The Forum Site - Join the conversation
Forums: Science:
Statistics

Hypothesis testing with multinomial distributions

Reply to Topic
AuthorMessage
Xeranox On January 10, 2010




Trondheim, Norway
#1New Post! Jan 06, 2010 @ 09:52:12
If you throw a die and record the outcomes, you can test if the die is 'fair' with a chi-test. But how can you use the same data to determine with a certain significance if, for instance, the probability of gatting a '6' is equal to the probability of getting a '1'.

More precisely: i want to test the null hypothesis that the probability of getting '6' is equal to the probability of getting '1'. How do i proceed?

Thanks a bunch in advance.
El_Tino On October 12, 2023
booyaka!





Albuquerque, New Mexico
#2New Post! Jan 06, 2010 @ 09:59:55
If your chi-square doesn't show anything then the probability of any one is the same as any other.
Xeranox On January 10, 2010




Trondheim, Norway
#3New Post! Jan 06, 2010 @ 10:02:48
My chi-test actually shows with a 95% confidence that the uniform distribution hypothesis in not correct. (Actually, this is not about throwing dice, but it is equivalent ).

So, any suggestions?
Xeranox On January 10, 2010




Trondheim, Norway
#4New Post! Jan 07, 2010 @ 14:22:35
Here's an update. I did some research:

I came across the following paper:

https://www.jstor.org/stable/2682861

The paper describes how to compare the ratios of two outcomes of a multinomial sample. The idea is to calculate a confidence interval of the odds ratio (rho) of two of the outcomes. The lower and upper limit of the confidence interval can be calculated as:

rho_ij_(low) = rho( alpha ; Xi ; Xj ) = {[( Xj + 1 ) / Xi ] * F( alpha/2 ; 2Xj + 2 ; 2Xi )}^(-1)
rho_ij_high = rho( alpha' ; Xi ; Xj ) = [( Xi + 1 ) / Xj ] * F( alpha'/2 ; 2Xi + 2 ; 2Xj )

where alpha is the significance level, Xi and Xj are the number of outcomes i and j, and F is the critical value of the F-distribution.

If the value '1', is not present in the confidence interval, we can discard the hypothesis that the ratios are equal. I'll provide an example as well:

Outcome 1: 13 times
Outcome 2: 2 times
Outcome 3: 9 times

The null hypothesis is that all three outcomes have the same probability (1/3). I calculate the lower and upper values of the odds ratio for all three cases with a 10% significance:

rho_12_min = 1.75
rho_12_max = 40.31
The null hypothesis is discarded with a 10% significance, and outcomes 1 and 2 do not have the same probability.

rho_13_min = 0.65
rho_13_max = 3.31
The null hypothesis is not discarded. Outcomes 1 and 3 have the same probability.

rho_23_min = 0.03
rho_23_max = 0.92
The null hypothesis is discarded with a 10% significance, and outcomes 2 and 3 do not have the same probability.

The corresponding 99% confidence intervals are:

rho_12 = [1.06 ; 139.7]
rho_13 = [0.44 ; 5.21]
rho_23 = [0.01 ; 1.65]

The outcomes 1 and 2 are different with a 1% significance.


Is this a feasible approach?


Here, I have tried the chi-test on the same data with two groups at a time:

X2_12 = [(13-7.5)^2 + (2-7.5)^2] = 8.067
Critical values:
10% significance: 6.64 ---> Discard H0
1% significance: 10.83 ---> Keep H0

X2_13 = [(13-11)^2 + (9-11)^2] = 0.72
10% significance: 6.64 ---> Keep H0

X2_23 = [(13-7.5)^2 + (2-7.5)^2] = 4.45
10% significance: 6.64 ---> Keep H0


As you can see, the conclusions are different depending on what test you choose. Is this because one (or both) of the tests are errenous, or are my calculations incorrect?

I would appreciate feedback on this matter.
Reply to Topic<< Previous Topic | Next Topic >>

1 browsing (0 members - 1 guest)

Quick Reply
Be Respectful of Others

      
Subscribe to topic prefs

Similar Topics
    Forum Topic Last Post Replies Views
New posts   Philosophy
Mon Sep 10, 2012 @ 23:27
54 4014
New posts   College Life
Mon Sep 26, 2011 @ 23:11
17 2521
New posts   Science
Sat Feb 21, 2009 @ 04:24
5 770
New posts   Statistics
Fri Dec 05, 2008 @ 04:37
0 1593
New posts   Math
Thu Jun 19, 2008 @ 02:39
5 986