Homework 6
My dataset is (loosely) based on an experiment I ran where I used centrifugation to get cells to uptake exogenous mitochondria. Thus, I have two groups, one that did not receive any extra mitochondria, and one that did. The metric I am using here is the total number of mitochondria per cell that were counted when the cells were imaged.
In my real dataset from this experiment, I have six replicates of each condition. The mean number of mitochondria in the the untreated cells was about 130 with a standard deviation of about 17, whereas the treated cells had about 230 mitochondria with a larger standard deviation of 54.
Using these numbers and the rnorm function, I made fake data sets for each of these conditions, then put the data into a data frame.
# make data
no_mito <- rnorm(6, mean=130, sd=17)
plus_mito <- rnorm(6, mean=230, sd=54)
# put data into a dataframe
Number_of_mitochondria <- c(no_mito,plus_mito)
Treatment_condition <- c(rep("no mito", length(no_mito)),rep("plus mito",length(plus_mito)))
mito_frame <- data.frame(Number_of_mitochondria, Treatment_condition)
print(mito_frame)## Number_of_mitochondria Treatment_condition
## 1 101.5865 no mito
## 2 146.5586 no mito
## 3 110.9962 no mito
## 4 131.0810 no mito
## 5 112.3225 no mito
## 6 120.9713 no mito
## 7 104.3664 plus mito
## 8 217.2279 plus mito
## 9 213.9424 plus mito
## 10 349.7303 plus mito
## 11 256.1358 plus mito
## 12 307.4515 plus mitoNext, I checked the mean and standard deviation of the data:
## [1] 120.586## [1] 16.16231# find mean and standard deviation of plus_mito
mean(mito_frame[(length(no_mito)+1):(length(no_mito)+length(plus_mito)),1])## [1] 241.4757## [1] 85.36015Since there are only two groups her I performed a T-test, as well as making a simple box plot to visualize the data.
# Perform a T-test
stat <- t.test(Number_of_mitochondria ~ Treatment_condition, data = mito_frame)
print(stat$p.value)## [1] 0.01711526#Make a boxplot
boxplot(Number_of_mitochondria~Treatment_condition,data=mito_frame, main="Mito Transfer Data",
xlab="Treatment", ylab="Number of Mitochondria")
Because I was already seeing a significant p-value at 6, I wrote a for loop to see if a lower replicate number with the same effect size would still be significant.
replicates <- 3:6
for (i in replicates){
no_mito <- rnorm(i, mean=130, sd=17)
plus_mito <- rnorm(i, mean=230, sd=54)
Number_of_mitochondria <- c(no_mito,plus_mito)
Treatment_condition <- c(rep("no mito", length(no_mito)),rep("plus mito",length(plus_mito)))
mito_frame <- data.frame(Number_of_mitochondria, Treatment_condition)
stat <- t.test(Number_of_mitochondria ~ Treatment_condition, data = mito_frame)
cat("Replicates=",i,"P-value=")
print(stat$p.value)
}## Replicates= 3 P-value=[1] 0.09907871
## Replicates= 4 P-value=[1] 0.04396656
## Replicates= 5 P-value=[1] 0.003518244
## Replicates= 6 P-value=[1] 0.00553439(Iām sure there is a more elegant way to do this, but this was what I was able to figure out)
After running this a few times, it appears that with this effect size, it takes at least four replicates for the data to consistently be significant, though down to 3 is occasionally significant.
Next, I wanted to check if a smaller effect size would decrease the significance. For this I decided to go back to six as the number of replicates, and keep the average number of mitochondria in the untreated the same while increasing the effect size.
mean_dif <- seq(from=10, to=100, by=10)
for (i in mean_dif){
no_mito <- rnorm(6, mean=130, sd=17)
plus_mito <- rnorm(6, mean=(130+i), sd=54)
Number_of_mitochondria <- c(no_mito,plus_mito)
Treatment_condition <- c(rep("no mito", length(no_mito)),rep("plus mito",length(plus_mito)))
mito_frame <- data.frame(Number_of_mitochondria, Treatment_condition)
stat <- t.test(Number_of_mitochondria ~ Treatment_condition, data = mito_frame)
cat("Mean Difference=",i,"P-value=")
print(stat$p.value)
}## Mean Difference= 10 P-value=[1] 0.8201889
## Mean Difference= 20 P-value=[1] 0.3272942
## Mean Difference= 30 P-value=[1] 0.01278585
## Mean Difference= 40 P-value=[1] 0.05829815
## Mean Difference= 50 P-value=[1] 0.0007931119
## Mean Difference= 60 P-value=[1] 0.1167165
## Mean Difference= 70 P-value=[1] 0.3311995
## Mean Difference= 80 P-value=[1] 0.004871051
## Mean Difference= 90 P-value=[1] 0.03312288
## Mean Difference= 100 P-value=[1] 0.002603946Running this for loop a few times suggested that the effect size must be an increase of at least 70 mitochondria per cell to be consistently significant.