Question

The scores on a placement test given to college freshmen for the past five years are approximately normally distributed with a mean μ = 74 and a variance σ2 = 8. Would you still consider σ2 = 8 to be a valid value of the variance if a random sample of 20 students who take the placement test this year obtain a value of s2 = 20?

Answer 1

Answer:

And on this case we have a 0.0303% of chance that the sample variance would be higher than 20. So on this case its not reasonable that the sample variance would be higher than 20.

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=20$ represent the sample size

$s^2 =20$ represent the sample variance obtained

$\sigma^2 =8$ represent the population variance

We can calculate the probability that $S^2$ would be higher than 20 and we will have an idea how to solve the problem.  

For this case we can use the following statistic:

$\chi^2 =\frac{n-1}{\sigma^2} s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =\frac{20-1}{8} 20 =47.5$

Calculate the probability

We can calculate the probability that $S^2$ would be higher than 20. The degrees of freedom are given by n-1=20-1=19

$P(S^2 >20) =P(\chi^2_{19} >47.5)=0.000303$

And hte reason of this is this one:

$P(S^2 >20)= P(\frac{\chi^2 \sigma^2}{n-1}>20)=P(\chi^2 >\frac{20(n-1)}{\sigma^2})=P(\chi^2 >\frac{20*19}{8})=P(\chi^2 >47.5)$

In order to calculate this probability we can use the following code in excel:

"==1-CHISQ.DIST(47.5,19,TRUE)"

And on this case we have a 0.0303% of chance that the sample variance would be higher than 20. So on this case its not reasonable that the sample variance would be higher than 20.