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Chapter 9: Q1 E (page 371)

1. An article in the November \(1983\) Consumer Reports compared various types of batteries. The average lifetimes of Duracell Alkaline \(AA\)batteries and Eveready Energizer Alkaline \(AA\) batteries were given as \(4.1\) hours and \(4.5\) hours, respectively. Suppose these are the population average lifetimes.

a. Let \(\bar X\) be the sample average lifetime of \(100\) Duracell batteries and \(\bar Y\) be the sample average lifetime of \(100\) Eveready batteries. What is the mean value of \(\bar X - \bar Y\) (i.e., where is the distribution of \({\bf{\bar X - \bar Y}}\) centered)? How does your answer depend on the specified sample sizes?

b. Suppose the population standard deviations of lifetime are \(1.8\) hours for Duracell batteries and \(2.0\) hours for Eveready batteries. With the sample sizes given in part (a), what is the variance of the statistic \(\bar X - \bar Y\), and what is its standard deviation?

c. For the sample sizes given in part (a), draw a picture of the approximate distribution curve of \(\bar X - \bar Y\) (include a measurement scale on the horizontal axis). Would the shape of the curve necessarily be the same for sample sizes of \(10\) batteries of each type? Explain

Short Answer

Expert verified

\(\begin{array}{l}a. E(\bar X - \bar Y) = - 0.4; \\b. Variance = 0.0724; Standard deviation = 0.2691\end{array}\)

c.No

Step by step solution

01

To find the mean value

a.

The mean value is

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

02

To find the variance and standard deviation

b.

The variance of the statistic is

\(\begin{array}{c}V(\bar X - \bar Y) = V(\bar X) + {( - 1)^2}V(\bar Y)\\ = \frac{{\sigma _1^2}}{m} + \frac{{\sigma _2^2}}{n}\\ = \frac{{{{1.8}^2}}}{{100}} + \frac{{{2^2}}}{{100}}\\ = 0.0724\end{array}\)

The standard deviation of \(\bar X - \bar Y\) is

\(\sqrt {0.0724} = 0.2691\)

03

Drawing the graph

c.

Both random variables \(\bar X\) and \(\bar Y\) would be nearly normal because the sample sizes are large enough (both are 100), and so the\(\bar X - \bar Y\) would have an approximately normal distribution. When \(m = n = 10\), however, determining the distribution of \(\bar X - \bar Y\) becomes difficult. This is due to the fact that the Central Limit Theorem cannot be applied to such a large sample size.

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