Step 1

The data compared various types of batteries. The average life times of Duracell Alkaline AA batteries and Eveready Energizer Alkaline AA batteries were given.

 
Step 2

a.

Let $\bar X$ be the sample average lifetime of 100 Duracell batteries and $\bar Y$ be the sample average lifetimes of Eveready batteries.

The population average lifetimes of Duracell Alkaline AA batteries is 4.1 hours and the population average lifetimes of Eveready Energizer Alkaline AA batteries is 4.5 hours are given, that is:

\[\begin{array}{l} E\left( {\bar X} \right) = 4.1\\ E\left( {\bar Y} \right) = 4.5 \end{array}\]

The mean value of $\bar X - \bar Y$ is calculated as:

\[\begin{array}{c} E\left( {\bar X - \bar Y} \right) = E\left( {\bar X} \right) - E\left( {\bar Y} \right)\\ = 4.1 - 4.5\\ = - 0.4 \end{array}\]

Thus, the mean value of $\bar X - \bar Y$ is -0.4 hours.

The distribution of difference remains same when the sample size increases. Thus, the expected difference does not change with the change in the sample size.

 
Step 3

b.

The population standard deviation of life time for Duracell batteries is 1.8 hours and The population standard deviation of life time for Eveready batteries is 2.0 hours.

\[\begin{array}{l} {\sigma _1} = 1.8\\ {\sigma _2} = 2.0\\ m = 100\\ n = 100 \end{array}\]

The variance of $\bar X - \bar Y$ is calculated as:

\[\begin{array}{c} V\left( {\bar X - \bar Y} \right) = V\left( {\bar X} \right) + V\left( {\bar Y} \right)\\ = \frac{{\sigma _1^2}}{m} + \frac{{\sigma _2^2}}{n}\\ = \frac{{{{\left( {1.8} \right)}^2}}}{{100}} + \frac{{{{\left( {2.0} \right)}^2}}}{{100}}\\ = \frac{{3.24}}{{100}} + \frac{4}{{100}} \end{array}\] \[\begin{array}{l} = 0.0324 + 0.04\\ = 0.0724 \end{array}\]

The variance of $\bar X - \bar Y$ is 0.0724.

Its standard deviation is calculated as:

\[\begin{array}{c} S.D.\left( {\bar X - \bar Y} \right) = \sqrt {V\left( {\bar X - \bar Y} \right)} \\ = \sqrt {0.0724} \\ = 0.2691 \end{array}\]

Thus, the standard deviation is 0.2691.

 
Step 4

c.

The normal probability curve of $\bar X - \bar Y$ for the sample size $m = n = 100$ is given using $E\left( {\bar X - \bar Y} \right) = - 0.4$ and $S.D\left( {\bar X - \bar Y} \right) = 0.2691$ is:

The normal curve is symmetric about mean. The probability curve for the sample size $m = n = 100$ is symmetric about mean. The central limit theorem implies that $\bar X - \bar Y$ has an approximately normal distribution. For the sample size $m = n = 10$, the sample size is not sufficient to use central limit theorem. Thus, the shape of $\bar X - \bar Y$ for sample size $m = n = 10$ is not same as the shape of $\bar X - \bar Y$ for sample size $m = n = 100$.