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From |
Thomas Speidel <[email protected]> |

To |
[email protected] |

Subject |
st: Binomial Sign Test for Two Dependent Samples |

Date |
Fri, 14 Jan 2005 15:18:53 -0700 |

I am trying to create little program for a "Binomial Sign Test for Two Dependant Samples".

This test is somewhere between a signrank test and a binomial. The computations are fairly simple:

1. Find the differences among the pairs

2. Ignore all null differences

3. Count the number of non-zero, non-missing differences (i.e. n)

4. Count the number of positive differences (call it x)

5. Compute: Binomial(n, k, 0.5)

I am trying to implement this into a little program so that I can loop it through a number of variables/groups. However, because I am not an expert programmer, I came to a roadblock... Following is a sample of my data:

NOTE: *posneg: is the sign of the difference (i.e. 0=no difference; 1=Positive; -1=Negative)

+--------------------------------------------------+

|liverid group mp1posneg mp2posneg mp3posneg|

|--------------------------------------------------|

| 929444 1 0 -1 -1 |

| 948579 1 -1 0 -1 |

| 953012 1 -1 0 -1 |

| 956482 1 -1 0 -1 |

| 986833 1 . . . |

| 9314571 1 . . . |

| 9314751 1 . -1 . |

| 9413243 1 -1 -1 -1 |

| 9510420 1 . . . |

| 9613027 1 . . . |

| 9614920 1 -1 0 -1 |

| 9820601 1 -1 0 -1 |

| 9821645 1 0 0 -1 |

| 9912795 1 0 0 -1 |

|--------------------------------------------------|

| 12896 2 0 0 -1 |

| 94787 2 . -1 0 |

| 98613 2 . 0 . |

| 111381 2 . 0 -1 |

| 936134 2 0 0 0 |

| 969968 2 . -1 . |

| 991853 2 . . . |

| 9617199 2 -1 0 -1 |

| 9810799 2 . 0 -1 |

|--------------------------------------------------|

| 9338 3 . -1 . |

| 953132 3 -1 -1 -1 |

| 954005 3 -1 -1 -1 |

+--------------------------------------------------+

I am trying to find the binomial for each variable by group

This is what I have thus far, but does not work:

foreach x of var mp1posneg mp2posneg mp3posneg {

foreach i in 1 2 3 {

qui count if `x'posneg==1 & group == `i'

local pos_`i'`x' = r(N)

qui count if `x'posneg==1 | `x'posneg==-1 & group == `i'

local n_`i'`x'=r(N)

display Binomial(n_`i'`x', pos_`i'`x', 0.5)

}

}

Thanks,

Thomas

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**Follow-Ups**:**st: RE: Binomial Sign Test for Two Dependent Samples***From:*"Scott Merryman" <[email protected]>

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