Question

Assume the sample variances to be continuous measurements. Find the probability that a random sample of 25observations, from a normal population with variance 02 = 6. will have a sample variance 52(a) greater than 9.1;(b) between 3.462 and 10.745.

Answer 1

Answer:

a) $P(4S^2 > 4*9.1) = P(\chi^2_{24} >36.4) = 0.0502$

b)$P(13.848$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 $S^2$ and we select a random sample of size n from a normal population with a population variance $\sigma^2$. And we define the following statistic:

$T = \frac{(n-1) S^2}{\sigma^2}$

And the distribution for this statistic is $T \sim \chi^2_{n-1}$

For this case we know that n =25 and $\sigma^2 = 6$ so then our statistic would be given by:

$\chi^2 = \frac{(n-1)S^2}{\sigma^2}=\frac{24 S^2}{6}= 4S^2$

With 25-1 =24 degrees of freedom.

Solution to the problem

Part a

For this case we want this probability:

$P(S^2 > 9.1)$

And we can multiply the inequality by 4 on both sides and we got:

$P(4S^2 > 4*9.1) = P(\chi^2_{24} >36.4) = 0.0502$

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:

$P(3.462 < S^2$

If we multiply the inequality by 4 on all the terms we got:

$P(3.462*4 < 4S^2 < 4*10.745)= P(13.848< \chi^2$And we can find this probability like this:$P(13.848$And we use the following code to find the answer in excel: "=CHISQ.DIST(42.98,24,TRUE)-CHISQ.DIST(13.848,24,TRUE)"