Question

The following observations are on time (h) for a AA 1.5volt alkaline battery to reach a \(0.8\)voltage ("Comparing the Lifetimes of Two Brands of Batteries," J. of Statistical Educ., 2013, online):

\(\begin{array}{*{20}{l}}{ Energizer: }&{8.65}&{8.74}&{8.91}&{8.72}&{8.85}\\{ Ultracell: }&{8.76}&{8.81}&{8.81}&{8.70}&{8.73}\\{ Energizer: }&{8.52}&{8.62}&{8.68}&{8.86}&{}\\{ Ultracell: }&{8.76}&{8.68}&{8.64}&{8.79}&{}\end{array}\)

Normal probability plots support the assumption that the population distributions are normal. Does the data suggest that the variance of the Energizer population distribution differs from that of the Ultra cell population distribution? Test the relevant hypotheses using a significance level of .05. (Note: The two-sample \(t\)test for equality of population means gives a \(P - \)value of .763.) The Energizer batteries are much more expensive than the Ultra cell batteries. Would you pay the extra money?

Short answer

There is sufficient evidence to support the claim that the variance of the Energizer population distribution differs from that of the Ultra cell population distribution. I would not pay the extra money, because the Energizer batteries contain also a lot more variability.

Step-by-step solution

Given information

The sample size

\(\begin{array}{l}{n_1} = {n_2} = 9\\\alpha = 0.05\end{array}\)

The mean is the sum of all values divided by the number of values:

\(\begin{array}{c}{{\bar x}_1} = \frac{{8.65 + 8.74 + 8.91 + \ldots + 8.62 + 8.68 + 8.86}}{9}\\ \approx 8.7278\end{array}\)

\(\begin{array}{c}{{\bar x}_2} = \frac{{8.76 + 8.81 + 8.81 + \ldots + 8.68 + 8.64 + 8.79}}{9}\\ \approx 8.7422\end{array}\)

The variance is the sum of squared deviations from the mean divided by \(n - 1.\)The standard deviation is the square root of the variance:

\(\begin{array}{c}{s_1} = \sqrt {\frac{{{{(8.65 - 8.7278)}^2} + \ldots . + {{(8.86 - 8.7278)}^2}}}{{9 - 1}}} \\ \approx 0.1270\end{array}\)

\(\begin{array}{c}{s_2} = \sqrt {\frac{{{{(8.76 - 8.7422)}^2} + \ldots . + {{(8.79 - 8.7422)}^2}}}{{9 - 1}}} \\ \approx 0.0595\end{array}\)

Writing hypothesis

Minimum of two steps are required.

Given claim: differs

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain an equality.

\(\begin{array}{l}{H_0}:\sigma _1^2 = \sigma _2^2\\{H_a}:\sigma _1^2 \ne \sigma _2^2\end{array}\)

Finding test statistic

Compute the value of the test statistic:

\(\begin{array}{c}F = \frac{{s_1^2}}{{s_2^2}}\\ = \frac{{{{0.1270}^2}}}{{{{0.0595}^2}}}\\ \approx 4.556\end{array}\)

Finding P value

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. The Pvalue is the number (or interval) in the column title of the F-distribution table containing the $F$-value in the column \(dfn = {n_1} - 1 = 9 - 1 = 8\)and in the row \(d{\rm{ }}f{\rm{ }}d = {n_2} - 1 = 9 - 1 = 8{\rm{ }}:\)

\(0.010 < P < 0.050\)

Decision rule and Conclusion

If the P-value is less than the significance level, then reject the null hypothesis.

\(P < 0.05 \Rightarrow {\rm{ Reject }}{H_0}\)

There is sufficient evidence to support the claim that the variance of the Energizer population distribution differs from that of the Ultra cell population distribution.

I would not pay the extra money, because the Energizer batteries contain also a lot more variability.