Respuesta :
Answer:
[tex]\frac{(49)(7.2)^2}{72.892} \leq \sigma^2 \leq \frac{(49)(7.2)^2}{30.036}[/tex]
[tex]34.848 \leq \sigma^2 \leq 84.57[/tex]
Taking square root in both sides we got the interval desired
[tex] 5.903 \leq \sigma \leq 9.196[/tex]
Step-by-step explanation:
Information provided
s=7.2 represent the sample standard deviation
[tex]\bar x=46.4[/tex] represent the sample mean
n=50 the sample size
Confidence=97% or 0.97
The confidence interval
We can use the following formula for the confidence interval for the true variance
[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 are given by:
[tex]df=n-1=50-1=49[/tex]
The confidence level is 0.97 or 97%, the significance [tex]\alpha=0.03[/tex] and [tex]\alpha/2 =0.015[/tex] and the critical values are:
The excel commands would be: "=CHISQ.INV(0.015,49)" "=CHISQ.INV(0.985,49)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=72.892[/tex]
[tex]\chi^2_{1- \alpha/2}=30.036[/tex]
Replacing the values we have:
[tex]\frac{(49)(7.2)^2}{72.892} \leq \sigma^2 \leq \frac{(49)(7.2)^2}{30.036}[/tex]
[tex]34.848 \leq \sigma^2 \leq 84.57[/tex]
Taking square root in both sides we got the interval desired
[tex]5.903 \leq \sigma \leq 9.196[/tex]
