0
$\begingroup$

I've been learning about the t-distribution for my stats class and while running some simulations in R to explore the concept, I've run into an issue. I don't know whether it arises from my code, or whether I have seriously misunderstood how t-distributions arise.

Below I've pasted my code and the graph it gives me. The dataset it works on is heights from the dslabs package. It's a dataframe that contains two normally distributed recordings of heights, one for men and one for women. In total there are 1050 observations. I ran a Monte Carlo simulation 1000 times, each time sampling 10 observations randomly from the data, got the sample mean $\bar{x}$ and sample standard deviation $s$. The t-value is then $t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}$ and it should have a t-distribution about 0. $\mu$ is the population mean of heights. I made a density plot of these t-values and also 1000 randomly generated normally distributed numbers with $sd= \frac{\sigma}{\sqrt{n}}$. This is the standard deviation of the distribution of sample means of samples of size n. It should be normally distributed.

Here's my code:

n <- 5
tvalues <- replicate(10000,{
    X <- heights[sample(1:1050,size=n),] %>% pull(height)
    Xbar <- mean(X)
    s <- sd(X)
    tvalue <- (Xbar-mu)/(s/sqrt(n))
})
df1 <- data.frame(values=tvalues,category="tvalues")
normal_values <- rnorm(1000,mean=0,sd=sigma/sqrt(n))
df2 <- data.frame(values=normal_values,category="normal sample mean distribution")
df <- rbind(df1,df2)
df$category <- as.factor(df$category)
df %>% ggplot(mapping=aes(values,color=category)) + geom_density(bw=1)

and the graph I get enter image description here

Shouldn't the t-distribution have fatter tails than the normal?

$\endgroup$

1 Answer 1

1
$\begingroup$

Yes, the t-distribution has heavier tails than the standard normal distribution. But, in your figure, you are comparing it with a normal distribution with standard deviation $\frac{\sigma}{\sqrt(n)} \ne 1$. If you change that to 1 and re-draw you should see what you expect.

$\endgroup$
1
  • $\begingroup$ Thanks, this makes sense, I was calculating t-values, and they need to be compared to the standard normal. t-values vs z-values. $\endgroup$
    – Vakarrian
    Commented Feb 3, 2021 at 12:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.