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UNSW MATH2089 2013 Q4c.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \usepackage{cancel} \begin{document} {\large UNSW MATH2089 2013 Q4c} \begin{align*} \text{Q4c}\quad&\text{An engineer was given a project to compare the abrasive wear of two difference laminated materials, $X$ and $Y$:}\\ &\text{Twelve pieces of randomly selected material $X$ were tested, by exposing each piece to a machine measuring wear.}\\ &\text{Similarly ten randomly selected pieces of material $Y$ were tested. The following summary gives the average wear}\\ &\text{and the standard deviation of each material:}\\ \\ &\text{For material $X$:}\quad\bar{x}=85;\quad s_x=4;\quad n_x=12.\\ &\text{For material $Y$:}\quad\bar{y}=82;\quad s_y=5;\quad n_y=10.\\ \\ \text{i) }&\text{What assumption(s) does the engineer need to make in order to construct a valid confidence interval?}\\ &\text{1. $X$ and $Y$ are independent.}\\ &\text{2. $X$ and $Y$ are normally distributed.}\\ &\text{3. The samples are random.}\\ &\text{4. The sample sizes are large enough.}\\ \\ \text{ii) }&\text{Assuming that $X$ and $Y$ have the same population variance, find the pooled estimate of this variance.}\\ &s_p^2 =\frac{(n_x-1)s_x^2+(n_y-1)s_y^2}{n_x+n_y-2} =\frac{(12-1)4^2+(10-1)5^2}{12+10-2}=20.05\quad\ldots\text{variance of the population.}\\ \\ \text{iii) }&\text{Find a 99\% confidence interval for the difference of the average wear of materials $X$ and $Y$.}\\ &\text{Degree of freedom }\nu=n_x+n_y-2=12+10-2=20. \quad t_{20,0.99}=2.845.\\ &t_{20,0.99}\cdot s_p\sqrt{\frac{1}{n_x}+\frac{1}{n_y}} =2.845\times\sqrt{20.05}\times\sqrt{\frac{1}{12}+\frac{1}{10}}=5.4546\\ &\bar{x}-\bar{y}\in\left[-t_{20,0.99}\cdot s_p\sqrt{\frac{1}{n_x}+\frac{1}{n_y}},~~t_{20,0.99}\cdot s_p\sqrt{\frac{1}{n_x}+\frac{1}{n_y}}~\right] =[-5.4546,5,4546].\\ \\ \\ &\text{By the way }\ldots\quad t_0=\frac{\bar{x}-\bar{y}}{s_p\sqrt{\frac{1}{n_x}+\frac{1}{n_y}}} =\frac{3}{\sqrt{20.05}\times\sqrt{\frac{1}{12}+\frac{1}{10}}}=1.5647. \quad|t_0|\ngtr t_{20,0.99}\ldots\text{accept.}\\ \end{align*} \end{document}