Question

9.2. A textile fiber manufacturer is investigating a new drapery yam, which the company elaims has a mean thread elongation of 12 kilograms with a standard deviation of 0.5 kilograms. The company wishes to test the hypothesis Ho:驰012 against Hi: μ < 12, using random sample of four specimens (a) What is the type I error probability if the critical region is defined as <\ 1.5 kilograms? (b) Find β for the case where the true mean elongation is 11.25 kilograms

Answer 1

Solution : 9.2

( a )

$\mathrm{\alpha =P(reject\:H_{0}\:when\:H_{0}\:is\:true)}$

$\mathrm{=P(\overline{X}\leq 11.5\:\:when\:\:\mu =12)}$

$\mathrm{=P\left ( \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{11.5-12}{\frac{0.5}{\sqrt{4}}} \right )}$

$\mathrm{=P\left ( \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{-0.5}{\frac{0.5}{\sqrt{4}}} \right )}$

$\mathrm{=P\left ( \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}\leq -\frac{0.5}{\frac{0.5}{\sqrt{4}}} \right )}$

$\mathrm{=P\left ( \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}\leq -\frac{\sqrt{4}*0.5}{0.5} \right )}$

$\mathrm{=P\left ( \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}\leq -\sqrt{4}\right )}$

$\mathrm{=P\left ( \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}\leq -2\right )}$

$\mathrm{=P\left ( Z\leq -2\right )}$

$\mathrm{=1-P\left ( Z\leq 2\right )}$

$=0.02275$

$\mathrm{The \:probability\: of \:rejecting\: the\: null\: hypothesis \:when\: it \:is \:true\: is\:\: 0.02275}$

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( b )

$\mathrm{\beta =P(accept\:H_{0}\:when\:\mu =11.25)}$

$\mathrm{=P(\overline{X}> 11.5\:\:when\:\:\mu =11.25)}$

$\mathrm{=P\left ( \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}> \frac{11.5-11.25}{\frac{0.5}{\sqrt{4}}} \right )}$

$\mathrm{=P\left ( \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}> \frac{0.25}{\frac{0.5}{\sqrt{4}}} \right )}$

$\mathrm{=P\left ( \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}> \frac{\sqrt{4}*0.25}{0.5} \right )}$

$\mathrm{=P\left ( \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}> \frac{2*0.25}{0.5} \right )}$

$\mathrm{=P\left ( \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}> \frac{0.5}{0.5} \right )}$

$\mathrm{=P\left ( \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}> 1 \right )}$

$\mathrm{=P\left ( Z> 1 \right )}$

$\mathrm{=1-P\left ( Z\leq 1 \right )}$

$\mathrm{=1-0.84134}$

$=0.15866$

$\mathrm{The \:probability\: of \:accepting\: the\: null\: hypothesis \:when\: it \:is \:false\: is\:\: 0.15866}$

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Solution : 9.47

( b )

$\mathrm{n=51,\:\:s=0.37,\:\:\alpha =0.05}$

$\mathrm{For\:\:\alpha =0.05\:\:and\:\:n=51,\:\chi_{\frac{\alpha }{2},n-1}^{2} =\chi_{0.025,50}^{2}=71.42\:\:and\:\:\:\chi_{1-\frac{\alpha }{2},n-1}^{2} =\chi_{0.975,50}^{2}=32.36}$

$\frac{50(0.37)^{2}}{71.42}\leq \sigma ^{2}\leq \frac{50(0.37)^{2}}{32.36}$

$\frac{50*0.1369}{71.42}\leq \sigma ^{2}\leq \frac{50*0.1369}{32.36}$

$\frac{6.845}{71.42}\leq \sigma ^{2}\leq \frac{6.845}{32.36}$

$0.09584\leq \sigma ^{2}\leq 0.21153$

$0.096\leq \sigma ^{2}\leq 0.2115$

$\mathrm{Taking\:the\:square\:root\:of\:the\:endpoints\:of\:this\:interval\:we\:obtain,}$

$\sqrt{0.096}\leq \sqrt{\sigma ^{2}}\leq \sqrt{0.2115}$

$0.30984\leq \sigma\leq 0.45989$

$0.31\leq \sigma \leq0.46$

$\mathrm{Hence,\quad\:95\%\:confidence\:interval\:for\:\sigma \::}$

$0.31\leq \sigma \leq0.46$