Running a T-test
Eli Davis
2025-06-07
1 Loading Libraries
# install any packages you have not previously used, then comment them back out.
#install.packages("car")
#install.packages("effsize")
library(psych) # for the describe() command
library(car) # for the leveneTest() command## Loading required package: carData##
## Attaching package: 'car'## The following object is masked from 'package:psych':
##
## logitlibrary(effsize) # for the cohen.d() command##
## Attaching package: 'effsize'## The following object is masked from 'package:psych':
##
## cohen.d2 Importing Data
d <- read.csv(file="Data/projectdata.csv", header=T)
# For the HW, you will import the project dataset you cleaned previously
# This will be the dataset you'll use for HWs throughout the rest of the semester3 State Your Hypothesis
There will be a significant difference in anxiety during the pandemic based on people’s age, between those under 18 and those 18-25 years old.
4 Check Your Variables
# you **only** need to check the variables you're using in the current analysis
## Checking the Categorical variable (IV)
str(d)## 'data.frame': 668 obs. of 7 variables:
## $ X : int 321 520 1422 1849 2247 2526 2609 2814 3053 3108 ...
## $ age : chr "1 under 18" "1 under 18" "1 under 18" "1 under 18" ...
## $ mhealth : chr "none or NA" "none or NA" "none or NA" "anxiety disorder" ...
## $ big5_neu : num 3.67 5.33 3.67 6 3.33 ...
## $ pas_covid : num 2.33 3 2.67 2.56 3 ...
## $ isolation_c: num 1.25 1 3.5 3.5 2 2.5 1.75 1 2 1.25 ...
## $ mfq_26 : num 4.3 2.7 2.95 2.4 4.7 4.35 4.25 4.55 4.95 4.95 ...# if the categorical variable you're using is showing as a "chr" (character), you must change it to be a ** factor ** -- using the next line of code (as.factor)
d$age<- as.factor(d$age)
str(d)## 'data.frame': 668 obs. of 7 variables:
## $ X : int 321 520 1422 1849 2247 2526 2609 2814 3053 3108 ...
## $ age : Factor w/ 2 levels "1 under 18","2 between 18 and 25": 1 1 1 1 1 1 1 1 1 1 ...
## $ mhealth : chr "none or NA" "none or NA" "none or NA" "anxiety disorder" ...
## $ big5_neu : num 3.67 5.33 3.67 6 3.33 ...
## $ pas_covid : num 2.33 3 2.67 2.56 3 ...
## $ isolation_c: num 1.25 1 3.5 3.5 2 2.5 1.75 1 2 1.25 ...
## $ mfq_26 : num 4.3 2.7 2.95 2.4 4.7 4.35 4.25 4.55 4.95 4.95 ...table(d$age, useNA = "always")##
## 1 under 18 2 between 18 and 25 <NA>
## 619 49 0## Checking the Continuous variable (DV)
# you can use the describe() command on an entire dataframe (d) or just on a single variable within your dataframe -- which we will do here
describe(d$pas_covid)## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 668 3.25 0.68 3.33 3.27 0.66 1 5 4 -0.3 0.17 0.03# also use a histogram to visualize your continuous variable
hist(d$pas_covid)
# use the describeBy() command to view the means and standard deviations by group
# it's very similar to the describe() command but splits the dataframe according to the 'group' variable
describeBy(d$pas_covid, group=d$age)##
## Descriptive statistics by group
## group: 1 under 18
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 619 3.24 0.69 3.33 3.26 0.66 1 5 4 -0.29 0.19 0.03
## ------------------------------------------------------------
## group: 2 between 18 and 25
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 49 3.33 0.6 3.44 3.36 0.66 2 4.44 2.44 -0.38 -0.63 0.09# lastly, use a boxplot to examine your chosen continuous and categorical variables together
boxplot(d$pas_covid~d$age)
5 Check Your Assumptions
5.1 T-test Assumptions
- IV must have two levels
- Data values must be independent (independent t-test only)
- Data obtained via a random sample
- Dependent variable must be normally distributed
- Variances of the two groups are approx. equal
# If the IV has more than 2 levels, you must DROP any additional levels in order to meet the first assumption of a t-test.
## NOTE: This is a FOUR STEP process!
## Repeat ALL THE STEPS ABOVE if your IV has more levels that need to be DROPPED. Copy the 4 lines of code, and replace the level name in the subset() command.5.2 Testing Homogeneity of Variance with Levene’s Test
We can test whether the variances of our two groups are equal using Levene’s test. The NULL hypothesis is that the variance between the two groups is equal, which is the result we WANT. So when running Levene’s test we’re hoping for a NON-SIGNIFICANT result!
# use the leveneTest() command from the car package to test homogeneity of variance
# it uses the same 'formula' setup that we'll use for our t-test: formula is y~x, where y is our DV and x is our IV
leveneTest(pas_covid~age, data = d)## Levene's Test for Homogeneity of Variance (center = median)
## Df F value Pr(>F)
## group 1 0.6023 0.438
## 666Levene’s test revealed that our data has equal variances between the two comparison groups, those 18 and under and those 18-25, on their levels of anxiety during the pandemic.
When running a t-test, we can account for heterogeneity in our variance by using the Welch’s t-test, which does not have the same assumption about variance as the Student’s t-test (the general default type of t-test in statistics). R defaults to using Welch’s t-test so this doesn’t require any changes on our part! Even if your data has no issues with homogeneity of variance, you’ll still use Welch’s t-test – it handles the potential issues around variance well and there are no real downsides. We’re using Levene’s test here to get into the habit of checking the homogeneity of our variance, even if we already have the solution for any potential problems.
5.3 Issues with My Data
My independent variable has only 2 two levels.
My data has no issue regarding homogeneity of variance, as Levene’s test was non-significant. Regardless, I will use Welchs’s t-test instead of Students’s t-test in my analysis.
6 Run a T-test
# Very simple! we use the same formula of y~x, where y is our DV and x is our IV
t_output <- t.test(d$pas_covid~d$age) # t_output will now show in your Global Environment7 View Test Output
t_output##
## Welch Two Sample t-test
##
## data: d$pas_covid by d$age
## t = -1.0114, df = 58.237, p-value = 0.316
## alternative hypothesis: true difference in means between group 1 under 18 and group 2 between 18 and 25 is not equal to 0
## 95 percent confidence interval:
## -0.27324784 0.08979783
## sample estimates:
## mean in group 1 under 18 mean in group 2 between 18 and 25
## 3.241608 3.3333338 Calculate Cohen’s d - Effect Size
# once again, we use the same formula, y~x, to calculate cohen's d
# We **only** calculate effect size if the test is SIG!
d_output <- cohen.d(d$pas_covid~d$age)
# View Effect Sized_output##
## Cohen's d
##
## d estimate: -0.1347479 (negligible)
## 95 percent confidence interval:
## lower upper
## -0.4262334 0.1567375## Remember to always take the ABSOLAUTE VALUE of the effect size value (i.e., it will never be negative)9 Write Up Results
To test our hypothesis that those under 18 in our sample would report significantly higher levels of anxiety during the pandemic than those 18-25, we used an independent samples t-test. This required us to not have to drop any levels from our sample, as we are limited to a two-group comparison when using this test. We tested the homogeneity of variance with Levene’s test and found signs of homogeneity (p > .001). This suggests that there is an increased chance of Type II error. To correct for this issue, we used Welch’s t-test, which does not assume homogeneity of variance. Our data met all other assumptions of an independent samples t-test.
Contrary to our prediction, we found that those under 18 (M = 3.24 , SD = .69 ) did not report significantly higher levels of anxiety during the pandemic than those 18-25 (M = 3.33, SD = .6); t(58.24) = -1.01 , p >.001 (see Figure 1). The effect size was calculated using Cohen’s d, with a value of 0.13 ( negligible effect; Cohen, 1988).
References
Cohen J. (1988). Statistical Power Analysis for the Behavioral Sciences. New York, NY: Routledge Academic.