Answer:
a) ![P(4S^2 > 4*9.1) = P(\chi^2_{24} >36.4) = 0.0502](https://tex.z-dn.net/?f=%20P%284S%5E2%20%3E%204%2A9.1%29%20%3D%20P%28%5Cchi%5E2_%7B24%7D%20%3E36.4%29%20%3D%200.0502)
b)
Step-by-step explanation:
Previous concepts
The Chi Square distribution is the distribution of the sum of squared standard normal deviates .
For this case we assume that the sample variance is given by
and we select a random sample of size n from a normal population with a population variance
. And we define the following statistic:
![T = \frac{(n-1) S^2}{\sigma^2}](https://tex.z-dn.net/?f=%20T%20%3D%20%5Cfrac%7B%28n-1%29%20S%5E2%7D%7B%5Csigma%5E2%7D%20)
And the distribution for this statistic is ![T \sim \chi^2_{n-1}](https://tex.z-dn.net/?f=%20T%20%5Csim%20%5Cchi%5E2_%7Bn-1%7D)
For this case we know that n =25 and
so then our statistic would be given by:
![\chi^2 = \frac{(n-1)S^2}{\sigma^2}=\frac{24 S^2}{6}= 4S^2](https://tex.z-dn.net/?f=%5Cchi%5E2%20%3D%20%5Cfrac%7B%28n-1%29S%5E2%7D%7B%5Csigma%5E2%7D%3D%5Cfrac%7B24%20S%5E2%7D%7B6%7D%3D%204S%5E2)
With 25-1 =24 degrees of freedom.
Solution to the problem
Part a
For this case we want this probability:
![P(S^2 > 9.1)](https://tex.z-dn.net/?f=%20P%28S%5E2%20%3E%209.1%29)
And we can multiply the inequality by 4 on both sides and we got:
![P(4S^2 > 4*9.1) = P(\chi^2_{24} >36.4) = 0.0502](https://tex.z-dn.net/?f=%20P%284S%5E2%20%3E%204%2A9.1%29%20%3D%20P%28%5Cchi%5E2_%7B24%7D%20%3E36.4%29%20%3D%200.0502)
And we can use the following excel code to find it: "=1-CHISQ.DIST(36.4,24,TRUE)"
Part b
For this case we want this probability:
![P(3.462 < S^2](https://tex.z-dn.net/?f=%20P%283.462%20%3C%20S%5E2%20%3C10.745%29)
If we multiply the inequality by 4 on all the terms we got:
And we can find this probability like this:
And we use the following code to find the answer in excel: "=CHISQ.DIST(42.98,24,TRUE)-CHISQ.DIST(13.848,24,TRUE)"