 99699 & 99592 \\ Average= & 996.45 & 996.84 & 996.87 & 996.99 & 995.92 \end{matrix}$

$\begin{matrix} \displaystyle\sum_{i}X_{ij}^2= \\ 99313355 & 99390488 & 99397147 & 99420471 & 99420471 \end{matrix}$

$\begin{matrix} \displaystyle St.Dev.= \\ 14.939 & 14.733 & 14.964 & 14.759 & 10.249 \end{matrix}$

$SS_i=(n_i-1)s_i^2$

$(100-1)14.939^2=22094.75$

$(100-1)14.733^2=21489.44$

$(100-1)14.964^2=22167.31$

$(100-1)14.759^2=21564.99$

$(100-1)10.249^2=10399.36$

$\begin{matrix} \displaystyle SS= \\ 22094.75 & 21489.44 & 22167.31 & 21564.99 & 10399.36 \end{matrix}$

$\begin{matrix} n= & 100 & 100 & 100 & 100 & 100 \\ \end{matrix}$

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

$=99645 +99684 + 99687 + 99699 + 99592=$

$=498307$

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

$=99313355+ 99390488 + 99397147+99420471 + 99420471=$

$=496717525$

$$SS_{total}=\displaystyle\sum_{i,j}X_{ij}^2-\dfrac{1}{N}(\displaystyle\sum_{i,j}X_{ij})^2=97792.502$$

$$SS_{within}=22094.75 + 21489.44 +22167.31+$$ $$+21564.99+ 10399.36=97715.85$$

$$SS_{between}=\displaystyle\sum_{j}n_j(\bar{x}_j-\bar{\bar{x}})^2=76.652$$ $$​$$ $$MS_{between}=\dfrac{SS_{between}}{df_{between}}=\dfrac{76.652}{4}=19.163$$ $$MS_{within}=\dfrac{SS_{within}}{df_{within}}=\dfrac{97715.85}{495}=197.406$$ $$F=\dfrac{MS_{between}}{MS_{within}}=\dfrac{19.163}{197.406}=0.097$$

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

$$F=\dfrac{MS_{between}}{MS_{within}}=\dfrac{19.163}{197.406}=0.097$$

Since it is observed that $F=0.097<2.39=F_c,$ 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 $p=0.9834,$ and since  $p=0.9834\geq0.05,$ 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.