Which hypothesis test should I use
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 This topic has 11 replies, 9 voices, and was last updated 15 years, 7 months ago by roadrage01.

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April 23, 2007 at 10:43 am #46794
I am having my Y and X both discrete. My Y is No. of denied claims and X’s are various reason of denials. I want to test my hypothesis which is ‘ These is no statistical between my Y and X. I would like to know which test should i choose and how should i apply that test.
For example
Denied claims ( Y) Reason of Denial1 (X’s) ROD2( X’s) ROD3(X)
25 10 10 5
30 20 10 0
15 7 5 3
So on and so forth. Which test should i choose ?0April 23, 2007 at 11:04 am #155107
indreshParticipant@indresh Include @indresh in your post and this person will
be notified via email.Chi
0April 23, 2007 at 11:04 am #155109
indreshParticipant@indresh Include @indresh in your post and this person will
be notified via email.Chi square
0April 23, 2007 at 11:05 am #155110It looks like your data would fall under a Poisson or hypergeometric distribution. In the casae of a Poisson distribution, you would define “unit” as the group of denied claims. You could compare the different reasons for denial against each other using a hypothesis test, or you could create a C chart for each type of reason for denial.
Read up on the assumptions for Poisson and Hypergemoetric. My foggy memory has something pertaining to sampling without replacement nagging at me, and hence the Hypergeometric would be applicable. I dont recall the test statistic to use in either case, but you could easily find them when you reseach on the Poisson and Hypergeometric.0April 23, 2007 at 11:07 am #155108
indreshParticipant@indresh Include @indresh in your post and this person will
be notified via email.Chi square
0April 23, 2007 at 11:28 am #155114As per theory is concern Chisquare will be suited here. But chiSquare always need Observed and Expected. But i am having observed data only.
0April 23, 2007 at 11:47 am #155118
Allthingsidiot OParticipant@AllthingsidiotO Include @AllthingsidiotO in your post and this person will
be notified via email.Then try to collect some Expected data?
0April 23, 2007 at 2:10 pm #155124
Calculate expectedParticipant@Calculateexpected Include @Calculateexpected in your post and this person will
be notified via email.Randy,
Calculate your expected data. It’s easy. You assume that there is no relationship to the variables and assign expectd frequencies proportanately.
http://www.georgetown.edu/faculty/ballc/webtools/web_chi_tut.html
Read through this link (first Google I looked at, others may be better yet) and increase your understanding of the Chi Squared test.
0April 23, 2007 at 2:28 pm #155127The null hypothesis of a chisquare test isH0: There is no difference between proportions
Ha: There is a difference between proportionsThe Expected count is the result of the calculated value if all proportions were the sameDo you have access to Minitab?0April 23, 2007 at 3:05 pm #155133
Dan ChaunceyParticipant@DanChauncey Include @DanChauncey in your post and this person will
be notified via email.I am in agreement with “sophos9” The expected value is a calculated value. Minitab does it for you. Look at the example in Minitab, it should lead you down the right path.
0April 23, 2007 at 6:31 pm #155148You are setting up the problem wrong. Your Y,as defined by you, is just the sum of the different reasons. Your hypothesis will there is no difference in the reasons for denial and it will be a simple oneway ANOVA.
0April 23, 2007 at 8:21 pm #155151
roadrage01Member@roadrage01 Include @roadrage01 in your post and this person will
be notified via email.Pareto was my first thought, but no statisticak reference from that. So Chi Square it is. To set this up you take the total opportunities for each type denial to occur (I extracted this from you data) 70, then list be each denial reason the number of times for Yes denied for this reason and No, not denied for this reason.
ChiSquare Test
Expected counts are printed below observed counts
Reason Yes No Total
1 37 33 70 (Actual Values)
23.33 46.67 (Expected values)
2 25 45 70
23.33 46.67
3 8 62 70
23.33 46.67
Total 70 140 210
ChiSq = 8.005+4.002+0.119+0.060+10.076+5.038 = 27.30
DF = 2, PValue = 0.000
Yes there is a statistical difference in the occurences of each type denial. From the formula you can determine that Reason 1 is used more frequently than the model expected and Reason 3 is used less frequently more than expected. The expecteed values are derived from the numbers, there were a total of 70 denials of which each of the 3 reasons could have been used, for a total of 210 (70*3=210)opportunities. There were 70 actual denials of the 210 opportunities and since each of the three reasons had an equal opportunity to occur, the models expects to see each reason having a 33.3% chance of occuring (70/210=33.3%). 33.3% of the 70 Yes =23.33 (the expected value for Y on each reason. For the No (140/210= 66.67%). 66.67% of the 140 No =46.670 
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