Suppose I had 4 treatments done to mice to determine when normal bowel movement would occur faster or as near to treatment (a).
a) no treatment, negative control
b) antibiotics only
c) antibiotics with probiotic A
d) antibiotics with probiotic B
The setup goes this way: I did an estimation of population mean to find a range, based on the (a) group, where my mean rate of droppings could be, and used that to determine whether a mice pooped within that range. The test will be a yes-no test. But what then? The number of mice that could normalize bowel movement would increase overtime and it may take long so I may have to cut short the experiment and make do with my temporal data, but (Q# 1) how do I go about analyzing said data. Sorry for the long explanation. But furthermore, taking into account that there is the (b) group which is antibiotic only, what would happen if this group normalizes bowel movement earlier than the with-probiotics groups (Q#2) will there still be further statistical analysis to be done or will that just automatically result to failure on the part of the assumption that the two probiotics had beneficial effects on the mice bowel movement.
I do not know if I understood exactly. I would ask some questions previously: are there repilicates of the treatments in the assay? I suppose not. Have you tested the normality of the data? (for example Kolmogorov-Smirnov) If the answer is yes, I suggest to use parametric tests, if is not you should use non-paramtetric tests. For parametric tests: a "t" Student test is correct to compare two dataset; the "t" test can be used also to compare paired data. For non-parametric tes you can use the Wilcoxon test for paired data or the Mann-Whitney test sum of ranks. The comparisons are two by two and the results say if there are differences or not between populations.
ISSUE: Effect measurement using group A as a reference group (control) and treatment group B, C and D
TWO GROUPS OF TESTS: There are two groups: (i) control vs. treatment; (ii) treatment antibiotic vs. antibiotic + probiotics
CONTROL (A) VS. TREATMENT B, C & D: Here use the control group as the reference group. use the mean of the control group as the point of reference to verify treatment efficacy.
X*A = mean of A where no treatment is used
XBi = xb1, xb2, ..., xbn
XCi = xc1, xc2, ..., xcn
XDi = xd1, xd2, ..., xdn
Set you significance level, say CI = 0.95; this means the critical Z(0.95) = 1.65. See attached Z table. Find the Z score for Group B, C and D using the standard score equation:
(1) ZB = (XBi - X*A) / SB
... where XBi are individual reading for each mouse in group B (antibiotics only); X*A = mean of value of mice in group A (no treatment); S = standard deviation of observed value for group B. this task could easily be accomplished by Excel spread sheet. Mark cell that has Z-score larger than 1.65 as significant. repeat the same process to obtain readings for ZC and ZD.
This reading tells you the significant result for the individual mice in the group. It does not tell you whether the treatment shows significant result for the group: B, C and D (as a whole group). In order to verify group significance, we need to score the significant individual reading as 1 (success) and 0 (failure). the Laplace Rule of Success follows.
LAPLACE RULE OF SUCCESS: With the significant individuals scored as 1, all all the 1's; this is total success or s. The Laplace probability is given by:
(2) p = (s + 10 / (n + 2)
The probability of failure;
(3) q = 1 - p
NOW, with the known probability of success (significant individual reading in a group), is that significant for the group, i.e. is the treatment significant for the group? To answer this question, we need to use de Moivre-Laplace theorem (this is assuming that we the sample is large enough):
(4) Z(obs) = (S - np) / SQRT(npq)
... where S = total success in the group; n = sample size; p = probability of success int he group; and q = probability of failure in the group. H0: Z(obs) < 1.65 means not significant; and HA: Z(obs) > 1.65 means that the treatment is significant for the group. For the number of success we had, is it significant? Let X = number of success, then:
(5) Z = (X / n) - p] / SQRT(npq)
The decision under hypothesis statements are the same as above.
GROUPS COMPARISON: Groups B, C and D could also be paired and verify if they are significant different after treatment. Here, the possible pairs are: BC, BD and CD. Let take BC pair as an example. B = b1, b2, ..., bn and C = c1, c2 ..., cn. Assume that the sample size is equal, the paired difference are;
d1: (b1 - c1)
d2 = (b2 - c2)
...
d3 = (bn - cn)
The mean of the difference is denoted as d* and the standard deviation of the series di is denoted as Sd. Then the significance difference of the two arrays is given by:
(6) Td = d* / [Sd / SQRT(n)]
Use the T-table for deciding if the treatments produces different result for group B and C. Repeat the procedure for other pairs: BD and CD.
This is a start. The design could be improve with actual data.
REFERENCES: See link and attached files.
http://fcc-statistics.wikispaces.com/file/view/100+Statistical+Tests.pdf