The probability density function of the random variable \(X\) is
$$\mathrm { f } ( x ) = \frac { x ^ { k - 1 } \mathrm { e } ^ { - x / \phi } } { \phi ^ { k } ( k - 1 ) ! } , x > 0$$
where \(k\) is a known positive integer and \(\phi\) is an unknown parameter ( \(\phi > 0\) ). Show that the moment generating function (mgf) of \(X\) is
$$\mathrm { M } _ { X } ( \theta ) = ( 1 - \phi \theta ) ^ { - k }$$
for \(\theta < \frac { 1 } { \phi }\).
Write down the mgf of the random variable \(W = \sum _ { i = 1 } ^ { n } X _ { i }\) where \(X _ { 1 } , X _ { 2 } , \ldots , X _ { n }\) are independent random variables each with the same distribution as \(X\).
Write down the mgf of the random variable \(Y = \frac { 2 W } { \phi }\). Given that the mgf of the random variable \(V\) having the \(\chi _ { m } ^ { 2 }\) distribution is \(\mathrm { M } _ { V } ( \theta ) = ( 1 - 2 \theta ) ^ { - m / 2 }\) (for \(\theta < \frac { 1 } { 2 }\) ), deduce the distribution of \(Y\).
Deduce that \(\mathrm { P } \left( l < \frac { 2 W } { \phi } < u \right) = 0.95\) where \(l\) and \(u\) are the lower and upper \(2 \frac { 1 } { 2 } \%\) points of the \(\chi _ { 2 n k } ^ { 2 }\) distribution. Hence deduce that a \(95 \%\) confidence interval for \(\phi\) is given by \(\left( \frac { 2 w } { u } , \frac { 2 w } { l } \right)\) where \(w\) is an observation on the random variable \(W\).
For the case \(k = 2\) and \(n = 10\), use percentage points of the \(\chi ^ { 2 }\) distribution to write down, in terms of \(w\), an expression for a \(95 \%\) confidence interval for \(\phi\). By considering the \(\operatorname { mgf }\) of \(W\), find in terms of \(\phi\) the expected length of this interval.
Explain the meaning of the following terms in the context of hypothesis testing: Type I error, Type II error, operating characteristic, power.
A chemical manufacturer is endeavouring to reduce the amount of a certain impurity in one of its bulk products by improving the production process. The amount of impurity is measured in a convenient unit of concentration, and this is modelled by the Normally distributed random variable \(X\). In the old production process, the mean of \(X\), denoted by \(\mu\), was 63 and the standard deviation of \(X\) was 3.7. Experimental batches of the product are to be made using the new process, and it is desired to examine the hypotheses \(\mathrm { H } _ { 0 } : \mu = 63\) and \(\mathrm { H } _ { 1 } : \mu < 63\) for the new process. Investigation of the variability in the new process has established that the standard deviation may be assumed unchanged.
The usual Normal test based on \(\bar { X }\) is to be used, where \(\bar { X }\) is the mean of \(X\) over \(n\) experimental batches (regarded as a random sample), with a critical value \(c\) such that \(\mathrm { H } _ { 0 }\) is rejected if the value of \(\bar { X }\) is less than \(c\). The following criteria are set out.
If in fact \(\mu = 63\), the probability of concluding that \(\mu < 63\) must be only \(1 \%\).
If in fact \(\mu = 60\), the probability of concluding that \(\mu < 63\) must be \(90 \%\).
Find \(c\) and the smallest value of \(n\) that is required. With these values, what is the power of the test if in fact \(\mu = 58.5\) ?