# Testing of the hypothesis of variance

• tzx9633
In summary, the null hypothesis is that the variance of IQ scores for professors is the same as the general population, with a mean IQ of 100 and standard deviation of 15. The alternative hypothesis is that the variance is different. Using a chi-square test with a 0.05 significance level, the calculated test statistic is 10.22 and the critical chi-square value is 38.076. Since this is a two-tailed test, we cannot reject or accept the null hypothesis.
tzx9633

## 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

## 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

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tzx9633 said:

## 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

## 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.

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tzx9633
Ray Vickson said:
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 ?

Last edited by a moderator:
tzx9633 said:
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

Last edited by a moderator:
tzx9633 said:

## 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"?

tzx9633 said:
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.

WWGD said:
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.

Mark44 said:
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.

WWGD said:
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.

WWGD said:
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).

WWGD

## 1. What is the purpose of testing the hypothesis of variance?

The purpose of testing the hypothesis of variance is to determine whether there is a significant difference between the variance of a sample and the expected variance of the population. This helps researchers to understand the variability of their data and make accurate conclusions about their findings.

## 2. How is the hypothesis of variance tested?

The hypothesis of variance is typically tested using a statistical test called the F-test. This test compares the variance of the sample to the expected variance of the population, and calculates a p-value which indicates the likelihood of obtaining the observed results by chance. If the p-value is below a predetermined threshold, the null hypothesis of no variance difference is rejected.

## 3. What is the null hypothesis in testing the hypothesis of variance?

The null hypothesis in testing the hypothesis of variance states that there is no significant difference between the variance of the sample and the expected variance of the population. This means that any observed variance is due to chance and not a true difference between the two.

## 4. What is the alternative hypothesis in testing the hypothesis of variance?

The alternative hypothesis in testing the hypothesis of variance states that there is a significant difference between the variance of the sample and the expected variance of the population. This means that there is a true difference between the two and it is not due to chance.

## 5. How is the results of a hypothesis of variance test interpreted?

If the p-value is below the predetermined threshold, the null hypothesis is rejected and it can be concluded that there is a significant difference between the variance of the sample and the expected variance of the population. If the p-value is above the threshold, the null hypothesis is not rejected and it can be concluded that there is no significant difference between the two. Additionally, the effect size of the variance difference can also be calculated to determine the magnitude of the difference.

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