Chapter 8.CR, Problem 9CR

To determine

To interpret:

A $95\%$ confidence interval for the population variance for the volumes of soda in all the soda cans that come off that particular assumbly line.

Answer to Problem 9CR

Solution:

The confidence interval for the population variance is given by (0.001, 0.003)

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

STEP 3:

Find the critical value ${\chi }_{\frac{\alpha }{2}}^2$ and ${\chi }_{(1-\frac{\alpha }{2})}^2$ 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

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

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

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

Calculation:

The confidence interval for the population variance is given by $(0.001, 0.003)$.

The volumes of soda in all the soda cans that come off that particular assumbly line will have the variance between 0.001 and 0.003.

Level confidence $=95\%$

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 ‘$\stackrel{-}{x}$’

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

$=\frac{\begin{array}{c}12.01+12.02+11.95+11.99+11.94+12.01+12.03+11.98+\\ 12.00+12.03+11.98+12.05+11.93+11.98\end{array}}{14}$

$\stackrel{-}{x}=\frac{167.9}{14}$

$\stackrel{-}{x}=11.9929$

$\stackrel{-}{x}=12$

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

Deviations and squared Deviations of the data
$x_i$	$d=x_i-\stackrel{-}{x}$	$d^2$
$12.01$	$12.01-12=0.01$	$0.0001$
$12.02$	$12.02-12=0.02$	$0.0004$
$11.95$	$11.95-12=-0.05$	$0.0025$
$11.99$	$11.99-12=-0.01$	$0.0001$
$11.94$	$11.94-12=-0.06$	$0.0036l$
$12.01$	$12.01-12=0.01$	$0.0001$
$12.03$	$12.03-12=0.03$	$0.0009$
$11.98$	$11.98-12=-0.02$	$0.0004$
$12.00$	$12.00-12=0$	$0$
$12.03$	$12.03-12=0.03$	$0.0009$
$11.98$	$11.98-12=-0.02$	$0.0004$
$12.05$	$12.05-12=0.05$	$0.0025$
$11.93$	$11.93-12=-0.07$	$0.0049$
$11.98$	$11.98-12=-0.02$	$0.0004$

$\sum {(x_i-\overline{x})}^2=0.0172$

Thus, we have

$\sum {\left(x_i-\overline{x}\right)}^2=0.0172$ (Squared deviation)

$n=14$ (Sample size)

Substituting the above values in

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

$=\frac{0.0172}{14-1}$

$=\frac{0.0172}{13}$

$s^2=0.0013$

Now, construct a confidence interval for the population variance with

$n=14$ (Sample size)

$s^2=0.0013$ (Sample variance)

$c=0.95$ (Level of confidence)

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

$s^2=0.0013$

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

$c=0.95$

We have,

$\alpha =1-c$

$\alpha =1-0.95$

$\alpha =0.05$

Thus,

$=\frac{\alpha }{2}$

$=\frac{0.05}{2}$

$=0.025$ and

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

$=1-0.025$

$=0.975$

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

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

$={\chi }_{0.025}^2$

$= 24.74$ and

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

$={\chi }_{\left(1-0.025\right)}^2$

$={\chi }_{\left(0.975\right)}^2$

$=5.009$

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

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

$\frac{\left(14-1\right)0.0013}{24.74}<{\sigma }^2<\frac{\left(14-1\right)0.0013}{5.009}$

$\frac{\left(13\right)0.0013}{24.74}<{\sigma }^2<\frac{\left(13\right)0.0013}{5.009}$

$\frac{0.0169}{24.74}<{\sigma }^2<\frac{0.0169}{5.009}$

$0.00068<{\sigma }^2<0.00337$

$0.001<{\sigma }^2<0.003$

Using, interval notation, the confidence interval can also be written as $(0.001, 0.003)$.

Final statement:

Therefore, the confidence interval for the population variance is given by $(0.001, 0.003)$.

The volumes of soda in all the soda cans that come off that particular assumbly line will have the variance between 0.001 and 0.003.