Comparing groups within the same experiment

Question

I've performed an experiment wherein I have stimulated cells (cell line) with a drug across different time-points - in its unmodified (Drug) or modified form (H-Drug or M-Drug).

All cells are seeded at the same concentration and all are harvested for analysis on the same day.

Method:

Cells were seeded on the same day. Cells were stimulated at 3:00 on a Wednesday (24 hour timepoint). Cells were then stimulated at 09:00 am on Thursday (6 hour timepoint) followed by 12:00 pm Thursday (3 hour timepoint) and so on until 3:00 pm on Thursday - at this time ALL cells were harvested and put into separate tubes according to what they were stimulated with and when e.g. 3 hour M-drug.

All tubes were analysed in sequence to measure the percentage of cells in that test tube that stained positive for that drug (i.e. as a means of measuring drug uptake)

Question: I have measured the percentage uptake of a drug at different time-points (0.5 - 24 hours) for the three drug variants. % uptake is taken as the % of cells that stain positive for that drug

What would the correct stat test be? Would a 3-way ANOVA make sense?

Answer 1

According to the updated description, I see two factors:

$\delta:$ Type of drug with three levels (U, M, H), and

$\tau:$ Time period with five levels (24, 6, 3, 1, 1/2).

Data are $Y_{ij}$ percentage stained positive.

Then a standard two-way ANOVA model (without interaction) would be $$Y_{ij} = \mu + \delta_i + \tau_j + e_{ij},$$ for $i = 1,2,3$ and $j = 1,2,3,4;$ and $e_{ij}$ are independently normal with mean $\mu$ and SD not depending on $i$ or $j,$ (estimated below as $\sigma = 9.8).$

Output from Minitab statistical software:

ANOVA: Resp versus Drg, Tim 

Factor  Type   Levels  Values
Drg     fixed       3  1, 2, 3
Tim     fixed       5  1, 2, 3, 4, 5

Analysis of Variance for Resp

Source  DF       SS      MS     F      P
Drg      2   596.93  298.47  3.14  0.099
Tim      4  2517.33  629.33  6.62  0.012
Error    8   761.07   95.13
Total   14  3875.33

S = 9.75363   R-Sq = 80.36%   R-Sq(adj) = 65.63%

Means

Drg  N    Resp
1    5  75.400
2    5  82.000
3    5  66.600

Tim  N    Resp
1    3  52.000
2    3  71.333
3    3  76.667
4    3  83.333
5    3  90.000

Time effect is significant at the 5% level (P-values < 0.05); Drug effect not significant. There is no 'Interaction' term in the model because we have only one percentage (Y-value) for each combination of Drug and Time. The data are sufficiently near normal because a points on a normal probability plot of the residuals (indicators of errors in the 15 'cells' of the data table) lie very nearly in a straight line.

The Fisher LSD ('least significant difference') method of comparing levels within Time factor is shown below:

Fisher Pairwise Comparisons: Response = Resp, Term = Tim 

Grouping Information Using Fisher LSD Method and 95% Confidence

Tim  N     Mean  Grouping
5    3  90.0000  A
4    3  83.3333  A   B
3    3  76.6667  A   B
2    3  71.3333      B
1    3  52.0000         C

Means that do not share a letter are significantly different.

That is, there is a clear trend of difference due to time. Specifically, Level 1 (.5hr) is significantly smaller than the others; Level 5 (24hr) is significantly larger than Level 2 (1hr) or 1.

Note: The Tukey HSD method of multiple comparisons shows fewer significant differences. It is 'more conservative' (less likely to declare significant differences). However, I believe the Fisher LSD method is appropriate for this experiment.