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Which hypothesis test should I use

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  • #46794

    Groll
    Participant

    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 ?

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    #155107

    indresh
    Participant

    Chi

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    #155109

    indresh
    Participant

    Chi square

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    #155110

    Craig
    Participant

    It 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.

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    #155108

    indresh
    Participant

    Chi square

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    #155114

    Groll
    Participant

    As per theory is concern Chi-square will be suited here. But chi-Square always need Observed and Expected. But i am having observed data only.

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    #155118

    Allthingsidiot O
    Participant

    Then  try  to  collect  some Expected  data?

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    #155124

    Calculate expected
    Participant

    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.
     

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    #155127

    sophos9
    Member

    The null hypothesis of a chi-square 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?

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    #155133

    Dan Chauncey
    Participant

    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.

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    #155148

    Mikel
    Member

    You 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 one-way ANOVA.

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    #155151

    roadrage01
    Member

    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.
    Chi-Square 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
    Chi-Sq = 8.005+4.002+0.119+0.060+10.076+5.038 = 27.30
    DF = 2, P-Value = 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 =46.67

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