PROTEUS

              

 

 

 

 

 

 

 

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#9
 
dates   couscous   local beer   river rafting   camels  
  ill well tot     ill well tot     ill well tot     ill well tot     ill well tot  
exp 8 10 18   exp 6 7 13   exp 3 4 7   exp 9 3 12   exp 8 10 18  
n 2 2 4   n 4 5 9   n 7 8 15   n 1 9 10   n 2 2 4  
tot 10 12 22   tot 10 12 22   tot 10 12 22   tot 10 12 22   tot 10 12 22  

OR: 0.80

(1.25 PROT)

   OR: 1.07  

OR: 0.86

(1.17 PROT)

   OR: 27  

OR: 0.80

(1.25 PROT)

  

The analysis must be done on the exposure having the highest odds ratio.  This is the river rafting (O:27).    ChiSq is not valid due to the expected cell sizes.   Fisher's Exact Test yields: P1: 0.0034 and P2: 0.0001 so overall P= 0.0035   The null hypothesis (of no association) can be rejected.  The probability of seeing the data arranged as presented in the absence of any association is 0.0035 or 0.35 percent or between 3 and 4 chances in a thousand.  Exposure to river rafting was therefore strongly associated with the illness. Those who were exposed to the river rafting had 27 times the risk of illness compared to those who were not so exposed.

NOTE: the question structured as an exam question only asks for a test of the null hypothesis on the exposure with the strongest odds ratio.   For practice, you are encouraged to complete the analysis for all the exposures.  The answers are given below:

                      dates: P=0.6316 ns     couscous: P=0.6393 ns    Local beer:   P=0.6161 ns    camels P=0.6316 ns

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I have posted (below) the exact keying sequence for the river rafting calculation.

step 1: determine that Chi-sq is not valid as I have at least one expected value of 5 or less (cell C: (10x10)/22 = 4.55

step 2: Calculate the P1:     10! x 12! x 10! x 12!      =    0.00340217

                                          9! x 3! x 1! x 9! x 22!

 

   (The exact keying sequence is 10 !  x  12 !  x  10 !  x  12 !  ∻ 9 !  ∻ 3 !  ∻ 1 !  ∻ 9!  ∻ 22 ! = 0.00340217

determine the dominant pair of cells: AD/BC = 81/3 = 27.  this demonstrated the A and D cells are "dominant" and will be increased by 1.  You have to leave the marginal totals unchanged, so that the non-dominant cells must be reduced by 1.   The calculation for the second probability now becomes  

                              P2:     10! x 12! x 10! x 12!      =    0.000102065

                                      10! x 2! x 0! x 10! x 22!

 

The overall probability is P1 + P2:  0.0035

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Note that I have not shown any canceling but this will simplify the calculation as follows

 

           P2:     10! x 12! x 10! x 12!      =       10! x 12! x 10! x 12!         =    0.000102065

                     10! x 2! x 0! x 10! x 22!          10! x 2! x 0! x 10! x 22!