(c) Given: n=30 and Q=(30widehat( sigma )^(2))/( sigma ^(2)) chi _(30)^(2) where widehat( sigma )^(2) is the maximum likelihood estimate of sigma ^(2) found in Question 1 (d) when n=30. Find a and b which satisfy 0.025=P(Q=b) and hence write down a 95% confidence interval for sigma ^(2). (d) How can the 95% confidence interval i

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(c) Given: n=30 and Q=(30widehat( sigma )^(2))/( sigma ^(2)) chi _(30)^(2) where widehat( sigma )^(2) is the maximum likelihood estimate of  sigma ^(2) found in Question 1 (d) when n=30. Find a and b which satisfy 0.025=P(Q<=a)=P(Q>=b) and hence write down a 95% confidence interval for  sigma ^(2). (d) How can the 95% confidence interval in part (c) above be used to test the hypotheses H_(0) :  sigma ^(2)=1 versus H_(1): sigma ^(2)!=1 ? (e) Evaluate your 95% confidence interval in part (b) at n=30 and compare it with that in part (c). Which is better? Justify you answer (4)

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Solution for - (c) Given: n=30 and Q=(30widehat( sigma )^(2))/( sigma ^(2)) chi _(30)^(2) where widehat( sigma )

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(Solved): (c) Given: n=30 and Q=(30widehat(\sigma )^(2))/(\sigma ^(2))\chi _(30)^(2) where widehat(\sigma ) ...

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