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In the following code I use bootstrapping to calculate the C.I. and the p-value under the null hypothesis that two different fertilizers applied to tomato plants have no effect in plants yields (and the alternative being that the "improved" fertilizer is better). The first random sample (x) comes from plants where a standard fertilizer has been used, while an "improved" one has been used in the plants where the second sample (y) comes from.
x <- c(11.4,25.3,29.9,16.5,21.1)
y <- c(23.7,26.6,28.5,14.2,17.9,24.3)
total <- c(x,y)
library(boot)
diff <- function(x,i) mean(x[i[6:11]]) - mean(x[i[1:5]])
b <- boot(total, diff, R = 10000)
ci <- boot.ci(b)
p.value <- sum(b$t>=b$t0)/b$R
What I don't like about the code above is that resampling is done as if there was only one sample of 11 values (separating the first 5 as belonging to sample x leaving the rest to sample y). Could you show me how this code should be modified in order to draw resamples of size 5 with replacement from the first sample and separate resamples of size 6 from the second sample, so that bootstrap resampling would mimic the “separate samples” design that produced the original data?
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Bootstrap comes in handy when there is no analytical form or normal theory to help estimate the distribution of the statistics of interest since bootstrap methods can apply to most random quantities, e.g., the ratio of variance and mean. There are at least two ways of performing case resampling.
Bootstrapping describes a situation in which an entrepreneur starts a company with little capital, relying on money other than outside investments. An individual is said to be bootstrapping when they attempt to found and build a company from personal finances or the operating revenues of the new company.
This process allows you to calculate standard errors, construct confidence intervals, and perform hypothesis testing for numerous types of sample statistics. Bootstrap methods are alternative approaches to traditional hypothesis testing and are notable for being easier to understand and valid for more conditions.
An entrepreneur who risks their own money as an initial source of venture capital is bootstrapping. For example, someone who starts a business using $100,000 of their own money is bootstrapping.
EDIT2 : Hack deleted as it was a wrong solution. Instead one has to use the argument strata of the boot function :
total <- c(x,y)
id <- as.factor(c(rep("x",length(x)),rep("y",length(y))))
b <- boot(total, diff, strata=id, R = 10000)
...
Be aware you're not going to get even close to a correct estimate of your p.value :
x <- c(1.4,2.3,2.9,1.5,1.1)
y <- c(23.7,26.6,28.5,14.2,17.9,24.3)
total <- c(x,y)
b <- boot(total, diff, strata=id, R = 10000)
ci <- boot.ci(b)
p.value <- sum(b$t>=b$t0)/b$R
> p.value
[1] 0.5162
How would you explain a p-value of 0.51 for two samples where all values of the second are higher than the highest value of the first?
The above code is fine to get a -biased- estimate of the confidence interval, but the significance testing about the difference should be done by permutation over the complete dataset.
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