Chapter 8.8, Problem 93E

a.

To determine

Construct the 95% confidence interval for $2{\mu }_1+{\mu }_2$ if ${\sigma }^2$ is known.

Answer to Problem 93E

The 95% confidence interval for $2{\mu }_1+{\mu }_2$ if ${\sigma }^2$ is known is $\left(2\overline{x}+\overline{y}-1.96\left(\sigma \sqrt{\left(\frac{4}{n}+\frac{3}{n}\right)}\right),\text{}2\overline{x}+\overline{y}+1.96\left(\sigma \sqrt{\left(\frac{4}{n}+\frac{3}{n}\right)}\right)\right)$.

Explanation of Solution

Here, ${\mu }_1$ is the mean of Population A and ${\mu }_2$ is the mean of Population B. The variance of two populations are ${\sigma }^2\text{ and 3}{\sigma }^2$. The two samples are independent.

The mean and variance of $2{\mu }_1+{\mu }_2$ are computed as follows:

$\begin{array}{c}E\left(\overline{x}\right)={\mu }_1\\ E\left(\overline{y}\right)={\mu }_2\\ E\left(2\overline{x}+\overline{y}\right)=2E\left(\overline{x}\right)+E\left(\overline{y}\right)\\ =2{\mu }_1+{\mu }_2\\ V\left(2\overline{x}+\overline{y}\right)=4V\left(\overline{x}\right)+V\left(\overline{y}\right)\\ =\frac{4{\sigma }^2}{n}+\frac{3{\sigma }^2}{n}\\ ={\sigma }^2\left(\frac{4}{n}+\frac{3}{n}\right)\end{array}$

From Table 4: Normal Curve Areas, the z value corresponding to 0.025$\left(=\frac{1-0.95}{2}\right)$ is 1.96. That is $z_{\frac{\alpha }{2}=0.025}=1.96$.

The 95% confidence interval for $2{\mu }_1+{\mu }_2$ if ${\sigma }^2$ is known is computed as follows:

$\begin{array}{c}0.95=P\left(-z_{\frac{\alpha }{2}}\le \frac{2\overline{x}+\overline{y}-\left(2{\mu }_1+{\mu }_2\right)}{\sqrt{{\sigma }^2\left(\frac{4}{n}+\frac{3}{n}\right)}}\le z_{\frac{\alpha }{2}}\right)\\ =P\left(-z_{\frac{\alpha }{2}}\left(\sqrt{{\sigma }^2\left(\frac{4}{n}+\frac{3}{n}\right)}\right)\le 2\overline{x}+\overline{y}-\left(2{\mu }_1+{\mu }_2\right)\le z_{\frac{\alpha }{2}}\left(\sqrt{{\sigma }^2\left(\frac{4}{n}+\frac{3}{n}\right)}\right)\right)\\ =P\left(\begin{array}{l}-\left(2\overline{x}+\overline{y}\right)-z_{\frac{\alpha }{2}}\left(\sqrt{{\sigma }^2\left(\frac{4}{n}+\frac{3}{n}\right)}\right)\le -\left(2{\mu }_1+{\mu }_2\right)\\ \le -\left(2\overline{x}+\overline{y}\right)+z_{\frac{\alpha }{2}}\left(\sqrt{{\sigma }^2\left(\frac{4}{n}+\frac{3}{n}\right)}\right)\end{array}\right)\\ =P\left(2\overline{x}+\overline{y}-z_{\frac{\alpha }{2}}\left(\sqrt{{\sigma }^2\left(\frac{4}{n}+\frac{3}{n}\right)}\right)\le \left(2{\mu }_1+{\mu }_2\right)\le 2\overline{x}+\overline{y}+z_{\frac{\alpha }{2}}\left(\sqrt{{\sigma }^2\left(\frac{4}{n}+\frac{3}{n}\right)}\right)\right)\\ =P\left(2\overline{x}+\overline{y}-1.96\left(\sigma \sqrt{\left(\frac{4}{n}+\frac{3}{n}\right)}\right)\le \left(2{\mu }_1+{\mu }_2\right)\le 2\overline{x}+\overline{y}+1.96\left(\sigma \sqrt{\left(\frac{4}{n}+\frac{3}{n}\right)}\right)\right)\end{array}$

Thus, the 95% confidence interval for $2{\mu }_1+{\mu }_2$ if ${\sigma }^2$ is known is $\left(2\overline{x}+\overline{y}-1.96\left(\sigma \sqrt{\left(\frac{4}{n}+\frac{3}{n}\right)}\right),\text{}2\overline{x}+\overline{y}+1.96\left(\sigma \sqrt{\left(\frac{4}{n}+\frac{3}{n}\right)}\right)\right)$.

b.

To determine

Construct the 95% confidence interval for $2{\mu }_1+{\mu }_2$ if ${\sigma }^2$ is not known.

Answer to Problem 93E

The 95% confidence interval for $2{\mu }_1+{\mu }_2$ if ${\sigma }^2$ is not known is $\left(2\overline{x}+\overline{y}-t_{\frac{\alpha }{2},n_1+n_2-2}s\sqrt{\left(\frac{4}{n_1}+\frac{3}{n_2}\right)},2\overline{x}+\overline{y}+t_{\frac{\alpha }{2},n_1+n_2-2}s\sqrt{\left(\frac{4}{n_1}+\frac{3}{n_2}\right)}\right)$.

Explanation of Solution

Here, ${\mu }_1$ is the mean of Population A and ${\mu }_2$ is the mean of Population B. The variance of two populations are ${\sigma }^2\text{ and 3}{\sigma }^2$. The two samples are independent.

As ${\sigma }^2$ is not known, the pooled sample variance is calculated as shown below:

$\begin{array}{c}{s}^2=\frac{{\widehat{\sigma }}_1^2+{\widehat{\sigma }}_2^2}{n_1+n_2-1}\\ =\frac{\left(n_1-1\right)s_1^2+\frac{\left(n_2-1\right)}{3}{s}_1^2}{n_1+n_2-1}\\ \text{where, }{s}_1^2=\frac{1}{n_1-1}{\displaystyle \sum _{i=1}^{n_1}{\left(x_i-\overline{x}\right)}^2}\\ s_2^2=\frac{1}{n_2-1}{\displaystyle \sum _{i=1}^{n_1}{\left(y_i-\overline{y}\right)}^2}\end{array}$

The 95% confidence interval for $2{\mu }_1+{\mu }_2$ if ${\sigma }^2$ is not known is computed as follows:

$\begin{array}{c}0.95=P\left(-t_{\frac{\alpha }{2},n_1+n_2-2}\le \frac{2\overline{x}+\overline{y}-\left(2{\mu }_1+{\mu }_2\right)}{\sqrt{s^2\left(\frac{4}{n_1}+\frac{3}{n_2}\right)}}\le t_{\frac{\alpha }{2},n_1+n_2-2}\right)\\ =P\left(-t_{\frac{\alpha }{2},n_1+n_2-2}\sqrt{s^2\left(\frac{4}{n_1}+\frac{3}{n_2}\right)}\le 2\overline{x}+\overline{y}-\left(2{\mu }_1+{\mu }_2\right)\le t_{\frac{\alpha }{2},n_1+n_2-2}\sqrt{s^2\left(\frac{4}{n_1}+\frac{3}{n_2}\right)}\right)\\ =P\left(\begin{array}{l}-\left(2\overline{x}+\overline{y}\right)-t_{\frac{\alpha }{2},n_1+n_2-2}\sqrt{s^2\left(\frac{4}{n_1}+\frac{3}{n_2}\right)}\le -\left(2{\mu }_1+{\mu }_2\right)\\ \le -\left(2\overline{x}+\overline{y}\right)+t_{\frac{\alpha }{2},n_1+n_2-2}\sqrt{s^2\left(\frac{4}{n_1}+\frac{3}{n_2}\right)}\end{array}\right)\\ =P\left(2\overline{x}+\overline{y}-t_{\frac{\alpha }{2},n_1+n_2-2}s\sqrt{\left(\frac{4}{n_1}+\frac{3}{n_2}\right)}\le \left(2{\mu }_1+{\mu }_2\right)\le 2\overline{x}+\overline{y}+t_{\frac{\alpha }{2},n_1+n_2-2}s\sqrt{\left(\frac{4}{n_1}+\frac{3}{n_2}\right)}\right)\end{array}$

Thus, the 95% confidence interval for $2{\mu }_1+{\mu }_2$ if ${\sigma }^2$ is not known is $\left(2\overline{x}+\overline{y}-t_{\frac{\alpha }{2},n_1+n_2-2}s\sqrt{\left(\frac{4}{n_1}+\frac{3}{n_2}\right)},2\overline{x}+\overline{y}+t_{\frac{\alpha }{2},n_1+n_2-2}s\sqrt{\left(\frac{4}{n_1}+\frac{3}{n_2}\right)}\right)$.