A consumer testing lab wants to compare the mean life of AA batteries produced by different manufacturers. Five brands of batteries are selected, and for each brand, 20 batteries are randomly sampled. The lab then tests for the lifetime of each battery (in hours) and compares the average battery life of different brands using ANOVA. Complete the following one-factor ANOVA summary table using $\alpha$= .05. Based on the results, do batteries of different brands have different lifetimes?

$\begin{array}{lccccc}
\hline \text { Source } & S S & d f & M S & F & \text { Critical Value and Decision } \
\hline \text { B etween } & & & & & \
\text { Within } & & & 10 & & \
\text { Total } & 1110 & & & & \
\hline
\end{array}$

Question

A consumer testing lab wants to compare the mean life of AA batteries produced by different manufacturers. Five brands of batteries are selected, and for each brand, 20 batteries are randomly sampled. The lab then tests for the lifetime of each battery (in hours) and compares the average battery life of different brands using ANOVA. Complete the following one-factor ANOVA summary table using

\alpha

= .05. Based on the results, do batteries of different brands have different lifetimes?

\begin{array}{lccccc}\hline \text { Source } & S S & d f & M S & F & \text { Critical Value and Decision } \\\hline \text { B etween } & & & & & \\\text { Within } & & & 10 & & \\\text { Total } & 1110 & & & & \\\hline\end{array}

$\begin{array}{lccccc}
\hline \text { Source } & S S & d f & M S & F & \text { Critical Value and Decision } \
\hline \text { B etween } & & & & & \
\text { Within } & & & 10 & & \
\text { Total } & 1110 & & & & \
\hline
\end{array}$

Quizplus · Accepted Answer

The Answer is There are 5 brands, so J = 5. Each brand has 20 batteries sampled, so n = 20. N = 5*20 = 100.

df_betw = J - 1 = 5 - 1 = 4, df_with = N - J = 100- 5 = 95, df_total = N - 1 = 100 - 1 = 99.

SS_with = MS_with*df_with = 10*95 = 950. SS_betw = SS_total - SS_with = 1100 - 950 = 160.

MS_betw = SS_betw/df_betw = 160/4 = 40.

F = MS_betw/MS_with = 40/10 = 4, critical value = _.05F₄_,₉₅ = 2.47.

Because F > critical F value, we reject H₀ and conclude that batteries of different brands have different lifetimes.

\begin{array}{lrrrcc}\hline ext { Source } & S S & {d f} & M S & F & ext { Critical Value and Decision } \\hline ext { B etween } & 160 & 4 & 40 & 4 & .05 F_{4,95}=2.47 \ ext { Within } & 950 & 95 & 10 & & ext { reject } H_{0} \ ext { Total } & 1110 & 99 & & & \\hline\end{array}

A Consumer Testing Lab Wants to Compare the Mean Life α\alphaα

A Consumer Testing Lab Wants to Compare the Mean Life $\alpha$