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Author Question: The chi-square test for independence makes use of the fact that the probability of the intersection ... (Read 58 times)

Katmoss16

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on: Feb 26, 2023
The chi-square test for independence makes use of the fact that the probability of the intersection of two independent events denoted A and B [i.e.,\(\style{font-family:Times New Roman;}{P(A\cap B)}\)] is:

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the sum of the probability of each event:\(\style{font-family:Times New Roman;}{P\left(A\right)+P(B)}\).


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the quotient of the probability of event A divided by B:\(\style{font-family:Times New Roman;}{P(A)/P(B)}\).


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the product of the probability of each event:\(\style{font-family:Times New Roman;}{P\left(A\right)\ast P(B)}\).


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the difference of the probability of the events:\(\style{font-family:Times New Roman;}{P\left(A\right)-P\left(B\right)}\).

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Marked as best answer by Katmoss16 on Feb 26, 2023

dsd1212

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Reply #1 on: Feb 26, 2023
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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Katmoss16

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Reply 2 on: Feb 26, 2023
Great answer, keep it coming :)

kishoreddi

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Reply 3 on: Yesterday
Thanks for the timely response, appreciate it

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