anyone understand chi squared test

Mr.Cookie

Mr.Cookie

Mr.Cookie

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Hi trying to help my wife with her stats data, but she's confusing me, I'm confusing her and excel is confusing both of us ;)

observed values are 587, 1962
expected values are 1274.5, 1274.5



my math is (o-e)2 over e
so (587-1274.5)2 equals 472656.25 over 1274.5
(1962-1274.5)2 equals 472656.25 over 1274.5

472656.25+472656.25 equals 945312.5 over 1274.5
which returns a p value of 741.71243625

HEEEEEEEEEEEEELP!!!
 
The critical value you calculated is correct and based on that you should reject the null hypothesis.

What's the problem?
 
Last edited:
Salsa said:
What is the total number of observations? Just two?

Not a big enough sample really and the test will be pretty meaningless...

Someone asked about chi sq a while back, the thread might help.

http://forums.overclockers.co.uk/showthread.php?t=18393471&highlight=chi+squared

Post your data and the question you are trying to ask perhaps?

/Salsa
Click to expand...

I think the sample size is irrelevant. He's just trying to find out if two numbers are significantly different or not. The test is not meaningless and perfectly legit.
 
Salsa said:
What is the total number of observations? Just two?

Not a big enough sample really and the test will be pretty meaningless...
Click to expand...

The number of observations does not necessarily equate to the sample size, and one degree of freedom is perfectly reasonable to give statistical significance.

For example, if you flipped a coin a thousand times to determine the statistical likelihood of returning a head, your sample size might be a thousand, but you would still only have 2 observation values (a total for heads and a total for tails) thereby having only 1 degree of freedom.
 
Mr.Cookie said:
Hi trying to help my wife with her stats data, but she's confusing me, I'm confusing her and excel is confusing both of us ;)

observed values are 587, 1962
expected values are 1274.5, 1274.5



my math is (o-e)2 over e
so (587-1274.5)2 equals 472656.25 over 1274.5
(1962-1274.5)2 equals 472656.25 over 1274.5

472656.25+472656.25 equals 945312.5 over 1274.5
which returns a p value of 741.71243625

HEEEEEEEEEEEEELP!!!
Click to expand...

Also, "741.71243625" is not your P value, its your Chi-sq value, you need to look this up on a Chisq table with your degrees of freedom (1 in your case) to find your P value.
 
cheers guys you rock, it's basically to see if there are behavioural differences in a herd when at pasture vs indoors, i think that was the grooming behaviour.
Sue is out at the moment so will post more when she gets back :)

Cheers :)
 
Sample size was a herd of 23 cattle which were scan sampled at 1 minute intervals for a total of 80 hours indoors and 80 outdoors.
23 different behaviours were sampled split into 7 categories (grooming, antagonistic, sedentary, locomotory, feeding, stereotypic and general), each category is being put through chi squared test to see if we reject or do not reject the hypothesis that there is a difference in behaviour at pasture and indoors.
Problem was that excels chitest came up with different numbers to me and can't understand why.

sum in excel =chisq.test(g3:h3,g4:h4)
 
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