Chapter 7.1, Problem 8E

Let α₁ > 0, α₂ > 0, with α₁ + α₂ = α. Then

$P\left(-z_{{\alpha }_1}\text{ < }\frac{\bar{X}\text{ - }\mu }{\sigma /\sqrt{n}}\text{ < }{z}_{{\alpha }_2}\right)\text{ = 1 - }\alpha$

1. a. Use this equation to derive a more general expression for a 100(1 - α)% CI for μ of which the interval (7.5) is a special case.
2. b. Let α = .05 and α₁ = α /4, α₂ = 3α/4. Does this result in a narrower or wider interval than the interval (7.5)?

To determine

Derive the general expression for $100\left(1-\alpha \right)\%$ confidence interval for $\mu$ in the given situation.

Answer to Problem 8E

The general expression for $100\left(1-\alpha \right)\%$ confidence interval for $\mu$ in the given situation is $\underset{\_}{\left(\overline{x}-z_{{\alpha }_2}\times \frac{\sigma }{\sqrt{n}},\overline{x}+z_{\alpha 1}\times \frac{\sigma }{\sqrt{n}}\right)}$.

Explanation of Solution

Given info:

For ${\alpha }_1>0$, ${\alpha }_2>0$ with ${\alpha }_1+{\alpha }_2=\alpha$ then $P\left(-z_{{\alpha }_1}<\frac{\overline{X}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}<z_{{\alpha }_2}\right)=1-\alpha$.

Calculation:

The usual $100\left(1-\alpha \right)\%$ confidence interval for population mean $\mu$ is,

$CI=\left(\overline{x}-z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}},\overline{x}+z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}\right)$.

Here, it is given that with probability $\left(1-\alpha \right)$, $-z_{{\alpha }_1}<\frac{\overline{X}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}<z_{{\alpha }_2}$.

General expression for confidence interval is obtained as follows:

$\begin{array}{c}P\left(-z_{{\alpha }_1}<\frac{\overline{X}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}<z_{{\alpha }_2}\right)=1-\alpha \\ P\left(-z_{{\alpha }_1}\times \left(\frac{\sigma }{\sqrt{n}}\right)\le \overline{X}-\mu \le z_{{\alpha }_2}\times \left(\frac{\sigma }{\sqrt{n}}\right)\right)=1-\alpha \\ P\left(-\overline{X}-z_{{\alpha }_1}\times \left(\frac{\sigma }{\sqrt{n}}\right)\le -\mu \le -\overline{X}+z_{{\alpha }_2}\times \left(\frac{\sigma }{\sqrt{n}}\right)\right)=1-\alpha \\ P\left(\overline{X}+z_{{\alpha }_1}\times \left(\frac{\sigma }{\sqrt{n}}\right)\ge \mu \ge \overline{X}-z_{{\alpha }_2}\times \left(\frac{\sigma }{\sqrt{n}}\right)\right)=1-\alpha \end{array}$

$P\left(\overline{X}-z_{{\alpha }_2}\times \left(\frac{\sigma }{\sqrt{n}}\right)\le \mu \le \overline{X}+z_{{\alpha }_1}\times \left(\frac{\sigma }{\sqrt{n}}\right)\right)=1-\alpha$

From the obtained result, there is evidence to infer that with probability $\left(1-\alpha \right)$, $\overline{X}-z_{{\alpha }_2}\times \left(\frac{\sigma }{\sqrt{n}}\right)\le \mu \le \overline{X}+z_{{\alpha }_1}\times \left(\frac{\sigma }{\sqrt{n}}\right)$

Therefore, the $100\left(1-\alpha \right)\%$ confidence interval for population mean $\mu$ in the given situation is $\underset{\_}{\left(\overline{x}-z_{{\alpha }_2}\times \frac{\sigma }{\sqrt{n}},\overline{x}+z_{\alpha 1}\times \frac{\sigma }{\sqrt{n}}\right)}$.

To determine

Find the width of the usual confidence interval about mean for $\alpha =0.5$.

Find the width of the confidence interval obtained in part (a) for ${\alpha }_1=\frac{\alpha }{4}$ and ${\alpha }_2=\frac{3\alpha }{4}$.

Compare the two confidence intervals.

Answer to Problem 8E

The width of the usual confidence interval about mean for $\alpha =0.05$ is $\underset{\_}{{\text{Width}}_1=3.92\times \frac{\sigma }{\sqrt{n}}}$.

The width of the confidence interval obtained in part (a) for ${\alpha }_1=\frac{\alpha }{4}$ and ${\alpha }_2=\frac{3\alpha }{4}$ is $\underset{\_}{{\text{Width}}_2=4.02\times \frac{\sigma }{\sqrt{n}}}$.

The confidence interval obtained in part (a) for ${\alpha }_1=\frac{\alpha }{4}$ and ${\alpha }_2=\frac{3\alpha }{4}$ is wider than the usual confidence interval about mean for $\alpha =0.05$.

Explanation of Solution

Calculation:

Width of usual confidence interval for $\alpha =0.05$:

Critical value:

The level of significance is $\alpha =0.05$

From Table A.3 of the standard normal distribution in Appendix , the critical value corresponding to 0.05 area to the right is 1.96.

Thus, the critical value is $\left(z_{\alpha }\right)=1.96$.

The usual $100\left(1-\alpha \right)\%$ confidence interval for population mean $\mu$ is,

$CI=\left(\overline{x}-z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}},\overline{x}+z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}\right)$.

The general formula to obtain width of the interval is,

$\begin{array}{c}{\text{Width}}_1=2\times z_{\alpha }\times \frac{\sigma }{\sqrt{n}}\\ =2\times 1.96\times \frac{\sigma }{\sqrt{n}}\\ =3.92\times \frac{\sigma }{\sqrt{n}}\end{array}$.

Thus, the width of the usual confidence interval about mean for $\alpha =0.05$ is $\underset{\_}{{\text{Width}}_1=3.92\times \frac{\sigma }{\sqrt{n}}}$.

Width of usual confidence interval obtained in part (a) for ${\alpha }_1=\frac{\alpha }{4}$and ${\alpha }_2=\frac{3\alpha }{4}$:

Critical value:

For ${\alpha }_1=\frac{\alpha }{4}$:

The level of significance is ${\alpha }_1=\frac{\alpha }{4}=\frac{0.05}{4}=0.0125$.

From Table A.3 of the standard normal distribution in Appendix , the critical value corresponding to 0.0125 area to the right is 2.24.

Thus, the critical value is $\left(z_{{\alpha }_1}\right)=2.24$.

For ${\alpha }_2=\frac{3\alpha }{4}$:

The level of significance is ${\alpha }_2=\frac{3\alpha }{4}=\frac{3\times 0.05}{4}=0.0375$.

From Table A.3 of the standard normal distribution in Appendix , the critical value corresponding to 0.0375 area to the right is 1.78.

Thus, the critical value is $\left(z_{{\alpha }_2}\right)=1.78$.

The $100\left(1-\alpha \right)\%$ confidence interval for population mean $\mu$ obtained in part (a) is $\left(\overline{x}-z_{{\alpha }_2}\times \frac{\sigma }{\sqrt{n}},\overline{x}+z_{\alpha 1}\times \frac{\sigma }{\sqrt{n}}\right)$.

The general formula to obtain width of the interval is,

$\begin{array}{c}{\text{Width}}_2=\left(z_{{\alpha }_1}+z_{{\alpha }_2}\right)\times \frac{\sigma }{\sqrt{n}}\\ =\left(2.24+1.78\right)\times \frac{\sigma }{\sqrt{n}}\\ =4.02\times \frac{\sigma }{\sqrt{n}}\end{array}$.

Thus, the width of the usual confidence interval for population mean $\mu$ obtained in part (a) is $\underset{\_}{{\text{Width}}_2=4.02\times \frac{\sigma }{\sqrt{n}}}$.

Comparison:

The width of the usual confidence interval about mean for $\alpha =0.05$ is $\underset{\_}{{\text{Width}}_1=3.92\times \frac{\sigma }{\sqrt{n}}}$.

The width of the confidence interval obtained in part (a) for ${\alpha }_1=\frac{\alpha }{4}$ and ${\alpha }_2=\frac{3\alpha }{4}$ is $\underset{\_}{{\text{Width}}_2=4.02\times \frac{\sigma }{\sqrt{n}}}$.

By observing the two widths it can be concluded that, confidence interval obtained in part (a) for ${\alpha }_1=\frac{\alpha }{4}$ and ${\alpha }_2=\frac{3\alpha }{4}$ is wider than the usual confidence interval about mean for $\alpha =0.05$.