Chapter 8.5, Problem 18E

To determine

To construct and interpret the specified confidence interval

A tire manufacture is testing the tire pressure in its new line of SUV tires. A random sample of $10$ tire pressure readings, in pounds per square inch (psi), yields a variance of $31.8$. Construct and interpret a $95\%$ confidence interval for in the pressures for all new SUV tires produced by the manufacturer.

$n=10, s^2=31.8, c=0.95$

Answer to Problem 18E

Solution:

The confidence interval for the population variance is given by

$(16.9, 106)$

The manufacturer can be 95% confident that the variance in pressure of all the new SUV tires is between 16.9 psi and 106 psi.

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

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

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:

Given,

A sample of $10$ tire pressure readings, in pounds per square inch (psi), yields a variance of $31.8$ and $95\%$ confidence interval for in the pressures for all new SUV tires.

Thus,

$n=10$ (Sample size)

$s^2=31.8$ (Population variance)

$c=0.95$ (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

$s^2=31.8$

STEP 2:

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

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

$=1-0.025$

$=0.975$

STEP 3:

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

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

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

$=16.92$ and

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

${=\chi }_{0.975}^2$

$=2.7$

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

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

$\frac{\left(10-s1\right)31.8}{16.92}<{\sigma }^2<\frac{\left(10-1\right)31.8}{2.7}$

$\frac{\left(9\right)31.8}{16.92}<{\sigma }^2<\frac{\left(9\right)31.8}{2.7}$

$\frac{286.2}{16.92}<{\sigma }^2<\frac{286.2}{2.7}$

$16.9<{\sigma }^2<106$

Using, interval notation, the confidence interval can also be written as $(16.9, 106)$

Final statement:

Therefore, the confidence interval for the population variance is given by

$(16.9, 106)$

The manufacturer can be 95% confident that the variance in pressure of all the new SUV tires is between 16.9 psi and 106 psi.