In cases we want to test for two population means and the standard
deviations are different between the two populations, the so-called
**non-pooled t-test** or

The **non-pooled t-test** is very similar to
the

\[t = \frac{(\bar x_1 - \bar x_2)}{ \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\text{.}\]

The denominator of the equation from above is the estimator of the standard deviation of \(\bar x_1 - \bar x_2\), given by

\[s_{\bar x_1 - \bar x_2} = \sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}}\text{.}\]

The test statistic \(t\) has a
*t*-distribution and the degrees of freedom \((df)\) are given by

\[ df=\frac{\left(\frac{s_1^2}{n_1}+\frac{s_1^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1}+\frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}}\text{.} \]

When using look-up tables, round down the degrees of freedom to the nearest integer!

The **non-pooled t-test** is robust to moderate
violations of the normal population assumption, but it is less robust
regarding outliers (Weiss, 2010).

The \(100(1-\alpha)\) % confidence interval for \(\mu_1 - \mu_2\) is

\[(\bar x_1 - \bar x_2) \pm t^* \times \sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}}\]

where the value of *t* is obtained from the
*t*-distribution for the given confidence level. The degrees of
freedom \((df)\) and obtained using the
equation above.

In order to get some hands-on experience we apply the
**non-pooled t-test** in an exercise. Therefore,
we load the

`students.csv`

file here
or import it directly into R:`students <- read.csv("https://userpage.fu-berlin.de/soga/data/raw-data/students.csv")`

The *students* data set consists of 8239 rows, each of them
representing a particular student, and 16 columns, each of them
corresponding to a variable/feature related to that particular student.
These self-explaining variables are: *stud.id, name, gender, age,
height, weight, religion, nc.score, semester, major, minor, score1,
score2, online.tutorial, graduated, salary*.

In order to showcase the **non-pooled t-test**
we examine the mean annual salary (in Euro) of female graduates with
respect to their major study subject. The first population consists of
female students with their major in Political Science and the second
population of female students with their major in Social Sciences.

We start with data preparation.

- We subset the data set based on the variables
`gender`

and`graduated`

.

- We split the subset into graduates of Political Science and Social
Sciences (variable
`major`

).

- We sample from each group 50 students and extract the variable of
interest, the mean annual salary (in Euro), which is stored in the
column
`salary`

. We assign those two vectors to the variables`PS`

and`SS`

.

```
female_graduates <- subset(students, graduated == 1 & gender == "Female")
subset_PS <- subset(female_graduates, major == "Political Science")
subset_SS <- subset(female_graduates, major == "Social Sciences")
n <- 50
PS <- sample(subset_PS$salary, n)
SS <- sample(subset_SS$salary, n)
```

Further, we check if the data is normally distributed by plotting a
Q-Q plot. In R we apply the `qqnorm()`

and the `qqline()`

functions for plotting Q-Q plots.

```
par(mfrow = c(1, 2), mar = c(5, 4, 4, 2))
qqnorm(PS, main = "Q-Q plot for female graduates of \nPolitical Science (sample data)", cex.main = 0.75)
qqline(PS, col = 4, lwd = 2)
qqnorm(SS, main = "Q-Q plot for female graduates of \n Social Sciences (sample data)", cex.main = 0.75)
qqline(SS, col = 3, lwd = 2)
```

We can see, that the data of both samples falls roughly onto a straight line.

Let us assume, that the data of the *students* data set is a
good approximation for the population. Then, we may check visually if
the standard deviations of the two populations actually differ from each
another by plotting a box plot.

```
boxplot(subset_PS$salary,
subset_SS$salary,
horizontal = TRUE,
names = c("Political Science", "Social Sciences"),
xlab = "Annual salary in EUR",
main = "Population data"
)
```

Based on the graphical evaluation approach we conclude, that the data is roughly normally distributed and that the standard deviations differ from each other.

Recall the research question: **Do the data provide sufficient
evidence to conclude, that the mean annual salary of female graduates
with a major in Political Science differs from the mean annual salary of
female graduates with a major in Social Sciences?**

In order to conduct the non-pooled *t*-test we follow the
step-wise implementation procedure for hypothesis testing.

**Step 1: State the null hypothesis \(H_0\) and alternative hypothesis \(H_A\)**

The null hypothesis states that the average annual salary of female graduates with a major in Political Science (\(\mu_1\)) is equal to the average annual salary of female graduates with a major in Social Sciences (\(\mu_2\)):

\[H_0: \quad \mu_1 = \mu_2\]

Alternative hypothesis:

\[H_A: \quad \mu_1 \ne \mu_2 \]

This formulation results in a two-sided hypothesis test.

**Step 2: Decide on the significance level, \(\alpha\)**

\[\alpha = 0.05\]

`alpha <- 0.05`

**Step 3 and 4: Compute the value of the test statistic and the
p-value**

For illustration purposes we manually compute the test statistic in R. Recall the equations for the test statistic from above:

\[t = \frac{(\bar x_1 - \bar x_2)}{ \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]

```
# compute the value of the test statistic
n1 <- length(PS)
n2 <- length(SS)
s1 <- sd(PS)
s2 <- sd(SS)
x1_bar <- mean(PS)
x2_bar <- mean(SS)
t <- (x1_bar - x2_bar) / (sqrt(s1^2 / n1 + s2^2 / n2))
t
```

`## [1] 2.871071`

The numerical value of the test statistic is 2.871071.

In order to calculate the *p*-value we apply the
`pt()`

function. Recall how to calculate the degrees of
freedom:

\[ df=\frac{\left(\frac{s_1^2}{n_1}+\frac{s_1^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1}+\frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}}\text{,} \]

```
# compute df
df_numerator <- (s1^2 / n1 + s2^2 / n2)^2
df_denominator <- (s1^2 / n1)^2 / (n1 - 1) + (s2^2 / n2)^2 / (n2 - 1)
df <- df_numerator / df_denominator
df
```

`## [1] 89.30555`

```
# compute the p-value
# recall we are applying a two-sided test
upper <- pt(abs(t), df = df, lower.tail = FALSE)
lower <- pt(-abs(t), df = df, lower.tail = TRUE)
p <- upper + lower
p
```

`## [1] 0.005107913`

\(p = 0.0051079\).

**Step 5: If \(p \le \alpha\),
reject \(H_0\); otherwise, do not
reject \(H_0\)**

`p <= alpha`

`## [1] TRUE`

The *p*-value is less than the specified significance level of
0.05; we reject \(H_0\). The test
results are statistically significant at the 5 % level and provide very
strong evidence against the null hypothesis.

**Step 6: Interpret the result of the hypothesis
test**

At the 5 % significance level the data provides very strong evidence to conclude, that the average annual salary of female graduates of Politcal Science differs from the average annual salary of female graduates of Social Sciences.

We just completed a non-pooled *t*-test in R manually. Now, we
make use of the full power of the R machinery to obtain the same result
as above using just one line of code!

Exercise: Repeat the above example by applying the`t.test()`

function in order to conduct a non-pooledt-test in R!

`### your code here`

We provide two vectors as input data. Further, we set
`var.equal = FALSE`

in order to explicitly state, that we
apply the non-pooled version of the *t*-test. We do not need to
set the `alternative`

argument, since the default value
corresponds to our alternative hypothesis \(H_A: \; \mu_1 \ne \mu_2\).

`t.test(x = PS, y = SS, var.equal = FALSE)`

```
##
## Welch Two Sample t-test
##
## data: PS and SS
## t = 2.8711, df = 89.306, p-value = 0.005108
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 1058.655 5816.534
## sample estimates:
## mean of x mean of y
## 33810.87 30373.27
```

Super powerful! Compare the output of the `t.test()`

function with our result from above. They match perfectly! Again, we may
conclude that at the 5 % significance level the data provides very
strong evidence to conclude, that the average annual salary of female
graduates of Politcal Science differs from the average annual salary of
female graduates of Social Sciences.