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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}
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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}
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