Help with statistics for treatment control analysis using R
1
0
Entering edit mode
5.2 years ago
Gene-ticks ▴ 10

Hi All, I have cancer and WT controls samples from two different groups for size comparison. I would like to do some statistics to calculate p-value and perhaps get some plots. I am not very familiar with statistical analysis and was pondering if someone could teach me how to analyze this type of data. Thanks for your time. my data:

df <- structure(list(Group = c(1L, 1L, 1L, 2L, 2L, 2L, 2L), cancer = c(0.7, 
0.7, 0.6, 0.65, 1, 0.75, 0.3), WTcontrol = c(1.1, 0.8, 0.7, 1.4, 
1, 1, 1.05)), .Names = c("Group", "cancer", "WTcontrol"
), class = "data.frame", row.names = c(NA, -7L))
R • 1.6k views
ADD COMMENT
4
Entering edit mode
5.2 years ago

What is your question, and what are you trying to analyze? I see a Group vector, a cancer vector and a control vector. Are you trying to identify if the values in the cancer vector are statistically different from the values in the control? Are each vectors supposed to be divided by "Group", as in a pre-drug treatment is Group 1 and after drug treatment is Group2? Are the samples paired between WT and cancer?

Those are important questions for analysis. I can probably suggested some tests and methods for you to look at, but you should provide more information about what you want to analyze.

There's a good chance what you want to do it a t-test, so I would take a look at: https://uc-r.github.io/t_test
and http://rstudio-pubs-static.s3.amazonaws.com/332835_b96d3bd2ce4b416f9ebfa8d7664e8e13.html

R has a lot of built-in plotting functions but I would recommend investing some time in learning ggplot2: https://ggplot2.tidyverse.org/ There's a learning curve but the pay-off of learning ggplot2 well is huge.

ADD COMMENT
0
Entering edit mode

Actually, Group 1 and Group 2 are the experiments performed at two time points. The measurement is the diameter of the lesions. Yes, the samples are paired.

ADD REPLY
1
Entering edit mode

So since your samples are paired, I assume you want to know if Group2 is any different from Group1 when comparing the cancer sample with the WT sample. I have no idea if that makes sense experimentally (which is why I suggest you come up with a clear answer to what is it that you want to analyze? What are you trying to quantify?), but you could compute a ratio of lesion between matched samples, and do a linear model to see if the ratio is affected by the groups.

> df$ratio = df$cancer / df$WTcontrol
> df$Group
[1] 1 1 1 2 2 2 2
Levels: 1 2
> summary( lm(formula = ratio ~ Group, data = df) )

Call:
lm(formula = ratio ~ Group, data = df)

Residuals:
       1        2        3        4        5        6        7 
-0.15314  0.08550  0.06764 -0.16071  0.37500  0.12500 -0.33929 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)   
(Intercept)   0.7895     0.1489   5.303  0.00319 **
Group2       -0.1645     0.1970  -0.835  0.44168   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.2579 on 5 degrees of freedom
Multiple R-squared:  0.1224,    Adjusted R-squared:  -0.05309 
F-statistic: 0.6975 on 1 and 5 DF,  p-value: 0.4417

If you look under coefficients for "Group2", you'll see an "estimate" (effect size of Group2 versus Group1), a t-value and a p-value. It seems like the difference is not significant (p=0.44168), when analyzed this way at least.

You should read more on linear regression in R.

Since you asked about visualization, you can plot your raw data quickly with this code:

library(ggplot2)
ggplot(df) + 
  geom_point(mapping=aes(x=cancer, y=WTcontrol, color=Group), size=5)

enter image description here

Visually it doesn't look like there's any correlation between Cancer and WT (i.e. control always more or less the same at time point 2 except for one outlier while cancer varies at lot. Group 1 seems a bit more consistent between the two pairs but with only 3 data points it's hard to say much about it.)

ADD REPLY

Login before adding your answer.

Traffic: 2635 users visited in the last hour
Help About
FAQ
Access RSS
API
Stats

Use of this site constitutes acceptance of our User Agreement and Privacy Policy.

Powered by the version 2.3.6