Respuesta :
Answer:
[tex]\chi^2 =\frac{15-1}{0.65} 0.721^2 =11.797[/tex]
We need to find in the chi square distribution with df = 14 a critical value who accumulates 0.025 of the area and we got the critical value on this case :
[tex]\chi^2_{\alpha/2}= 5.629, \chi^2_{1-\alpha/2}=26.119[/tex]
If we compare the calculated value is between the two critical values so then we don't have enough evidence to reject the null hypothesis that the true variance is different from 0.65 on this case.
Step-by-step explanation:
Data given
[tex]n=12[/tex] represent the sample size
[tex]\alpha=0.05[/tex] represent the confidence level
[tex]s^2 =0.721^2 =0.520 [/tex] represent the sample variance obtained
[tex]\sigma^2_0 =0.65[/tex] represent the value that we want to test
Null and alternative hypothesis
On this case we want to check if the population variance has changed, so the system of hypothesis would be:
Null Hypothesis: [tex]\sigma^2 = 0.65[/tex]
Alternative hypothesis: [tex]\sigma^2 \neq 0.65[/tex]
Calculate the statistic
For this test we can use the following statistic:
[tex]\chi^2 =\frac{n-1}{\sigma^2_0} s^2[/tex]
And replacing we got:
[tex]\chi^2 =\frac{15-1}{0.65} 0.721^2 =11.797[/tex]
Calculate the critical value
We need to find in the chi square distribution with df = 14 a critical values who accumulates 0.025 of the area and we got the critical values on this case :
[tex]\chi^2_{\alpha/2}= 5.629, \chi^2_{1-\alpha/2}=26.119[/tex]
Conclusion
If we compare the calculated value is between the two critical values so then we don't have enough evidence to reject the null hypothesis that the true variance is different from 0.65 on this case.
