Question

Let X1,X2, variance ơ2. Suppose that n < 30 (and the CLT is not applicable). Use Chebyshev's inequality for X to construct a 90% CI for μ. Discuss a disadvantage of this method when it is additionally known that the population is normal. 9. , Xn be a random sample from a population with an unknown μ and known

Answer 1

$From Chebyshev's~inequality,\\ P(|\bar{X}-\mu|\leq t\sigma/\sqrt{n})\geq 1-\frac{1}{t^2}......(1)\\ According~to~the~problem,\\ 1-\frac{1}{t^2}=0.9\Rightarrow t^2=10~or~t=3.1623~since~t>0\\ Therefore,~90\%~C.I.~for~\mu~using~Chebychev's~inequality:\\ \left[\bar{X}-\frac{3.1623\sigma}{\sqrt{n}},~\bar{X}&plus;\frac{3.1623\sigma}{\sqrt{n}} \right ].\\ From~(1),~we~see~that~we~are~at~least~90\%~confident~that~the~\mu~is\\ lies~inside~\left[\bar{X}-\frac{3.1623\sigma}{\sqrt{n}},\bar{X}&plus;\frac{3.1623\sigma}{\sqrt{n}} \right ].~However~if~we~use~normal~distribution,\\ then~we~are~more~than~99\%~confident~that~the~\mu~is~ lies~inside~\left[\bar{X}-3.1623\sigma,\bar{X}&plus;3.1623\sigma \right ]\\ Since~P\left(-3.1623\leq\frac{\sqrt{n}(\bar{X}-\mu)}{\sigma}\leq 3.1623\right)=2\Phi(3.1623)-1=0.9984.~Hence~if~we~use~normal\\ ~distribution~then~we~have~more~confident~than~previous.$