Timeline for Performing a chi-square goodness of fit test
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| May 11, 2020 at 15:58 | comment | added | Nicholas G | My calculation of degrees of freedom was off. The correct formula is pearsonTest[obs_List, exp_List] /;Dimensions[obs] == Dimensions[exp] := Block[{t}, t = Total[(Flatten@obs - Flatten@exp)^2/Flatten@exp] // N;{Rule["chisqr", t], Rule[pval, SurvivalFunction[ ChiSquareDistribution[ Times @@Table[Dimensions[exp][[i]] - 1, {i, Length@Dimensions@exp}]],t]]}]. | |
| May 11, 2020 at 15:05 | comment | added | Nicholas G | Because I want to compare 2-dimensional lists, I adjusted this code to pearsonTest[obs_List, exp_List]/;Dimensions[obs] == Dimensions[exp] :=Block[{t}, t = Total[(Flatten@obs - Flatten@exp)^2/Flatten@exp] // N;{Rule["chisqr", t],Rule[pval,SurvivalFunction[ChiSquareDistribution[Length[Flatten@exp] - 1],t]]}]. However, comparing 2x2 outcomes in Excel and this function gives different values, example observed {{4,6},{6,4}} and expected {{5,5},{5,5}}. Any idea what I am missing? | |
| May 15, 2012 at 19:07 | comment | added | Andy Ross | @FrederikBrinckJensen you are welcome. I recommend looking at the following post | |
| May 15, 2012 at 18:56 | comment | added | Frederik Brinck Jensen | Thanks for this wonderful reply Andy! Can you give me any reference on where to read about the functions and the syntax of mathematica more deeply? | |
| May 15, 2012 at 18:55 | vote | accept | Frederik Brinck Jensen | ||
| May 15, 2012 at 15:18 | history | edited | J. M.'s missing motivation | CC BY-SA 3.0 | cleaned up links |
| May 15, 2012 at 15:12 | comment | added | J. M.'s missing motivation | Hah. I was scratching my head trying to reconcile the stuff in my textbook with the stuff in Mathematica's docs. Thanks for this! | |
| May 15, 2012 at 15:08 | history | edited | Andy Ross | CC BY-SA 3.0 | added 23 characters in body |
| May 15, 2012 at 15:01 | history | answered | Andy Ross | CC BY-SA 3.0 |