Two Sample T-Test
Two-Sample T-Test
In the paired t-test topic note we examined the difference in means between paired data. To compare the difference in means between two unpaired groups we will use the two-sample t-test. The two-sample t-test can be used to ask the following:
- Does the mean of group 1, \(\mu_1\), equal the mean of our group 2, \(\mu_2\)?
- \(H_0: \mu_1 = \mu_2\)
- \(H_a: \mu_1 \neq \mu_2\)
- Is the mean of group 1, \(\mu_1\), less than the mean of group 2, \(\mu_2\)?
- \(H_0: \mu_1 \le \mu_2\)
- \(H_a: \mu_1 > \mu_2\)
- Is the mean of group 1, \(\mu_1\), greater than the mean of group 2, \(\mu_2\)?
- \(H_0: \mu_1 \ge \mu_2\)
- \(H_a: \mu_1 < \mu_2\)
Tip
When we ask if mean of group 1 is equal to the mean of group 2 we are conducting a two-sided test. When we ask if the mean of group 1 is less than or greater than the mean of group 2, we are conducting a one-sided test.
Test Statistic
Our one-sample t-test statistic can be calculated by:
Explanation of Terms
- \(\mu_1\) : mean of group 1
- \(\mu_2\) : mean of group 2
- \(\sigma_1\) : standard deviation of group 1
- \(\sigma_2\) : standard deviation of group 2
- \(n_1\) : size of group 1
- \(n_2\) : size of group 2
- \(d.f.\) : degrees of freedom
Normal Distribution
Just like in the one-sample t-test topic note we will be comparing our test statistic to the normal distribution with a probability density function of:
Explanation of Terms
- \(\sigma\) : standard deviation
- \(\mu\) : mean
Confidence Interval
Just like our proportion tests, we also have a confidence interval around our sample parameter, in this case the sample mean. So for a test statistic, \(t\), at an \(\alpha\) level of 0.05, our confidence interval would be:
Explanation of Terms
- \(\mu_1\) : mean of group 1
- \(\mu_2\) : mean of group 2
- \(t\) : test statistic for an \(\alpha\) of 0.05
- \(\sigma_1\) : standard deviation of group 1
- \(\sigma_2\) : standard deviation of group 2
- \(n_1\) : size of group 1
- \(n_2\) : size of group 2
Running the Two-Sample T-Test
Putting this all together let's ask: Is mean age of males equal to the mean age of females in our glioblastoma data? We will start by manually calculating the mean age and standard deviation of males and females:
# what is mean of each group
meta %>%
group_by(SEX) %>%
summarise(
n=n(),
standard_deviation=sd(AGE),
mean=mean(AGE)
)
# A tibble: 2 × 4
SEX n standard_deviation mean
<chr> <int> <dbl> <dbl>
1 Female 44 13.7 57.9
2 Male 55 11.6 57.9
Remarkably we notice that the mean age of males and the mean age of females are the same! We will now use the two-sample t-test to ask if the mean age of males equals the mean age of females:
# run the two-sample t-test to determine if the
# mean age of males is equal to the mean
# age of females
t.test(
AGE ~ SEX,
data = meta,
method = "two.sided",
var.equal = FALSE)
Welch Two Sample t-test
data: AGE by SEX
t = 0.014062, df = 84.559, p-value = 0.9888
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-5.105621 5.178348
sample estimates:
mean in group Female mean in group Male
57.90909 57.87273
Explanation
- our test statistic is
0.014062 - The pvalue is below
0.9888 - our alternative hypothesis is that the true difference in means is not equal to 0
- the mean female is is
57.90909and the mean male age is57.87273 - the 95% confidence interval for our difference in means is
-5.105621to5.178348 - So we see that we do not have enough evidence to reject the null hypothesis that the true difference in means is equal to 0
Assumptions
So now that we have conducted our test, we should assess the test's assumptions:
- the values are independent of one another
- the variances in each group are equal
- the data in each group normally distributed
Here we note that our age values should be independent of one another (the age of one patient should not affect the age of another). Now for the assumption that variances of each group are equal - we avoided this by setting the var.equal argument equal to false. If we were to set it to true, we could check our variances with the F-Test:
# use the f test to determine if the variance in
# group 1 is the same as the variance in group 2
var.test(
AGE ~ SEX,
data = meta
)
F test to compare two variances
data: AGE by SEX
F = 1.3849, num df = 43, denom df = 54, p-value = 0.2559
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.7879316 2.4804317
sample estimates:
ratio of variances
1.384904
Here we will just note that the p-value is above 0.05 and thus there isn't a significant difference in the variance of group 1 and the variance of group 2. Now to check if the data in each group are normally distributed we will use the Shapiro-Wilk test:
# check the normality of male ages and female ages
shapiro.test(meta$AGE[meta$SEX == "Male"])
shapiro.test(meta$AGE[meta$SEX == "Female"])
Shapiro-Wilk normality test
data: meta$AGE[meta$SEX == "Male"]
W = 0.98809, p-value = 0.8604
Shapiro-Wilk normality test
data: meta$AGE[meta$SEX == "Female"]
W = 0.95578, p-value = 0.09048
Given our p-value is above 0.05 we do not have enough evidence to reject the null hypothesis of the Shapiro-Wilk Test; that the data are normally distributed. In other words, if the p-value is above 0.05 your data are normally distributed. However, it should be noted that the Shapiro-Wilk test p-value for female ages is rather close to 0.05 so there does seem to be some skey in female ages.
Non-Parametric Alternative
A non-parametric test is often used when either the assumptions about the distribution are not met. Additionally, these tests do not depend on the parameter they are assessing. Here, if the assumptions above are not met we can use the non-parametric equivalent, the Wilcoxon signed rank test:
# run the non-parametric alternative to the unpaired
# t-test the Wilcoxon signed rank test
wilcox.test(AGE ~ SEX,
data = meta,
alternative = "two.sided")
Wilcoxon rank sum test with continuity correction
data: AGE by SEX
W = 1278.5, p-value = 0.6318
alternative hypothesis: true location shift is not equal to 0