Posted by **Brandon** on Thursday, December 27, 2007 at 11:37am.

Use a chi-squared test to test the claim about the population variance or standard deviation at the given level of significance and using the given sample statistics.

Claim: ơ2 ≤ 60; α = 0.025. Sample statistics: s2 = 72.7, n = 15

- Statistics -
**MathGuru**, Thursday, December 27, 2007 at 9:28pm
You might try this formula for your chi-square test:

Chi-square = (n - 1)(sample variance)/(value specified in the null hypothesis)

n = 15

sample variance = 72.7

value specified in the null hypothesis = 60

Chi-square = (15 - 1)(72.7)/60

Finish the calculation for your chi-square test statistic.

Degrees of freedom is equal to n - 1, which is 14.

Check a chi-square distribution table using alpha = 0.025, with 14 degrees of freedom to determine your critical or cutoff value.

If the test statistic does not exceed the critical or cutoff value from the table, then the null is not rejected. If the test statistic exceeds the critical or cutoff value from the table, then the null is rejected.

I hope this will help.

## Answer this Question

## Related Questions

- Statistics - Use a x squared-test to test the claim about the population ...
- statistics - Given a sample size of 38, with sample mean 660.3 and sample ...
- statistics - Prep Courses? Scores for men (nationwide) on the verbal portion of ...
- statistics - it is stated that the standard deviation sample of the is 7.6 (n = ...
- statistics - it is stated that the standard deviation sample of the is 7.6 (n = ...
- Statistics - This is the first question..... For a population that has a ...
- statistics - A random sample of 49 credit ratings is obtained, and it is found ...
- statistics - an auditor wishes to test the assumption that the mean value of all...
- statistics - For each of the following samples that were given an experimental ...
- statistics - For each of the following samples that were given an experimental ...