Question

Refer to Exercise 6.94. For each part, a–d, form a 90% confidence interval for $\sigma$

Short answer

a.For 50 degrees of freedom, the 90% confidence interval for $\sigma$is $\left(2.1486,3.0043\right)$.

b.For 15 degrees of freedom, the 90% confidence interval for $\sigma$is $\left(0.0155,0.0292\right)$.

c.For 22 degrees of freedom, the 90% confidence interval for $\sigma$is $\left(25.3348,42.5334\right)$.

d. For 5 degrees of freedom, the 90% confidence interval for $\sigma$is $\left(0.9740,3.5585\right)$.

Step-by-step solution

Given information

For each part, values of the sample mean $\left(\overline{x}\right)$, sample standard deviation (s) and degrees of freedom (n) are given.

(a) Calculating the 90% confidence interval for 50 degrees of freedom

Given $\overline{x}=21,s=2.5,n=50$

The 90% confidence interval can be calculated using the formula,

$\sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{\chi }}_{\frac{\mathbf{\alpha }}{\mathbf{2}}}^{\mathbf{2}}}}\mathbf{⩽}\mathit{\sigma }\mathbf{⩽}\sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{\chi }}_{\left(1-\frac{\alpha }{2}\right)}^{\mathbf{2}}}}$

From the table values, at the 0.10 level of significance and at 49 degrees of freedom, the value for ${\chi }_{\frac{\alpha }{2}}^2$is 66.3387, and the value for ${\chi }_{\left(1-\frac{\alpha }{2}\right)}^2$is 33.9303.

Substitute the values to get the required confidence interval.

$$\begin{gathered} \left(\sqrt{\frac{\left(50-1\right){2.5}^2}{66.3387}}⩽\sigma ⩽\sqrt{\frac{\left(50-1\right){2.5}^2}{33.9303}}\right)=\sqrt{4.6165}⩽\sigma ⩽\sqrt{9.0259} \\ =2.1486⩽\sigma ⩽3.0043 \end{gathered}$$

Therefore, the 90% confidence interval for $\sigma$is $\left(2.1486,3.0043\right)$.

(b) Calculating the 90% confidence interval for 15 degrees of freedom

Given $\overline{x}=1.3,s=0.02,n=15$

The 90% confidence interval can be calculated using the formula,

$\sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{\chi }}_{\frac{\mathbf{\alpha }}{\mathbf{2}}}^{\mathbf{2}}}}\mathbf{⩽}\mathit{\sigma }\mathbf{⩽}\sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{\chi }}_{\left(1-\frac{\alpha }{2}\right)}^{\mathbf{2}}}}$

From the table values, at the 0.10 level of significance and at 14 degrees of freedom, the value for ${\chi }_{\frac{\alpha }{2}}^2$is 23.6848, and the value for ${\chi }_{\left(1-\frac{\alpha }{2}\right)}^2$is 6.5706.

Substitute the values to get the required confidence interval.

$$\begin{gathered} \left(\sqrt{\frac{\left(15-1\right){0.02}^2}{23.6848}}⩽\sigma ⩽\sqrt{\frac{\left(14-1\right){0.02}^2}{6.5706}}\right)=\sqrt{0.00024}⩽\sigma ⩽\sqrt{0.00085} \\ =0.0155⩽\sigma ⩽0.0292 \end{gathered}$$

Therefore, the 90% confidence interval for $\sigma$is$\left(0.0155,0.0292\right)$.

(c) Calculating the 90% confidence interval for 22 degrees of freedom

Given$\overline{x}=167,s=31.6,n=22$

The 90% confidence interval can be calculated using the formula,

role="math" localid="1668660233980" $\sqrt{\frac{\left(n-1\right)s^2}{{\chi }_{\frac{\alpha }{2}}^2}}⩽\sigma ⩽\sqrt{\frac{\left(n-1\right)s^2}{{\chi }_{\left(1-\frac{\alpha }{2}\right)}^2}}$

From the table values, at the 0.10 level of significance and at 21 degrees of freedom, the value for ${\chi }_{\frac{\alpha }{2}}^2$is 32.6706, and the value for ${\chi }_{\left(1-\frac{\alpha }{2}\right)}^2$is 11.5913.

Substitute the values to get the required confidence interval.

$$\begin{gathered} \left(\sqrt{\frac{\left(22-1\right){31.6}^2}{32.6706}}⩽\sigma ⩽\sqrt{\frac{\left(22-1\right){31.6}^2}{11.5913}}\right)=\sqrt{641.85}⩽\sigma ⩽\sqrt{1809.09} \\ =25.3348⩽\sigma ⩽42.5334 \end{gathered}$$

Therefore, the 90% confidence interval for $\sigma$is $\left(25.3348,42.5334\right)$.

(d) Calculating the 90% confidence interval for 5 degrees of freedom

Given$\overline{x}=9.4,s=1.5,n=5$

The 90% confidence interval can be calculated using the formula,

$\sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{\chi }}_{\frac{\mathbf{\alpha }}{\mathbf{2}}}^{\mathbf{2}}}}\mathbf{⩽}\mathit{\sigma }\mathbf{⩽}\sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{\chi }}_{\left(1-\frac{\alpha }{2}\right)}^{\mathbf{2}}}}$

From the table values, at the 0.10 level of significance and at 4 degrees of freedom, the value for${\chi }_{\frac{\alpha }{2}}^2$ is 9.4877, and the value for ${\chi }_{\left(1-\frac{\alpha }{2}\right)}^2$is 0.7107.

Substitute the values to get the required confidence interval.

$$\begin{gathered} \left(\sqrt{\frac{\left(5-1\right){1.5}^2}{9.4877}}⩽\sigma ⩽\sqrt{\frac{\left(5-1\right){1.5}^2}{0.7107}}\right)=\sqrt{0.9486}⩽\sigma ⩽\sqrt{12.6632} \\ =0.9740⩽\sigma ⩽3.5585 \end{gathered}$$

Therefore, the 90% confidence interval for $\sigma$is $\left(0.9740,3.5585\right)$.