Respuesta :
Answer:
[tex]\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}[/tex]
[tex] 4.525 \leq \sigma^2 \leq 7.602[/tex]
Now we just take square root on both sides of the interval and we got:
[tex] 2.127 \leq \sigma \leq 2.757[/tex]
Step-by-step explanation:
Information given
[tex]\bar X=32.1[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=2.4 represent the sample standard deviation
n=83 represent the sample size
Confidence interval
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 degrees of freedom given by:
[tex]df=n-1=8-1=7[/tex]
The confidence level is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical values.
The excel commands would be: "=CHISQ.INV(0.05,82)" "=CHISQ.INV(0.95,82)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=104.139[/tex]
[tex]\chi^2_{1- \alpha/2}=62.132[/tex]
The confidence interval is given by:
[tex]\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}[/tex]
[tex] 4.525 \leq \sigma^2 \leq 7.602[/tex]
Now we just take square root on both sides of the interval and we got:
[tex] 2.127 \leq \sigma \leq 2.757[/tex]
