Chapter 8.5, Problem 24E

To determine

To build and to interpret:

A $98\%$ confidence interval for the population variance for the weights of all boxes of crackers that come off the assembly line.

Answer to Problem 24E

Solution:

The confidence interval for the population variance is given by (0.0019, 0.0128).

Thus, we are 98% confident that the population variance is between 0.0019 and 0.0128.

Explanation of Solution

Procedure:

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 =98%

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 ‘$\stackrel{-}{x}$’

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

$=\frac{\begin{array}{c}16.87+16.92+17.01+16.98+16.99+16.92+16.91+17.00+\\ 17.01+16.96+16.95+16.94+17.00+16.92\end{array}}{14}$

$\stackrel{-}{x}=\frac{237.38}{14}$

$\stackrel{-}{x}=16.9557$

$\stackrel{-}{x}=17$

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

Deviations and squared Deviations of the data
$x_i$	$x_i-\overline{x}$	${(x_i-\overline{x})}^2$
$16.87$	$16.87-17=-0.13$	$0.0169$
$16.92$	$16.92-17=-0.08$	$0.0064$
$17.01$	$17.01-17=0.01$	$0.0001$
$16.98$	$16.98-17=-0.02$	$0.0004$
$16.99$	$16.99-17=-0.01$	$0.0001$
$16.92$	$16.92-17=-0.08$	$0.0064$
$16.91$	$16.91-17=-0.09$	$0.0081$
$17.00$	$17.00-17=0$	$0$
$17.01$	$17.01-17=0.01$	$0.0001$
$16.96$	$16.96-17=-0.04$	$0.0016$
$16.95$	$16.95-17=-0.05$	$0.0025$
$16.94$	$16.94-17=-0.06$	$0.0036$
$17.00$	$17.00-17=0$	$0$
$16.92$	$16.92-17=-0.08$	$0.0064$

$\sum {(x_i-\overline{x})}^2=0.0526$

Thus, we have

$\sum {(x_i-\overline{x})}^2=0.0526$ (Squared deviation)

$n=14$ (Sample size)

Substituting the above values in

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

$=\frac{0.0526}{14-1}$

$=\frac{0.0526}{13}$

$s^2=0.00405$

Now, construct a confidence interval for the population variance with

$n=14$ (Sample size)

$s^2=0.00405$ (Sample variance)

$c=0.98$ (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 $s^2=0.00405$

STEP 2:

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

$c=0.98$

We have,

$\sigma =1-c$

$\sigma =1-0.98$

$\sigma =0.02$

Thus,

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

$=\frac{0.02}{2}$

$=0.01$ and

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

$=1-0.01$

$=0.99$

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,

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

$={\chi }_{0.01}^2$

$=27.69$ and

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

$={\chi }_{\left(1-0.01\right)}^2$

$={\chi }_{\left(0.99\right)}^2$

$=4.107$

STEP 4:

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

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

$\frac{\left(14-1\right)0.00405}{27.69}<{\sigma }^2<\frac{\left(14-1\right)0.00405}{4.107}$

$\frac{\left(13\right)0.00405}{27.69}<{\sigma }^2<\frac{\left(13\right)0.00405}{4.107}$

$\frac{0.05265}{27.69}<{\sigma }^2<\frac{0.05265}{4.107}$

$0.0019<{\sigma }^2<0.0128$

Using, interval notation, the confidence interval can also be written as (0.0019, 0.0128)

Final statement:

Therefore, the confidence interval for the population variance is given by (0.0019, 0.0128).

Thus, we are 98% confident that the population variance is between 0.0019 and 0.0128.