R Function 获得均值差的置信区间

Question

I am trying find a function that allows me two easily get the confidence interval of difference between two means.

I am pretty sure t.test has this functionality, but I haven't been able to make it work. Below is a screenshot of what I have tried so far:

Image

This is the dataset I am using

   Indoor Outdoor
1    0.07    0.29
2    0.08    0.68
3    0.09    0.47
4    0.12    0.54
5    0.12    0.97
6    0.12    0.35
7    0.13    0.49
8    0.14    0.84
9    0.15    0.86
10   0.15    0.28
11   0.17    0.32
12   0.17    0.32
13   0.18    1.55
14   0.18    0.66
15   0.18    0.29
16   0.18    0.21
17   0.19    1.02
18   0.20    1.59
19   0.22    0.90
20   0.22    0.52
21   0.23    0.12
22   0.23    0.54
23   0.25    0.88
24   0.26    0.49
25   0.28    1.24
26   0.28    0.48
27   0.29    0.27
28   0.34    0.37
29   0.39    1.26
30   0.40    0.70
31   0.45    0.76
32   0.54    0.99
33   0.62    0.36

and I have been trying to use t.test function that has been installed from

install.packages("ggpubr")

I am pretty new to R, so sorry if there is a simple answer to this question. I have searched around quite a bit and haven't been able to find anything that I am looking for.

Note: The output I am looking for is Between -1.224 and 0.376

Edit:

The CI of difference between means I am looking for is if a random 34th datapoint was added to the chart by picking a random value in the Indoor column and a random value in the Outdoor column and duplicating it. Running the t.test will output the correct CI for the difference of means for the given sample size of 33.

How can I go about doing this pretending the sample size is 34?

Answer 1

there's probably something more convenient in the standard library, but it's pretty easy to calculate. given your df variable, we can just do:

# calculate mean of difference
d_mu <- mean(df$Indoor) - mean(df$Outdoor)
# calculate SD of difference
d_sd <- sqrt(var(df$Indoor) + var(df$Outdoor))

# calculate 95% CI of this
d_mu + d_sd * qt(c(0.025, 0.975), nrow(df)*2)

giving me: -1.2246 0.3767

mostly for @AkselA: I often find it helpful to check my work by sampling simpler distributions, in this case I'd do something like:

a <- mean(df$Indoor) + sd(df$Indoor) * rt(1000000, nrow(df)-1)
b <- mean(df$Outdoor) + sd(df$Outdoor) * rt(1000000, nrow(df)-1)
quantile(a - b, c(0.025, 0.975))

which gives me answers much closer to the CI I gave in the comment

Answer 2

Even though I always find the approach of manually calculating the results, as shown by @Sam Mason, the most insightful, there are some who want a shortcut. And sometimes, it's also ok to be lazy:)

So among the different ways to calculate CIs, this is imho the most comfortable:

DescTools::MeanDiffCI(Indoor, Outdoor)

Here's a reprex:

IV <- diamonds$price
DV <- rnorm(length(IV), mean = mean(IV), sd = sd(IV))
DescTools::MeanDiffCI(IV, DV)

gives

 meandiff    lwr.ci    upr.ci 
-18.94825 -66.51845  28.62195 

This is calculated with 999 bootstrapped samples by default. If you want 1000 or more, you can just add that in the argument R :

DescTools::MeanDiffCI(IV, DV, R = 1000)