Chapter 9.5, Problem 20E

To determine

To construct:

Each specified confidence interval and interpret the interval.

Answer to Problem 20E

Solution:

The confidence interval is $\left(0.4126,2.4637\right)$.

Explanation of Solution

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 ${\sigma }_2^2$.

Given Information:

From the given information, Amy’s practice run is $n_1=16$ and Veronica’s practice run is $n_2=15$. Their population variances are $s_1^2=2.560$ and $s_2^2=2.519$ at 90% level of confidence.

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 2.560 for $s_1^2$ and 2.519 for $s_2^2$ in the point estimate formula as,

$\begin{array}{rll}\frac{s_1^2}{s_2^2}& =& \frac{2.560}{2.519}\\ & =& 1.0163\end{array}$

At 90% level of confidence or $c=0.90$, the level of significance is,

$\begin{array}{rll}\alpha & =& 1-0.90\\ & =& 0.10\end{array}$

The right endpoint for $F$-distribution is $\frac{\alpha }{2}$. Substitute $0.10$ for $\alpha$ in $\frac{\alpha }{2}$.

$\begin{array}{rll}\frac{\alpha }{2}& =& \frac{0.10}{2}\\ & =& 0.05\end{array}$

Similarly the left endpoint for $F$-distribution is $1-\frac{\alpha }{2}$. Substitute $0.10$ for $\alpha$ in $\frac{\alpha }{2}$.

$\begin{array}{rll}1-\frac{\alpha }{2}& =& 1-0.05\\ & =& 0.95\end{array}$

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 $n_1=16$ and $n_2=15$.

Substitute 16 for $n_1$ in equation $\left(1\right)$.

$\begin{array}{rll}d{f}_1& =& n_1-1\\ & =& 16-1\\ & =& 15\end{array}$

Substitute 15 for $n_2$ in equation $\left(2\right)$.

$\begin{array}{rll}d{f}_2& =& n_2-1\\ & =& 15-1\\ & =& 14\end{array}$

 For the column of $d{f}_1=15$ and the row of $d{f}_2=14$ in the $F$-distribution table under the area of $\frac{\alpha }{2}$, the value of $F_{\frac{\alpha }{2}}$ is $2.4630$.

For the column of $d{f}_1=15$ and the row of $d{f}_2=14$ in the $F$-distribution table under the area of $1-\frac{\alpha }{2}$, the value of $F_{1-\frac{\alpha }{2}}$ is $0.4125$.

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 $1.0163$ for $\frac{s_1^2}{s_2^2}$, 2.4630 for $F_{\alpha /2}$ and 0.4125 for $F_{(1-\alpha /2)}$ in the above formula of confidence interval,

$\begin{array}{rll}\text{Confidence}\text{interval}& =& \left(1.0163\times \frac{1}{2.4630}\right)<\frac{{\sigma }_1^2}{{\sigma }_2^2}<\left(1.0163\times \frac{1}{0.4125}\right)\\ & =& \left(0.4126\right)<\frac{{\sigma }_1^2}{{\sigma }_2^2}<\left(2.4637\right)\end{array}$

Thus, the confidence interval is $\left(0.4126,2.4637\right)$.

Interpretation:

Since the value 1 is in the interval $\left(0.4126,2.4637\right)$, the data do not provide evidence at 90% level of confidence that, the population variances of Amy’s run time and Veronika’s run time are unequal.