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September 15, 2014

September 15, 2014

Posted by **Rob** on Monday, March 28, 2011 at 11:43am.

Any help on figuring this out would be greatly appreciated. I need the mean to figure this out, but I feel like the question is leaving something short. Thanks for your time

- Stats. -
**MathGuru**, Thursday, March 31, 2011 at 12:24pmNull and alternative hypotheses:

Ho: The population standard deviation is = 505.6

Ha: The population standard deviation is > 505.6

Data given:

sample standard deviation = 840 (Note: standard deviation is the square root of the variance; therefore, variance is standard deviation squared).

level of significance = 0.01

n = 41

The test statistic used in making a decision is the chi-square.

Here's the formula:

chi-square = (sample size - 1)(sample variance)/(value specified in the null hypothesis)

Note: Convert both standard deviations to variances (505.6 and 840) when working with the formula.

Degrees of freedom is equal to n - 1, which is 40.

Checking a chi-square table using alpha = 0.01 with 40 degrees of freedom, find the critical value. If the critical value exceeds the test statistic, the null is rejected and there is a difference. If the critical value does not exceed the test statistic, then the null cannot be rejected and there is not enough evidence to support the claim that the population standard deviation is greater than 505.6 grams.

I hope this will help get you started.

- Stats. -
**MathGuru**, Thursday, March 31, 2011 at 12:35pmIf the test statistic exceeds the critical value from the chi-square table, then the null is rejected. Also, if the test statistic does not exceed the critical value from the table, then the null cannot be rejected.

The above corrects the last few sentences. Sorry for any confusion.

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