Question

light bulb manufacturer wants to compare the mean lifetimes of two of its light bulbs, model A and model B. Independent random samples of the two models were taken. Analysis of 9 bulbs of model A showed a mean lifetime of 1234 hours and a standard deviation of 81 hours. Analysis of 15 bulbs of model B showed a mean lifetime of 1391 hours and a standard deviation of 110 hours. Assume that the populations of lifetimes for each model are normally distributed and that the variances of these populations are equal. Construct a 90 % confidence interval for the difference u-u, between the mean lifetime u, of model A bulbs and the mean lifetime u of model bulbs. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your responses to at least two decimal places. What is the lover limit of the 90% confidence interval? What is the upper limit of the 90% confidence interval? ? X

please answer neatly and correctly!

Answer 1

We know that the 90% confidence interval for difference of means is given by:

$[(\overline{x_{1}}-\overline{x_{2}})\pm t_{n_{1}+n_{2}-2}(0.10/2)\sqrt{\frac{s_{1}^2}{n_{1}}+\frac{s_{2}^2}{n_{2}}}]$

$[(1234-1391)\pm t_{9+15-2}(0.05)\sqrt{\frac{81^2}{9}+\frac{110^2}{15}}]$

$[(1234-1391)\pm t_{22}(0.05)\sqrt{\frac{81^2}{9}+\frac{110^2}{15}}]$

$[(1234-1391)\pm (1.717)\sqrt{\frac{81^2}{9}+\frac{110^2}{15}}]$

$[-157\pm 67.2851]$

$[-224.2851,-89.7149]$

Lower limit = -224.2851

Upper limit = -89.7149