Question

The standard deviation of math test scores at one high school is 16.1. A teacher claims that the standard deviation of the girls' test scores is smaller than 16.1. A random sample of 22 girls results in scores with a standard deviation of 12.9. Use a significance level of 0.01 to test the teacher's claim.

Answer 1

`\chi^2 =(22-1)/(259.21) 166.41 =13.482`

`p_v =P(\chi^2 <13.482)=0.1092`

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(13.482,21,TRUE)"

If we compare the p value and the significance level provided we see that
`p_v >\alpha` so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population deviation is not significantly lower than 16.1 at 1% of significance.

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=22` represent the sample size

`\alpha=0.01` represent the confidence level

`s^2 =12.9^2=166.41` represent the sample variance obtained

`\sigma^2_0 =16.1^2 =259.21` represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population deviation is smaller than 16.1 (that's equivalent to check if the population variance is lower than 259.21:

Null Hypothesis:
`\sigma^2 \geq 259.21`

Alternative hypothesis:
`\sigma^2 <259.21`

Calculate the statistic

For this test we can use the following statistic:

`\chi^2 =(n-1)/(\sigma^2_0) s^2`

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

`\chi^2 =(22-1)/(259.21) 166.41 =13.482`

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 21. And since is a left tailed test the p value would be given by:

`p_v =P(\chi^2 <13.482)=0.1092`

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(13.482,21,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that
`p_v >\alpha` so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population deviation is not significantly lower than 16.1 at 1% of significance.