Question

The following data represent the repair cost for a low-impact collision in a simple random sample of mini and micro-vehicles (such as the Chevrolet Aveo or Mini Cooper), it was verified that the data come from a population that is normally distributed with no outliers and s = $1007.4542.
Construct and interpret a 90% confidence interval for the standard deviation repair cost of a low-impact collision involving mini- and micro-vehicles.
$3148 $1758 $1071 $3345 $743
$2057 $773 $1370 $663 $2637

Answer 1

Answer:

$\frac{(9)(1007.4542)^2}{16.919} \leq \sigma^2 \leq \frac{(9)(1007.4542)^2}{3.325}$

$539906.407 \leq \sigma^2 \leq 2747271.128$

Now we just take square root on both sides of the interval and we got:

$734.783 \leq \sigma \leq 1657.489$

For this case we can conclude that the true deviation at 90% of confidence is between 734.783 and 1657.489

Step-by-step explanation:

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The sample deviation is s=1007.4542

The degrees of freedom are:

$df=n-1=10-1=9$

The confidence is 0.90 or 90%, and the significance would be $\alpha=0.1$ and $\alpha/2 =0.05$, and the critical values are:

$\chi^2_{\alpha/2}=16.919$

$\chi^2_{1- \alpha/2}=3.325$

Replacing the info given we got:

$\frac{(9)(1007.4542)^2}{16.919} \leq \sigma^2 \leq \frac{(9)(1007.4542)^2}{3.325}$

$539906.407 \leq \sigma^2 \leq 2747271.128$

Now we just take square root on both sides of the interval and we got:

$734.783 \leq \sigma \leq 1657.489$

For this case we can conclude that the true deviation at 90% of confidence is between 734.783 and 1657.489