Welch's t-Test and Paired Data
-
Suppose that $X_{1,1},X_{1,2},…,X_{1,n_1}$ is a random sample of size $n_1$ from the normal distribution with mean $\mu_1$ and variance $\sigma_1^2$.
-
Suppose that $X_{2,1},X_{2,2},…,X_{2,n_2}$ is a random sample of size $n_2$ from the normal distribution with mean $\mu_2$ and variance $\sigma_1^2$.
-
Suppose that $\sigma_1^2$ and $\sigma_2^2$ are unknown and that the samples are independent.
-
Don't assume
that $\sigma_1^2$ and $\sigma_2^2$ are equal!
-
There is no known exact test.
-
This is known as the Behrens-Fisher problem.
- The most popular approximates solution is given by Welch's t-test
Welch says that:
\[\frac{\overline{X}_1-\overline{X}_2-(\mu_1-\mu2)}{\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}}\]has an approximate t-distribution with r degrees of freedom where
\[r = \frac{S_1^2/n_1+S_2^2/n_2}{\frac{(S_1^2/n_1)^2}{n1-1}+\frac{(S_2^2/n_2)^2}{n2-1}}\]rounded down.
Example: (In R)
x <- c(1.2, 3.2, 2.7, 1.6, 2.1)
y <- c(4.2, 0.8, 2.2, 2.3, 1.5, 3.0)
t.test(x,y)
Welch Two Sample t-test
data: x and y
t = -0.28741, df = 8.742, p-value = 0.7805
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.543768 1.197102
sample estimates:
mean of x mean of y
2.160000 2.333333
# Calulate P-value
2*pt(-0.28741, 8.742)
0.780494693057154
Example
A random sample of 6 students’ grades were recorded for Midterm 1 and Midterm 2.
Assuming exmam scores are nomally distributed, test whether the true (total population of students) average grade on Midterm 2 is greater than Midterm 1.
The Data
| Student | Midterm 1 Grade | Midterm 2 Grade |
|---|---|---|
| 1 | 72 | 81 |
| 2 | 93 | 89 |
| 3 | 85 | 87 |
| 4 | 77 | 84 |
| 5 | 91 | 100 |
| 6 | 84 | 82 |
The data are "paired".
First we compute the differences:
| Student | Midterm 1 Grade | Midterm 2 Grade | Differences:</font> M1-M2 |
|---|---|---|---|
| 1 | 72 | 81 | 9 |
| 2 | 93 | 89 | -4 |
| 3 | 85 | 87 | 2 |
| 4 | 77 | 84 | 7 |
| 5 | 91 | 100 | 9 |
| 6 | 84 | 82 | -2 |
The Hypotheses:
Let $\mu$ be the true average difference for all students.
\[H_0:\mu=0\\ H_1:\mu>0\]Data: $9, -4, 2, 7, 9, -2$
\[\sum x_i = 23\qquad\sum x_i^2=267\qquad n=6\]This is simply a one sample t-test on the differences.
$\overline{x}=3.5$
\[s^2=\frac{\sum x_i^2-(\sum x_i)^2/n}{n-1}=32.3\]# QQ Plot
data = c(9, -4, 2, 7, 9, -2)
qqnorm(data)
qqline(data)
The QQ Plot looks pretty linear!
$t_{\alpha,n-1}=t_{0.05,5}=2.01$
Reject $H_0$, in favor of $H_1$, if
\[\underbrace{\overline{X}}_{3.5}\gt\underbrace{\mu_0+t_{\alpha,n-1}\frac{S}{\sqrt{n}}}_{4.66}\]Conclusion
We fail to reject $H_0$, in favor of $H_1$, at 0.05 level of significance.
These data do not indicate that Midterm 2 scores are higher than Midterm 1 scores.