91Ó°ÊÓ

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:

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

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

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.

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

Compute the value of the test statistic:

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

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 \(d{\rm{ }}f{\rm{ }}d = {n_2} - 1 = 9 - 1 = 8{\rm{ }}:\)

\(0.010 < P < 0.050\)

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.