Of course you will have decimals. That is not an issue, just go about the computation as normal.
Just sum ((o-e)^2/e) and use that test-stat to find the p-value.
Expected:
....DISORDER................. .....SEASO...
.........................Summ er...........
.......no..............23.333 33...........
.......yes............11.6666 7........1...
Thus, sum the (observed - expected )^ 2 / expected to get the test stat
(26-23.333333)^2 / 23.33333 + (24-23.33333)^2 / 23.333333 +........ = 3.621428572 (approximately)
The degrees of freedom = (r-1)(c-1) = (2-1)(4-1) = 3 (there are 2 disorder options and 4 seasons)
Then go to a chi-square table and try to find the 3.62 value for 3 degrees of freedom
The test stat is larger than the p-value for the Chi-Square of .25. This means it is NOT significant at the .05 level. (p-value = .3053).
You can also do this in your TI-83 or higher calculator.
Just enter the observed into a 2x4 MATRIX A
Just enter the expected into a 2x4 MATRIX B
Then STAT, TESTS, scroll down to chi-Square(option C) and hit enter 3 times.
Should give you the exact stuff I did.
Thank so much for your help .. the SPSS program I was using is on my desktop that crashed on me and I have two weeks left to go.
Hope this helps...
I also corrected the title of the thread for you so that it's in the correct format of a question.
Thank you so much (so I am to basically post the first question in the title) I am starting to get it now. I am just panicked because my computer is down and I have two weeks left.
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