\documentclass[a4paper,12pt]{article} \newcommand\ds{\displaystyle} \begin{document} \parindent=0pt QUESTION It is known from experience that the mean breaking strength of a particular brand of fibre is 9.25N with a standard deviation of 1.20N, where N denotes Newtons. A sample of 36 fibres was recently selected at random from the production and found to have a mean breaking strength of 8.90N. \begin{description} \item[(i)] State the standard error of the sample mean. \item[(ii)] Construct a 95\% confidence interval for the mean breaking strength foe the fibre currently being produced, and determine how large a sample would have been required to obtain an interval with length less then 0.5N. \item[(iii)] Using a 5\% test of significance, can we conclude from the sample measurements that the strength of the fibre has changed? \end{description} \bigskip ANSWER We are given that $\mu=9.25N,\ \sigma=1.20N,\ n=36$ \begin{description} \item[(i)] Standard error of sample mean is $\ds\frac{\sigma}{\sqrt{n}}=\frac{1.20}{\sqrt{36}}=0.20N$ \item[(ii)]The confidence interval is double sided, so the interval is $\overline{x}\pm(1.96)(0.20)=8.90\pm0.392$ i.e. $8.508$ to $9.292$\\ The interval length is $\ds2(1.96)\frac{\sigma}{\sqrt{n}}=2(1.96)\frac{(1.2)}{\sqrt{n}}=\frac{4.704}{\sqrt{n}}$\\ Length$<$0.5 if $\ds\frac{4.704}{\sqrt{n}}<0.5,$ i.e. $\ds\sqrt{n}>\frac{4.704}{0.5}=9.408$ i.e. $n>(9.408)^2=88.5$ therefore sample size must be at least 89. \item[(iii)] Hypothesis $H_0:\mu=9.25N,\ H_1:\mu\neq9.25N$ Standard deviation $ =$ 1.2N\\ This is a two sided hypothesis\\ Test procedure: accept $H_0$ if $\ds\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}>-1.96$ or $<1.96$ for a 5\% significance test. Now $\ds\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{8.90-9.25}{0.20}=-\frac{0.35}{0.20}=-1.75$ This is inside the interval $(-1.96,1.96),$ hence sample results suggest that the strength of the fibre has not changed. \end{description} \end{document}