Chapter 8, Problem 95SE

To determine

Obtain the interval for the case ${\alpha }_1={\alpha }_2=\alpha /2=0.025$.

Obtain the interval for the case ${\alpha }_1=0.01$ and ${\alpha }_2=0.04$.

Find the shorter interval and explain whether there is any advantage to a “symmetric” confidence interval.

Answer to Problem 95SE

The interval for ${\alpha }_1={\alpha }_2=\alpha /2=0.025$ is $\overline{x}-1.96\frac{\sigma }{\sqrt{n}}\le \mu \le \overline{x}+1.96\frac{\sigma }{\sqrt{n}}$.

The interval for ${\alpha }_1=0.01$ and ${\alpha }_2=0.04$ is $\overline{x}-2.33\frac{\sigma }{\sqrt{n}}\le \mu \le \overline{x}+1.75\frac{\sigma }{\sqrt{n}}$.

The significance level is not affected by the symmetrical interval.

Explanation of Solution

Given info:

The confidence interval for $\mu$ with known standard deviation $\sigma$ is

$\overline{x}-z_{{\alpha }_1}\frac{\sigma }{\sqrt{n}}\le \mu \le \overline{x}+z_{{\alpha }_2}\frac{\sigma }{\sqrt{n}}$ where ${\alpha }_1+{\alpha }_2=\alpha$. And consider $\alpha =0.05$.

Calculation:

For ${\alpha }_1={\alpha }_2=\alpha /2=0.025$ the interval is,

$\overline{x}-z_{{\alpha }_1}\frac{\sigma }{\sqrt{n}}\le \mu \le \overline{x}+z_{{\alpha }_2}\frac{\sigma }{\sqrt{n}}=\overline{x}-z_{0.025}\frac{\sigma }{\sqrt{n}}\le \mu \le \overline{x}+z_{0.025}\frac{\sigma }{\sqrt{n}}$

From “Table III: Cumulative standard normal distribution”, the required critical value is $z_{0.025}=1.96$.

$\overline{x}-z_{{\alpha }_1}\frac{\sigma }{\sqrt{n}}\le \mu \le \overline{x}+z_{{\alpha }_2}\frac{\sigma }{\sqrt{n}}=\overline{x}-1.96\frac{\sigma }{\sqrt{n}}\le \mu \le \overline{x}+1.96\frac{\sigma }{\sqrt{n}}$

That is,

$\Phi \left(1.96\right)=P\left(Z<1.96\right)$

From “Table III: Cumulative standard normal distribution”, the required critical value is

• Locate the value 1.9 in the first column of the table.
• Locate the value 0.06 in the first row of the table.
• The intersecting value of row and column is 0.975.

$\Phi \left(1.96\right)=0.975$

The confidence interval is 95%.

For ${\alpha }_1=0.01$ and ${\alpha }_2=0.04$ the interval is,

$\overline{x}-z_{{\alpha }_1}\frac{\sigma }{\sqrt{n}}\le \mu \le \overline{x}+z_{{\alpha }_2}\frac{\sigma }{\sqrt{n}}=\overline{x}-z_{0.01}\frac{\sigma }{\sqrt{n}}\le \mu \le \overline{x}+z_{0.04}\frac{\sigma }{\sqrt{n}}$

From “Table III: Cumulative standard normal distribution”, the required critical value is $z_{0.01}=2.36$ and $z_{0.04}=1.75$

That is,

$\Phi \left(2.36\right)=P\left(Z<2.36\right)$ and $\Phi \left(1.75\right)=P\left(z\le 1.75\right)$

Thus, $\overline{x}-z_{{\alpha }_1}\frac{\sigma }{\sqrt{n}}\le \mu \le \overline{x}+z_{{\alpha }_2}\frac{\sigma }{\sqrt{n}}=\overline{x}-2.33\frac{\sigma }{\sqrt{n}}\le \mu \le \overline{x}+1.75\frac{\sigma }{\sqrt{n}}$

Here $\alpha =0.05$ therefore the confidence interval remains same.

Hence the significance level is not affected by the symmetrical interval but it affect the width of the interval and the symmetrical interval is limited.