Chapter 8, Problem 128SE

a.

To determine

Verify that $Z^{*}=\frac{\left({\overline{Y}}_1-{\overline{Y}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{{\sigma }_1\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}}$ has a standard normal distribution.

Explanation of Solution

Consider two independent random samples of size $n_1\text{ and }{n}_2$ from the normal distribution with parameters ${\mu }_1,{\sigma }_1^2$ and ${\mu }_2,{\sigma }_2^2$.

The mean and variance for ${\overline{Y}}_1$ and ${\overline{Y}}_2$ are obtained as follows:

$\begin{array}{c}E\left({\overline{Y}}_1\right)=\frac{1}{n_1}{\displaystyle \sum _{i=1}^{n_1}{Y}_{1i}}\\ ={\mu }_1\\ V\left({\overline{Y}}_1\right)=\frac{{\sigma }_1^2}{n_1}\\ E\left({\overline{Y}}_2\right)=\frac{1}{n_2}{\displaystyle \sum _{i=1}^{n_2}{Y}_{2i}}\\ ={\mu }_2\\ V\left({\overline{Y}}_2\right)=\frac{{\sigma }_2^2}{n_2}\end{array}$

The mean and variance for $Z^{*}$ is given as follows:

$\begin{array}{c}\text{Note that }E\left(Y_i-{\mu }_i\right)=0\\ E\left(Z^{*}\right)=\frac{E\left({\overline{Y}}_1-{\mu }_1\right)-E\left({\overline{Y}}_2-{\mu }_2\right)}{{\sigma }_1\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}}\\ =0\\ V\left(Z^{*}\right)=\frac{V\left[\left({\overline{Y}}_1-{\mu }_1\right)+\left({\overline{Y}}_2-{\mu }_2\right)\right]}{{\left({\sigma }_1\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}\right)}^2}\\ =\frac{V\left({\overline{Y}}_1\right)+V\left({\overline{Y}}_2\right)}{{\left({\sigma }_1\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}\right)}^2}\\ =\frac{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}{{\left({\sigma }_1\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}\right)}^2}\\ =\frac{\frac{{\sigma }_1^2}{n_1}+\frac{k{\sigma }_1^2}{n_2}}{{\sigma }_1^2\left(\frac{1}{n_1}+\frac{k}{n_2}\right)}\\ =1\end{array}$

Therefore, it is verified that $Z^{*}=\frac{\left({\overline{Y}}_1-{\overline{Y}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{{\sigma }_1\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}}$ has a standard normal distribution.

b.

To determine

Verify that $W^{*}=\frac{\left(n_1-1\right)S_1^2+\left(n_2-1\right)S_2^2/k}{{\sigma }_1^2}$ has a ${\chi }^2$ distribution with $n_1+n_2-2\text{df}$.

Explanation of Solution

Note that the statistic $\frac{\left(n-1\right)S^2}{{\sigma }^2}$ for the random samples (with n observations) from normal distribution with mean $\mu$ and variance ${\sigma }^2$ follows ${\chi }^2$ distribution with $n-1$ degrees of freedom.

Consider the following for two independent random samples of size $n_1\text{ and }{n}_2$ from the normal distribution with parameters ${\mu }_1,{\sigma }_1^2$ and ${\mu }_2,k{\sigma }_1^2\left(={\sigma }_2^2\right)$:

$\begin{array}{l}\frac{\left(n_1-1\right)S_1^2}{{\sigma }_1^2}~{\chi }_{n_1-1}^2\\ \frac{\left(n_1-1\right)S_2^2}{k{\sigma }_1^2}~{\chi }_{n_2-1}^2\\ W^{*}=\frac{\left(n_1-1\right)S_1^2+\frac{\left(n_2-1\right)S_2^2}{k}}{{\sigma }_1^2}\end{array}$

It is known that the above expression is a linear combination of square of independent variables from normal distribution. Thus, by definition of chi-square distribution, the sum of the two statistics has the ${\chi }^2$ distribution with df of  $n_1+n_2-2$.

Therefore, it is verified that $W^{*}=\frac{\left(n_1-1\right)S_1^2+\left(n_2-1\right)S_2^2/k}{{\sigma }_1^2}$ has a ${\chi }^2$ distribution with df of  $n_1+n_2-2$.

c.

To determine

Verify that $T*=\frac{\left({\overline{Y}}_1-{\overline{Y}}_2\right)+\left({\mu }_1-{\mu }_1\right)}{S_p^2\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}}$ has a $t$ distribution with $n_1+n_2-2\text{df}$.

Explanation of Solution

Note that the statistic $\frac{X}{\sqrt{\frac{Y}{k}}}$ follows $t_k(*)$.

Thus, consider the following:

$\frac{Z*}{\sqrt{\frac{W*}{n_1+n_2-2}}}~t_{n_1+n_2-2}$

It is solved as follows:

$\begin{array}{c}\frac{Z*}{\sqrt{\frac{W*}{n_1+n_2-2}}}=\frac{\frac{\left({\overline{Y}}_1-{\overline{Y}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{{\sigma }_1\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}}}{\sqrt{\frac{\frac{\left(n_1-1\right)S_1^2+\left(n_2-1\right)S_2^2/k}{{\sigma }_1^2}}{n_1+n_2-2}}}\\ =\frac{\frac{\left({\overline{Y}}_1-{\overline{Y}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{{\sigma }_1\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}}}{\frac{1}{{\sigma }_1^2}\sqrt{\frac{\left(n_1-1\right)S_1^2+\left(n_2-1\right)S_2^2/k}{n_1+n_2-2}}}\\ =\frac{\left({\overline{Y}}_1-{\overline{Y}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{S_p^{*}\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}}\end{array}$

The above expression follows Student t distribution with $n_1+n_2-2\text{df}$. Thus, it is verified that $T*=\frac{\left({\overline{Y}}_1-{\overline{Y}}_2\right)+\left({\mu }_1-{\mu }_1\right)}{S_p^2\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}}$ has a $t$ distribution with $n_1+n_2-2\text{df}$.

d.

To determine

Compute  $100\left(1-\alpha \right)\%$ confidence interval for ${\mu }_1-{\mu }_2$, assuming that ${\sigma }_2^2=k{\sigma }_1^2$.

Answer to Problem 128SE

The $100\left(1-\alpha \right)\%$ confidence interval for ${\mu }_1-{\mu }_2$ is $\left(\left({\overline{Y}}_1-{\overline{Y}}_2\right)-t_{n_1+n_2-2,\frac{\alpha }{2}}{S}_p^{*}\sqrt{\frac{1}{n_1}+\frac{k}{n_2}},\left({\overline{Y}}_1-{\overline{Y}}_2\right)-t_{n_1+n_2-2,\frac{\alpha }{2}}{S}_p^{*}\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}\right)$.

Explanation of Solution

The $100\left(1-\alpha \right)\%$ confidence interval for ${\mu }_1-{\mu }_2$, assuming that ${\sigma }_2^2=k{\sigma }_1^2$ is computed as follows:

$\begin{array}{c}1-\alpha =P\left(-t_{n_1+n_2-2,\frac{\alpha }{2}}\le \frac{{\overline{Y}}_1-{\overline{Y}}_2-\left({\mu }_1-{\mu }_2\right)}{S_p^{*}\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}}\le t_{n_1+n_2-2,\frac{\alpha }{2}}\right)\\ =P\left(-t_{n_1+n_2-2,\frac{\alpha }{2}}{S}_p^{*}\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}\le {\overline{Y}}_1-{\overline{Y}}_2-\left({\mu }_1-{\mu }_2\right)\le t_{n_1+n_2-2,\frac{\alpha }{2}}{S}_p^{*}\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}\right)\\ =P\left(-\left({\overline{Y}}_1-{\overline{Y}}_2\right)-t_{n_1+n_2-2,\frac{\alpha }{2}}{S}_p^{*}\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}\le -\left({\mu }_1-{\mu }_2\right)\le -\left({\overline{Y}}_1-{\overline{Y}}_2\right)-t_{n_1+n_2-2,\frac{\alpha }{2}}{S}_p^{*}\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}\right)\\ =P\left(\left({\overline{Y}}_1-{\overline{Y}}_2\right)-t_{n_1+n_2-2,\frac{\alpha }{2}}{S}_p^{*}\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}\le \left({\mu }_1-{\mu }_2\right)\le \left({\overline{Y}}_1-{\overline{Y}}_2\right)-t_{n_1+n_2-2,\frac{\alpha }{2}}{S}_p^{*}\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}\right)\end{array}$

Therefore, the $100\left(1-\alpha \right)\%$ confidence interval for ${\mu }_1-{\mu }_2$ is $\left(\left({\overline{Y}}_1-{\overline{Y}}_2\right)-t_{n_1+n_2-2,\frac{\alpha }{2}}{S}_p^{*}\sqrt{\frac{1}{n_1}+\frac{k}{n_2}},\left({\overline{Y}}_1-{\overline{Y}}_2\right)-t_{n_1+n_2-2,\frac{\alpha }{2}}{S}_p^{*}\sqrt{\frac{1}{n_1}+\frac{k}{n_2}}\right)$

e.

To determine

Explain the Parts (a)–(d), by substituting k=1.

Explanation of Solution

If k=1, Part (a) becomes $Z^{*}=\frac{\left({\overline{Y}}_1-{\overline{Y}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{{\sigma }_1\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$, which is still a standard normal distribution.

If k=1, Part (b) becomes $W^{*}=\frac{\left(n_1-1\right)S_1^2+\left(n_2-1\right)S_2^2}{{\sigma }_1^2}$. It still has a ${\chi }^2$ distribution with $n_1+n_2-2\text{df}$.

If k=1, Part (c) becomes $T*=\frac{\left({\overline{Y}}_1-{\overline{Y}}_2\right)+\left({\mu }_1-{\mu }_1\right)}{S_p^2\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$. It remains having $t$ distribution with $n_1+n_2-2\text{df}$.

If k=1, the $100\left(1-\alpha \right)\%$ confidence interval for ${\mu }_1-{\mu }_2$ in Part (d) becomes $\left(\left({\overline{Y}}_1-{\overline{Y}}_2\right)-t_{n_1+n_2-2,\frac{\alpha }{2}}{S}_p^{*}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}},\left({\overline{Y}}_1-{\overline{Y}}_2\right)-t_{n_1+n_2-2,\frac{\alpha }{2}}{S}_p^{*}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\right)$