Timeline for Solving 5 eqns with 6 unknowns in a 2x3 contingency matrix, is there a unique solution?
Current License: CC BY-SA 2.5
8 events
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| Jan 27, 2011 at 18:14 | vote | accept | David LeBauer | ||
| Jan 25, 2011 at 3:25 | comment | added | Will Jagy | Thanks, Gottfried. Comments can continue on closed questions. | |
| Jan 25, 2011 at 2:16 | comment | added | Gottfried Helms | @Will: "contingency table" is just a bivariate table of frequencies. And yes, a 2x3-contingency-table has 2 "degrees of freedom" , so 2 parameters (2 frequencies) are free and 6-4=2 frequencies are then determined. (But the question is closed, so this comment will likely not appear...; the question should have occured in stat.exchange instead) | |
| Jan 25, 2011 at 1:11 | comment | added | Gottfried Helms | @David: I don't understand your last question. A distribution is a property of a random-variable, that means something which can have many realizations. But here you have 6 variables in an arrangement of a contingency-table with given margins, you have 2 degrees of freedom: 2 variables (not in the same column) can vary continuously - (with some range-restriction when negative values are forbidden). If -for the varying a,b,c,..., - the resp. chi-square-value $\Chi$ is computed you'll compare it with a "chi-square-distribution for 2 degrees of freedom" | |
| Jan 25, 2011 at 0:40 | comment | added | David LeBauer | @Gottfried Thank you for your answer - now I understand @Will's point better. Perhaps I misused the term contingency table: the values do not have to be integers. Are you saying that if I assume that the variable has a chi-square distribution? | |
| Jan 24, 2011 at 23:25 | history | edited | Gottfried Helms | CC BY-SA 2.5 |
Formatting
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| Jan 24, 2011 at 23:13 | history | edited | Gottfried Helms | CC BY-SA 2.5 |
improving explanation; added 62 characters in body
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| Jan 24, 2011 at 23:05 | history | answered | Gottfried Helms | CC BY-SA 2.5 |