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An investigator wishes to estimate the difference between two population mean lifetimes of two different brands of batteries under specified conditions. If the population standard deviations are both roughly 2 hr and the sample size from the first brand will be twice the sample size from the second brand, what values of the sample sizes will be necessary to estimate the difference to within $0.5$ hours with $99 \%$ confidence?
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$\sigma_{1}=\sigma_{2}=2$

$n_{1}=2 n_{2}$

$C=99 \% \Rightarrow z^{*}=2.576$

two-sample means $c I$ for $u_{1}-\mu_{2}$

$\left(\bar{x}_{1}-\bar{x}_{2}\right) \pm \underbrace{z^{*} \sqrt{\frac{\sigma_{2}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}_{\text {margin of error }}$

$$\begin{array}{rl}2.576 \times \sqrt{\frac{2^{2}}{2 n_{2}}+\frac{2^{2}}{n_{2}}}=0.5 \\ 2.576 & x \sqrt{\frac{6}{n_{2}}}=0.5 \\ n_{2} & =6 \times\left(\begin{array}{c}2.576 \\ 0.5\end{array}\right)^{2} \\ & =159.2 \quad y=160\end{array}$$

$n_{1}=2 n_{2}=320$

$n_{1}=320, \quad n_{2}=160$
by Diamond (88,180 points)

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