Chapter 9.5, Problem 21E

To determine

To construct:

Each specified confidence interval and interpret the interval.

Answer to Problem 21E

Solution:

The confidence interval is $\left(0.1824,6.9828\right)$.

Explanation of Solution

Given Information:

From the given information the total number of flour bags filled by Machine A is $n_1=11$ and by Machine B is $n_2=10$. Their population variances are $s_1^2=0.584$ and $s_2^2=0.499$ at 99% level of confidence.

Concept:

Confidence interval is used to compare two population variances, estimate the value of the ratio or fraction, $\frac{{\sigma }_1^2}{{\sigma }_2^2}$.

1. If the confidence interval contains 1 that means the two population variances are unequal at a particular level of confidence.

2. If both endpoints of the interval are greater than 1 it concludes that ${\sigma }_1^2$ is larger than ${\sigma }_2^2$.

3. If both endpoints of the interval are less than 1 it concludes that ${\sigma }_1^2$ is smaller than ${\sigma }_2^2$

Formula used:

The formula to calculate the confidence interval is,

$\left(\frac{s_1^2}{s_2^2}.\frac{1}{F_{\frac{\alpha }{2}}}\right)<\frac{{\sigma }_1^2}{{\sigma }_2^2}<\left(\frac{s_1^2}{s_2^2}.\frac{1}{F_{\left(1-\frac{\alpha }{2}\right)}}\right)$

The formula to calculate the point estimate is,

$\frac{s_1^2}{s_2^2}$

Where, $s_1$ and $s_2$ are the sample variances for the respective samples.

For the given level of confidence $c$, the level of significance is $\alpha =1-c$.

The right and left endpoints for $F$-distribution are $\frac{\alpha }{2}$ and $1-\frac{\alpha }{2}$ respectively. The critical values are $F_{\frac{\alpha }{2}}$ and $F_{1-\frac{\alpha }{2}}$ that can be determined in the $F$-distribution table using the column for $d{f}_1=n_1-1$ and row for $d{f}_2=n_2-1$.

Calculation:

Substitute 0.584 for $s_1^2$ and 0.499 for $s_2^2$ in the point estimate formula as,

$\begin{array}{rll}\frac{s_1^2}{s_2^2}& =& \frac{0.584}{0.499}\\ & =& 1.1703\end{array}$

At 99% level of confidence or $c=0.99$, the level of significance is,

$\begin{array}{rll}\alpha & =& 1-0.99\\ & =& 0.01\end{array}$

The right endpoint for $F$-distribution is $\frac{\alpha }{2}$. Substitute $0.01$ for $\alpha$ in $\frac{\alpha }{2}$.

$\begin{array}{rll}\frac{\alpha }{2}& =& \frac{0.01}{2}\\ & =& 0.005\end{array}$

Similarly the left endpoint for $F$-distribution is $1-\frac{\alpha }{2}$. Substitute $0.01$ for $\alpha$ in $\frac{\alpha }{2}$.

$\begin{array}{rll}1-\frac{\alpha }{2}& =& 1-0.005\\ & =& 0.995\end{array}$

The degrees of freedom are,

$d{f}_1=n_1-1$……$\left(1\right)$

And,

$d{f}_2=n_2-1$……$\left(2\right)$

The sample sizes are given as $n_1=11$ and $n_2=10$.

Substitute 11 for $n_1$ in equation $\left(1\right)$.

$\begin{array}{rll}d{f}_1& =& n_1-1\\ & =& 11-1\\ & =& 10\end{array}$

Substitute 10 for $n_2$ in equation $\left(2\right)$.

$\begin{array}{rll}d{f}_2& =& n_2-1\\ & =& 10-1\\ & =& 9\end{array}$

For the column of $d{f}_1=10$ and the row of $d{f}_2=9$ in the $F$-distribution table under the area of $\frac{\alpha }{2}$, the value of $F_{\frac{\alpha }{2}}$ is $6.4172$.

For the column of $d{f}_1=10$ and the row of $d{f}_2=9$ in the $F$-distribution table under the area of $1-\frac{\alpha }{2}$, the value of $F_{1-\frac{\alpha }{2}}$ is $0.1676$.

The formula to calculate the confidence interval is,

$\left(\frac{s_1^2}{s_2^2}.\frac{1}{F_{\frac{\alpha }{2}}}\right)<\frac{{\sigma }_1^2}{{\sigma }_2^2}<\left(\frac{s_1^2}{s_2^2}.\frac{1}{F_{\left(1-\frac{\alpha }{2}\right)}}\right)$.

Substitute $1.1703$ for $\frac{s_1^2}{s_2^2}$, $6.4172$ for $F_{\alpha /2}$ and $0.1676$ for $F_{(1-\alpha /2)}$ in the above formula of confidence interval,

$\begin{array}{rll}\text{Confidence interval}& =& \left(1.1703\times \frac{1}{6.4172}\right)<\frac{{\sigma }_1^2}{{\sigma }_2^2}<\left(1.1703\times \frac{1}{0.1676}\right)\\ & =& \left(0.1824\right)<\frac{{\sigma }_1^2}{{\sigma }_2^2}<\left(6.9828\right)\end{array}$

Thus, the confidence interval is $\left(0.1824,6.9828\right)$.

Interpretation:

Since the value 1 is in the interval $\left(0.1824,6.9828\right)$, the data do not provide evidence at 99% level of confidence that, the population variances of the amount of flour per bag for Machine A and Machine B are unequal.

If it is important that the two machines have the same variances, the inspector needs to adjust Machine A as it has higher variance than Machine B, so the variance will be 0.499 for both the machines.