Basic - T-Test -> Grouping Factor Must have Exactly 2 Levels

Question

I am relatively new to R. For my assignment I have to start by conducting a T-Test by looking at the effect of a politician's (Conservative or Labour) wealth on their real gross wealth and real net wealth. I have to attempt to estimate the effect of serving in office wealth using a simple t-test.

The dataset is called takehome.dta

Labour and Tory are binary where 1 indicates that they serve for that party and 0 otherwise.

The variables for wealth are lnrealgross and lnrealnet.

I have imported and attached the dataset, but when I attempt to conduct a simple t-test. I get the following message "grouping factor must have exactly 2 levels." Not quite sure where I appear to be going wrong. Any assistance would be appreciated!

Magnitogorsk

Please add sample data and show your code (see these guidelines for making a reproducible example. – Yachting

Answer 1

are you doing this:

t.test(y~x)

when you mean to do this

t.test(y,x)

In general use the ~ then you have data like

y <- 1:10
x <- rep(letters[1:2], each = 5)

and the , when you have data like

y <- 1:5
x <- 6:10

I assume you're doing something like:

y <- 1:10
x <- rep(1,10)
t.test(y~x) #instead of t.test(y,x)

because the error suggests you have no variation in the grouping factor x

Exmoor

I just tried it with t.test(y,x) as opposed to t.test(y~x), and it worked! Thanks! – Magnitogorsk

what is the difference between ~ and , operators for this example? – Televise

hey @cuneytyvz to see the difference try: x <- rnorm(10); y <- 1:10; par(mfrow=c(1,2)); plot(y,x);plot(y ~ x) – Exmoor

Answer 2

The differences between ~ and , is the type of statistical test you are running. ~ gives you the mean differences. This is for dependent samples (e.g. before and after). , gives you the difference in means. This is for independent samples (e.g. treatment and control). These two tests are not interchangeable.

Dionysius

This is incorrect. The switch paired in the function call is what achieves the dependent/independent samples distinction, it's unrelated to whether you indicate your samples through two numeric vectors t.test(x, y, paired=T) or through a longer vector with a factor vector with two levels df <- data.frame(z = c(x,y), f = rep("a", "b", each = length(x))); t.test("z ~ f", paired = T, data = df). See this sthda easy guide on the topic – Donn

Answer 3

I was having a similar problem and did not realize given the size of my dataset that one of my y's had no values for one of my levels. I had taken a series of gene readings for two groups and one gene had readings only for group 2 and not group 1. I hadn't even noticed but for some reason this presented with the same error as what I would get if I had too many levels. The solution is to remove that y or in my case gene from my analysis and then the error is solved.

Drysalter

Answer 4

You have to have one independent variable with only two groups, for example group 1 and group 2. The independent variable is nominal. For example, you call the independent variable groups.

Then you also have the dependent variable, which is metric. For example, you have a dependent variable, which is weight. That is metric. The data frame is called numbers.

In R you type:

t.test(dataframe$dependent variable~dataframe$independent variable, var.equal = TRUE, alternative = "two.sided")

t.test(numbers$weight~numbers$groups, var.equal = TRUE, alternative = "two.sided")

If the variances are not equal you write:

var.equal = FALSE

If you have a directed hypothesis you write:

alternative = "greater"

or

alternative = "less"

Piroshki