Weighted Means: Part 2

Let’s look at the Bland and Kerry procedure, giving that we have calculated the weighted mean as reported in the previous post. We will see the direct application of the procedure, without details about theory that can be read directly at the bibliographic reference.

Suppose that we have the same data recalled in the previous post.  We use only the difference between Controls and C1, which has the lowest significance.

The weighted mean of controls is 69.8, that of C1 is 89.6.

Firstly, we calculate a term called “Weighted sum of observations squared” (s²)  which is dependent on the number of experiments (k):

s² = k*(S_{i=1 to k} X_i²*W_i)/ (S_{i=1 to k} W_i)

In our cases:

s² (C)=3*(62²*0.0682+72.6²*0.0693+83.6²*0.0243)/(0.0682+0.0693+0.0243)=14782.25

s² (C1)=3*(85.6²*0.0266+86.8²*0.109+99.6²*0.0425)/(0.0266+0.109+0.0425)=24218.09

Then, we have to calculate the correction term, Corr, simply calculated as:

Corr=k*X²_weighted

Corr(C)=3*69.8²=14616.12 and Corr(C1)=3*89.6²=24084.48

Then, we define “Sum of squares about the mean” (S²) as:

S²=s²-Corr =>

S²(C)=166.13 and S²(C1)=133.61

The weighted estimation of SD of each condition is simply:

SD_best=sqrt[S²/(K-1)] =>

SD_best (C) = 9.11 and SD_best(C1) = 8.17

Note that in this case we have some comments to do:

1)  The values are independent on the number of replicates, but depend on the SD values of the single experiments, being it a SD weighted mean;

2) The relationship between SD calculated here and those calculated with the previously presented method was approximately sqrt(5) [some rounding differences];

3)  How many degrees of freedoms we have? The most immediate thought is k-1 df, which is 2 in our case. However, we should take into account that each experiment has n samples (n-1 df). Therefore, in my opinion the correct way to take into account of df is: df=(k-1)*(n-1) if n is the same in all the experiments. In our case, df=8. Why the multiplication? Because we want that each experiment has a constraint, and therefore we introduce an AND among the assumptions on our experiments, and therefore the probabilities are multiplicative. It solves also the point 1.

Therefore,  we have to compare C=69.8 (SD: 9.11) and C1=89.6 (SD: 8.17) with 8 df per group, which implies a significance of p=0.0002, lower than that obtained with the calculations reported in my previous post. On the contrary, if we retain that n-1 should not be considered as df, we may use only k-1 df (2 in this case) with a very lower significance (p=0.049). In this case, the number of experiments should be well > 3 to control the beta error.

In the next post, we will summarize all the results.