Testing of the hypothesis of variance

Homework Statement

for randomly selected IQ adult scores are normally distributed with mean of 100 and standarrd deviation of 15 ,. a sample of 24 randomly selected professors resulted in the IQ scores having a standard deviation of 10 . Test the claim that the IQ scores for the professors is same as the general population of 1.5 . Use 0.05 significance of level

Homework Equations

The Attempt at a Solution

H0 = standard deviation = 15

H0 = standard deviation not equal to 15

So , chi square test = (24-1)(10^2) / (15^2) = 10.22

From table , critical chi square value = 38.076 I am not sure to reject or accept the H0 , since this is 2 tailed test . I was told that for chi square test , there's only one tailed test exist

 

[Ray Vickson]

Homework Statement

for randomly selected IQ adult scores are normally distributed with mean of 100 and standarrd deviation of 15 ,. a sample of 24 randomly selected professors resulted in the IQ scores having a standard deviation of 10 . Test the claim that the IQ scores for the professors is same as the general population of 1.5 . Use 0.05 significance of level

Homework Equations

The Attempt at a Solution

H0 = standard deviation = 15

H0 = standard deviation not equal to 15

So , chi square test = (24-1)(10^2) / (15^2) = 10.22

From table , critical chi square value = 38.076 I am not sure to reject or accept the H0 , since this is 2 tailed test . I was told that for chi square test , there's only one tailed test exist

You have written two different things for the same symbol "H0". Can I suppose one of them is H1? If the second one is H1, you need to be more careful about whether to take it to be "$\neq 15$", "$< 15$" or "$> 15$". Those three different possible H1s will produce 3 different significant tests.

 

You have written two different things for the same symbol "H0". Can I suppose one of them is H1? If the second one is H1, you need to be more careful about whether to take it to be "$\neq 15$", "$< 15$" or "$> 15$". Those three different possible H1s will produce 3 different significant tests.

Yes , sorry for my mistake .

H0 = standard deviation = 15

H1 = standard deviation not equal to 15

So , chi square test = (24-1)(10^2) / (15^2) = 10.22

From table , critical chi square value = 38.076 I am not sure to reject or accept the H0 , since this is 2 tailed test . I was told that for chi square test , there's only one tailed test exist

Can you help me on this ?

 

[Ray Vickson]

Yes , sorry for my mistake .

H0 = standard deviation = 15

H1 = standard deviation not equal to 15

So , chi square test = (24-1)(10^2) / (15^2) = 10.22

From table , critical chi square value = 38.076 I am not sure to reject or accept the H0 , since this is 2 tailed test . I was told that for chi square test , there's only one tailed test exist

Can you help me on this ?

Please stop using a bold font; it looks like you are yelling at us.
Mod edit: I removed the extra bolding in this and the previous posts.
Anyway, all you need is given in http://www.itl.nist.gov/div898/handbook/eda/section3/eda358.htm

 

[WWGD]

Homework Statement

for randomly selected IQ adult scores are normally distributed with mean of 100 and standarrd deviation of 15 ,. a sample of 24 randomly selected professors resulted in the IQ scores having a standard deviation of 10 . Test the claim that the IQ scores for the professors is same as the general population of 1.5 . Use 0.05 significance of level

I am confused: are you testing for IQ or for variance in the distribution? don't you mean "Test the claim that the standard deviation for professors is the same as that for the general population"?

 

[Mark44]

Test the claim that the IQ scores for the professors is same as the general population of 1.5

The figure 1.5 is a typo. It should be 15, which is what was used in the calculation.

I am confused: are you testing for IQ or for variance in the distribution? don't you mean "Test the claim that the standard deviation for professors is the same as that for the general population"?

I'm pretty sure that what you wrote is what he meant; i.e., the test is for the standard deviation of IQ scores of this subpopulation.

 

[WWGD]

The figure 1.5 is a typo. It should be 15, which is what was used in the calculation.

I'm pretty sure that what you wrote is what he meant; i.e., the test is for the standard deviation of IQ scores of this subpopulation.

But then one thing that is confusing is the mention of IQ of the general population, when what is being tested is the variance. The profs' IQ may also be different, but the general pop's IQ being 100 is never used and has no effect on the results.

 

[Mark44]

But then one thing that is confusing is the mention of IQ of the general population, when what is being tested is the variance. The profs' IQ may also be different, but the general pop's IQ being 100 is never used and has no effect on the results.

The implicit assumption that is behind the null hypothesis is that the distrubutions of the professors' IQs is the same as the population. In the problem in this thread, the goal is to test whether the variance of the professors' IQs is the same as that of the general population. The mean isn't used in a $\chi ^2$ test.

 

[Ray Vickson]

But then one thing that is confusing is the mention of IQ of the general population, when what is being tested is the variance. The profs' IQ may also be different, but the general pop's IQ being 100 is never used and has no effect on the results.

Right, and not only that, the "professor" group may have a different mean IQ as well. As long as the professors' sample mean is used when computing the professors' sample variance, that will not matter at all (assuming normality, of course).