Chapter 10, Problem 126SE

a.

To determine

Obtain the standard error of the estimator $\widehat{\theta }$.

Answer to Problem 126SE

The standard error of the estimator $\widehat{\theta }$ is $\underset{\_}{S.E\left(\widehat{\theta }\right)=\sigma \sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}}$.

Explanation of Solution

Here, it is known that the random variables X, Y, and W follow the normal distribution with the means ${\mu }_1,{\mu }_2,{\mu }_3$ and the common variances ${\sigma }_1^2={\sigma }_2^2={\sigma }_3^2=\sigma$, respectively.

In this context, it is supposed to estimate a linear function of the means $\theta =a_1{\mu }_1+a_2{\mu }_2+a_3{\mu }_3$; and also, MLE of a function of parameters is the function of MLEs of the parameters. Thus, MLE of $\theta$ is $\widehat{\theta }=a_1\overline{X}+a_2\overline{Y}+a_3\overline{W}$.

The variance of the estimator $\widehat{\theta }$ is given below:

$\begin{array}{c}\text{Var}\left(\widehat{\theta }\right)=\text{Var}\left(a_1\overline{X}+a_2\overline{Y}+a_3\overline{W}\right)\\ =a_1^2\text{Var}\overline{X}+a_2^2\text{Var}\overline{Y}+a_3^2\text{Var}\overline{W}\\ =a_1^2\frac{{\sigma }^2}{n_1}+a_2^2\frac{{\sigma }^2}{n_2}+a_3^2\frac{{\sigma }^2}{n_3}\\ ={\sigma }^2\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)\end{array}$

The standard error of the estimator $\widehat{\theta }$ is obtained below:

$\begin{array}{c}\text{S}\text{.E}\left(\widehat{\theta }\right)\text{=}\sqrt{\text{Var}\left(\widehat{\theta }\right)}\\ =\sqrt{{\sigma }^2\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}\\ =\sigma \sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}\end{array}$

Therefore, the standard error of the estimator $\widehat{\theta }$ is $\underset{\_}{S.E\left(\widehat{\theta }\right)=\sigma \sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}}$.

b.

To determine

Identify the distribution of the estimator $\widehat{\theta }$.

Explanation of Solution

From Part (a), it is known that the random variables X, Y, and W follow the normal distribution with the means ${\mu }_1,{\mu }_2,{\mu }_3$ and the common variances ${\sigma }_1^2={\sigma }_2^2={\sigma }_3^2=\sigma$. As all the means follow the normal distribution and are independent of each other, their linear combination also follows the normal distribution. Therefore, $\widehat{\theta }$ follows the normal distribution with the mean $\theta$ and standard deviation $S.E\left(\widehat{\theta }\right)=\sigma \sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}$.

c.

i.

To determine

Identify the distribution of $\frac{S_P^2\left(n_1+n_2+n_3-3\right)}{{\sigma }^2}$.

Explanation of Solution

From Part (b), it is known that all the random variables are mutually independent.

Here, it is known that $S_P^2=\frac{\left(n_1-1\right)S_1^2+\left(n_2-1\right)S_2^2+\left(n_3-1\right)S_3^2}{n_1+n_2+n_3-3}$.

On simplifying the pooled variance,

$\begin{array}{c}{S}_P^2=\frac{\left(n_1-1\right)S_1^2+\left(n_2-1\right)S_2^2+\left(n_3-1\right)S_3^2}{n_1+n_2+n_3-3}\\ \frac{\left(n_1+n_2+n_3-3\right)S_P^2}{{\sigma }^2}=\frac{\left(n_1-1\right)S_1^2+\left(n_2-1\right)S_2^2+\left(n_3-1\right)S_3^2}{{\sigma }^2}\end{array}$

It is clear that the right-hand side of the expression is the sum of 3 independent ${\chi }^2$ random variables. Similarly, the left-hand side also follows a ${\chi }^2$ distribution with the sum of degrees of freedom of 3 random variables.

Therefore, $\frac{\left(n_1+n_2+n_3-3\right)S_P^2}{{\sigma }^2}~{\chi }_{n_1+n_2+n_3-3}^2$.

ii.

To determine

Identify the distribution of $T=\frac{\widehat{\theta }-\theta }{S_P\sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}}$.

Explanation of Solution

Normally, it is known that $Z=\frac{\widehat{\theta }-\theta }{\sigma \sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}}~N\left(0,1\right)$.

On simplifying,

$\begin{array}{c}T=\frac{Z}{\sqrt{\frac{S_P^2}{{\sigma }^2}}}\\ =\frac{Z}{\sqrt{\frac{{\chi }_{n_1+n_2+n_3-3}^2}{n_1+n_2+n_3-3}}}\end{array}$

As the ratio of independent normal and square root of ${\chi }^2$ distribution follows t-distribution:

Therefore, $T~t_{n_1+n_2+n_3-3}$.

d.

To determine

Provide the confidence interval for $\theta$ with confidence coefficient $1-\alpha$.

Answer to Problem 126SE

The required confidence interval for $\theta$ with confidence coefficient $\left(1-\alpha \right)\%$ is $\underset{\_}{\left(\widehat{\theta }-t_{n_1+n_2+n_3-3,\frac{\alpha }{2}}\left[S_P\sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}\right],\widehat{\theta }+t_{n_1+n_2+n_3-3,\frac{\alpha }{2}}\left[S_P\sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}\right]\right)}$.

Explanation of Solution

The confidence interval for $\theta$ with confidence coefficient $1-\alpha$ is explained below:

From previous Part (c, ii), it is known that $T~t_{n_1+n_2+n_3-3}$.

$\begin{array}{l}P\left(-t_{n_1+n_2+n_3-3,\frac{\alpha }{2}}\le T\le t_{n_1+n_2+n_3-3,\frac{\alpha }{2}}\right)=1-\alpha \\ P\left(-t_{n_1+n_2+n_3-3,\frac{\alpha }{2}}\le \frac{\widehat{\theta }-\theta }{S_P\sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}}\le t_{n_1+n_2+n_3-3,\frac{\alpha }{2}}\right)=1-\alpha \\ P\left(-t_{n_1+n_2+n_3-3,\frac{\alpha }{2}}\left[S_P\sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}\right]\le \widehat{\theta }-\theta \le t_{n_1+n_2+n_3-3,\frac{\alpha }{2}}\left[S_P\sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}\right]\right)=1-\alpha \\ P\left(\widehat{\theta }-t_{n_1+n_2+n_3-3,\frac{\alpha }{2}}\left[S_P\sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}\right]\le \theta \le \widehat{\theta }+t_{n_1+n_2+n_3-3,\frac{\alpha }{2}}\left[S_P\sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}\right]\right)=1-\alpha \end{array}$

Therefore, the required confidence interval for $\theta$ with confidence coefficient $\left(1-\alpha \right)\%$ is

$\left(\widehat{\theta }-t_{n_1+n_2+n_3-3,\frac{\alpha }{2}}\left[S_P\sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}\right],\widehat{\theta }+t_{n_1+n_2+n_3-3,\frac{\alpha }{2}}\left[S_P\sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}\right]\right)$.

e.

To determine

Delineate a test for $H_0:\theta ={\theta }_0$ versus $H_a:\theta \ne {\theta }_0$.

Explanation of Solution

The test hypotheses are given below:

Null hypothesis: $H_0:\theta ={\theta }_0$

Alternative hypothesis: $H_a:\theta \ne {\theta }_0$

From previous Part (c, ii), it is known that $T~t_{n_1+n_2+n_3-3}$.

Similarly, $\frac{\widehat{\theta }-\theta }{S_P\sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}}~t_{n_1+n_2+n_3-3}$

The test statistics under $H_0$ is given below:

$T_0=\frac{\widehat{\theta }-{\theta }_0}{S_P\sqrt{\left(\frac{a_1^2}{n_1}+\frac{a_2^2}{n_2}+\frac{a_3^2}{n_3}\right)}}~t_{n_1+n_2+n_3-3}$

Decision rule:

If $\left|T_0\right|>t_{n_1+n_2+n_3-3,\frac{\alpha }{2}}$, then reject the null hypothesis $H_0$.