Posted by Brandon on Thursday, December 27, 2007 at 11:37am.
Use a chisquared 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 chisquare test:
Chisquare = (n  1)(sample variance)/(value specified in the null hypothesis)
n = 15
sample variance = 72.7
value specified in the null hypothesis = 60
Chisquare = (15  1)(72.7)/60
Finish the calculation for your chisquare test statistic.
Degrees of freedom is equal to n  1, which is 14.
Check a chisquare 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.