Respuesta :
Answer:
[tex]\frac{(9)(1007.4542)^2}{16.919} \leq \sigma^2 \leq \frac{(9)(1007.4542)^2}{3.325}[/tex]
[tex] 539906.407 \leq \sigma^2 \leq 2747271.128[/tex]
Now we just take square root on both sides of the interval and we got:
[tex] 734.783 \leq \sigma \leq 1657.489[/tex]
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:
[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]
The sample deviation is s=1007.4542
The degrees of freedom are:
[tex]df=n-1=10-1=9[/tex]
The confidence is 0.90 or 90%, and the significance would be [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and the critical values are:
[tex]\chi^2_{\alpha/2}=16.919[/tex]
[tex]\chi^2_{1- \alpha/2}=3.325[/tex]
Replacing the info given we got:
[tex]\frac{(9)(1007.4542)^2}{16.919} \leq \sigma^2 \leq \frac{(9)(1007.4542)^2}{3.325}[/tex]
[tex] 539906.407 \leq \sigma^2 \leq 2747271.128[/tex]
Now we just take square root on both sides of the interval and we got:
[tex] 734.783 \leq \sigma \leq 1657.489[/tex]
For this case we can conclude that the true deviation at 90% of confidence is between 734.783 and 1657.489
