We’ll take some simple data from Motulsky, enter it into R by hand, and do a t-test. This will require making two R “objects” to hold the data for us.
Motulsky 2nd Ed, Chapter 30, page 220, Table 30.1. Maximal relaxaction of muscle strips of old and young rat bladders stimualted w/ high concentrations of nonrepinephrine (Frazier et al 2006). Response variable is %E.max
## Warning: package 'knitr' was built under R version 3.3.2
| old | young |
|---|---|
| 20.8 | 45.5 |
| 2.8 | 55 |
| 50.0 | 60.7 |
| 33.3 | 61.5 |
| 29.4 | 61.1 |
| 38.9 | 65.5 |
| 29.4 | 42.9 |
| 52.6 | 37.5 |
| 14.3 |
Load each column of data into a seperate R object
HINTS: You’ll need need to use:
#TYpe your attempt here:
HINT: The basic form of the code is: * R.object <- c(datum1, )
#Type your attempt here:
HINT: To make an object with just the first “old” datum do this
old.E.max <- c(20.8)
HINT: To make another object with just the first “young” datum do this
young.E.max <- c(45.5)
#Old data
old.E.max <- c(20.8,2.8,50.0,33.3,29.4,38.9, 29.4,52.6,14.3)
#Young data
young.E.max <- c(45.5,55.0, 60.7, 61.5, 61.1, 65.5,42.9,37.5)
Perform a t-test to compare the means of these 2 samples. This requires
Scroll down for the answer…
#Type your attempt here:
t.test(old.E.max, young.E.max)
##
## Welch Two Sample t-test
##
## data: old.E.max and young.E.max
## t = -3.6242, df = 13.778, p-value = 0.002828
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -37.501081 -9.590586
## sample estimates:
## mean of x mean of y
## 30.16667 53.71250