Chapter 8.5, Problem 23E

To determine

To build and to interpret:

A $99\%$ confidence interval for the population variance of the sample data.

Answer to Problem 23E

Solution:

The confidence interval for the population variance at the level of confidence 99% is given by (0.02, 0.15).

Thus, we are 99% confident that the population variance is between 0.02 and 0.15.

Explanation of Solution

Steps need to be followed while calculating the confidence interval.

STEP 1:

Find the point estimate, $s^2$ for confidence interval for the population variance and $s$ for confidence interval for the population standard deviation.

STEP 2:

Calculate $\frac{\sigma }{2}$ and $1-\frac{\sigma }{2}$, based on the level of confidence $c$ given.

STEP 3:

Find the critical value ${\chi }_{\frac{\sigma }{2}}^2$ and ${\chi }_{\left(1-\frac{\sigma }{2}\right)}^2$ for the distribution with $n-1$ degrees of freedom using the ${\chi }^2$-distribution table.

STEP 4:

Find the confidence interval for the population variance by substituting the necessary values in the formula

$\frac{\left(n-1\right)s^2}{{\chi }_{\frac{\sigma }{2}}^2}<{\sigma }^2<\frac{\left(n-1\right)s^2}{{\chi }_{\left(1-\frac{\sigma }{2}\right)}^2}$

Find the confidence interval for the population standard deviation by substituting the necessary values in the formula

$\sqrt{\frac{\left(n-1\right)s^2}{{\chi }_{\frac{\sigma }{2}}^2}}<\sigma <\sqrt{\frac{\left(n-1\right)s^2}{{\chi }_{\left(1-\frac{\sigma }{2}\right)}^2}}$

Calculation:

Level confidence = 99%

First let us calculate the sample variance for the given data.

The sample variance of a data having ‘n’ number of data values in the sample with mean ‘$\stackrel{-}{x}$’ is given by

$s^2=\frac{\sum {\left(x_i-\stackrel{-}{x}\right)}^2}{n-1}$

Here, we need to find the mean ‘$\stackrel{-}{x}$’

$\stackrel{-}{x}=\frac{\sum x_i}{n}$

$=\frac{3.2+3.6+2.9+3.0+3.0+3.1+3.2+3.3+2.9+3.3+2.9+3.1+3.4+3.3+3.0}{15}$

$\stackrel{-}{x}=\frac{47.2}{15}$

$\stackrel{-}{x}=3.1466$

$\stackrel{-}{x}=3.1$

Now, construct a table of deviations and squared deviations of the data.

Deviations and squared Deviations of the data
$x_i$	$x_i-\overline{x}$	${(x_i-\overline{x})}^2$
$3.2$	$3.2-3.1=0.1$	$0.01$
$3.6$	$3.6-3.1=0.5$	$0.25$
$2.9$	$2.9-3.1=-0.2$	$0.04$
$3.0$	$3.0-3.1=-0.1$	$0.01$
$3.0$	$3.0-3.1=-0.1$	$0.01$
$3.1$	$3.1-3.1=0$	$0$
$3.2$	$3.2-3.1=0.1$	$0.01$
$3.3$	$3.3-3.1=0.2$	$0.04$
$2.9$	$2.9-3.1=-0.2$	$0.04$
$3.3$	$3.3-3.1=0.2$	$0.04$
$2.9$	$2.9-3.1=-0.2$	$0.04$
$3.1$	$3.1-3.1=0$	$0$
$3.4$	$3.4-3.1=0.3$	$0.09$
$3.3$	$3.3-3.1=0.2$	$0.04$
$3.0$	$3.0-3.1=-0.1$	$0.01$

$\sum {(x_i-\overline{x})}^2=0.63$

Thus, we have

$\sum {(x_i-\overline{x})}^2=0.63$ (Squared deviation)

$n=15$ (Sample size)

Substituting the above values in

$s^2=\sum \frac{{\left(x_i-\overline{x}\right)}^2}{n-1}$

$=\frac{0.63}{15-1}$

$=\frac{0.63}{14}$

$s^2=0.045$

Now, construct a confidence interval for the population variance with

$n=15$ (Sample size)

$s^2=0.045$ (Sample variance)

$c=0.99$ (Level of confidence)

STEP 1:

Find the point estimate:

Here, we need to construct the confidence interval for the population variance, thus the point estimate is the value of $s^2$ and is given as $s^2=0.045$

STEP 2:

Calculate $\frac{\sigma }{2}$ and $1-\frac{\sigma }{2}$, based on the level of confidence given, for the level of confidence given,

$c=0.99$

We have,

$\sigma =1-c$

$\sigma =1-0.99$

$\sigma =0.01$

Thus,

$=\frac{\sigma }{2}$

$=\frac{0.01}{2}$

$=0.005$ and

$1-\frac{\sigma }{2}$

$=1-0.005$

$=0.995$

STEP 3:

Find the critical value ${\chi }_{\frac{\sigma }{2}}^2$ and ${\chi }_{\left(1-\frac{\sigma }{2}\right)}^2$ for the distribution with $n-1$ degrees of freedom using the ${\chi }^2$-distribution table,

$={\chi }_{\frac{\sigma }{2}}^2$

$={\chi }_{0.005}^2$

$=31.32$ and

${\chi }_{\left(1-\frac{\sigma }{2}\right)}^2$

$={\chi }_{\left(0.995\right)}^2$

$=4.075$

STEP 4:

Find the confidence interval by substituting the necessary values in the formula:

$\frac{\left(n-1\right)s^2}{{\chi }_{\frac{\sigma }{2}}^2}<{\sigma }^2<\frac{\left(n-1\right)s^2}{{\chi }_{\left(1-\frac{\sigma }{2}\right)}^2}$

$\frac{\left(15-1\right)0.045}{31.32}<{\sigma }^2<\frac{\left(15-1\right)0.045}{4.075}$

$\frac{\left(14\right)0.045}{31.32}<{\sigma }^2<\frac{\left(14\right)0.045}{4.075}$

$\frac{0.63}{31.32}<{\sigma }^2<\frac{0.63}{4.075}$

$0.02<{\sigma }^2<0.15$

Using, interval notation, the confidence interval can also be written as (0.02, 0.15)

Final statement:

Therefore, the confidence interval for the population variance at the level of confidence 99% is given by (0.02, 0.15).

Thus, we are 99% confident that the population variance is between 0.02 and 0.15.