Running a T-test
Emery Tanksley
2025-03-26
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## Warning: package 'psych' was built under R version 4.4.3library(car) # for the leveneTest() command## Warning: package 'car' was built under R version 4.4.3## Loading required package: carData## Warning: package 'carData' was built under R version 4.4.3##
## Attaching package: 'car'## The following object is masked from 'package:psych':
##
## logitlibrary(effsize) # for the cohen.d() command## Warning: package 'effsize' was built under R version 4.4.3##
## 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
We predict that individuals with a graduate degree or higher will report significantly lower levels of depression compared to those who are equivalent to completing high school only.
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': 979 obs. of 7 variables:
## $ X : int 321 401 520 1390 1422 1849 2247 2526 2609 2689 ...
## $ age : chr "1 under 18" "4 between 36 and 45" "1 under 18" "5 over 45" ...
## $ education: chr "1 equivalent to not completing high school" "6 graduate degree or higher" "1 equivalent to not completing high school" "5 undergraduate degree" ...
## $ phq : num 1.89 2.44 1.56 1.22 4 ...
## $ gad : num 1 2.14 1.14 1 1.57 ...
## $ covid_neg: int 0 0 0 0 0 0 0 0 0 0 ...
## $ pas_covid: num 2.33 4 3 2.89 2.67 ...# 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$education <- as.factor(d$education)
str(d)## 'data.frame': 979 obs. of 7 variables:
## $ X : int 321 401 520 1390 1422 1849 2247 2526 2609 2689 ...
## $ age : chr "1 under 18" "4 between 36 and 45" "1 under 18" "5 over 45" ...
## $ education: Factor w/ 7 levels "1 equivalent to not completing high school",..: 1 6 1 5 1 1 1 2 7 6 ...
## $ phq : num 1.89 2.44 1.56 1.22 4 ...
## $ gad : num 1 2.14 1.14 1 1.57 ...
## $ covid_neg: int 0 0 0 0 0 0 0 0 0 0 ...
## $ pas_covid: num 2.33 4 3 2.89 2.67 ...table(d$education, useNA = "always")##
## 1 equivalent to not completing high school
## 284
## 2 equivalent to high school completion
## 282
## 3 equivalent to vocational/technical program completion
## 21
## 4 equivalent to AP/IB completion
## 114
## 5 undergraduate degree
## 123
## 6 graduate degree or higher
## 93
## prefer not to say
## 62
## <NA>
## 0## Checking the Continuous variable (DV)
# you can use the describe() command on an entire dataframe (d) or just on a single variable within your dataframe -- which we will do here
describe(d$phq)## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 979 2.05 0.86 1.89 1.96 0.99 1 4 3 0.68 -0.58 0.03# also use a histogram to visualize your continuous variable
hist(d$phq)
# use the describeBy() command to view the means and standard deviations by group
# it's very similar to the describe() command but splits the dataframe according to the 'group' variable
describeBy(d$phq, group=d$education)##
## Descriptive statistics by group
## group: 1 equivalent to not completing high school
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 284 2.15 0.85 2 2.08 0.99 1 4 3 0.61 -0.63 0.05
## ------------------------------------------------------------
## group: 2 equivalent to high school completion
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 282 2.27 0.84 2.22 2.23 0.99 1 4 3 0.31 -0.88 0.05
## ------------------------------------------------------------
## group: 3 equivalent to vocational/technical program completion
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 21 1.72 0.78 1.33 1.62 0.49 1 3.67 2.67 0.94 -0.32 0.17
## ------------------------------------------------------------
## group: 4 equivalent to AP/IB completion
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 114 2.26 0.91 2.22 2.21 1.15 1 4 3 0.34 -1.08 0.09
## ------------------------------------------------------------
## group: 5 undergraduate degree
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 123 1.46 0.47 1.22 1.38 0.33 1 3.33 2.33 1.54 2.4 0.04
## ------------------------------------------------------------
## group: 6 graduate degree or higher
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 93 1.5 0.55 1.33 1.41 0.49 1 3.56 2.56 1.52 2.35 0.06
## ------------------------------------------------------------
## group: prefer not to say
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 62 2.26 1.01 2 2.2 1.07 1 4 3 0.47 -1.27 0.13# lastly, use a boxplot to examine your chosen continuous and categorical variables together
boxplot(d$phq~d$education)
5 Check Your Assumptions
5.1 T-test Assumptions
- IV must have two levels
- Data values must be independent (independent t-test only)
- Data obtained via a random sample
- Dependent variable must be normally distributed
- Variances of the two groups are approximately equal
# If the IV has more than 2 levels, you must DROP any additional levels in order to meet the first assumption of a t-test.
## NOTE: This is a FOUR STEP process!
d <- subset(d, education != "1 equivalent to not completing high school") # use subset() to remove all participants from the additional level
table(d$education, useNA = "always") # verify that now there are ZERO participants in the additional level##
## 1 equivalent to not completing high school
## 0
## 2 equivalent to high school completion
## 282
## 3 equivalent to vocational/technical program completion
## 21
## 4 equivalent to AP/IB completion
## 114
## 5 undergraduate degree
## 123
## 6 graduate degree or higher
## 93
## prefer not to say
## 62
## <NA>
## 0d$education <- droplevels(d$education) # use droplevels() to drop the empty factor
table(d$education, useNA = "always") # verify that now the entire factor level is removed##
## 2 equivalent to high school completion
## 282
## 3 equivalent to vocational/technical program completion
## 21
## 4 equivalent to AP/IB completion
## 114
## 5 undergraduate degree
## 123
## 6 graduate degree or higher
## 93
## prefer not to say
## 62
## <NA>
## 0d <- subset(d, education != "3 equivalent to vocational/technical program completion") # use subset() to remove all participants from the additional level
table(d$education, useNA = "always") # verify that now there are ZERO participants in the additional level##
## 2 equivalent to high school completion
## 282
## 3 equivalent to vocational/technical program completion
## 0
## 4 equivalent to AP/IB completion
## 114
## 5 undergraduate degree
## 123
## 6 graduate degree or higher
## 93
## prefer not to say
## 62
## <NA>
## 0d$education <- droplevels(d$education) # use droplevels() to drop the empty factor
table(d$education, useNA = "always") # verify that now the entire factor level is removed##
## 2 equivalent to high school completion 4 equivalent to AP/IB completion
## 282 114
## 5 undergraduate degree 6 graduate degree or higher
## 123 93
## prefer not to say <NA>
## 62 0d <- subset(d, education != "4 equivalent to AP/IB completion") # use subset() to remove all participants from the additional level
table(d$education, useNA = "always") # verify that now there are ZERO participants in the additional level##
## 2 equivalent to high school completion 4 equivalent to AP/IB completion
## 282 0
## 5 undergraduate degree 6 graduate degree or higher
## 123 93
## prefer not to say <NA>
## 62 0d$education <- droplevels(d$education) # use droplevels() to drop the empty factor
table(d$education, useNA = "always") # verify that now the entire factor level is removed##
## 2 equivalent to high school completion 5 undergraduate degree
## 282 123
## 6 graduate degree or higher prefer not to say
## 93 62
## <NA>
## 0d <- subset(d, education != "5 undergraduate degree") # use subset() to remove all participants from the additional level
table(d$education, useNA = "always") # verify that now there are ZERO participants in the additional level##
## 2 equivalent to high school completion 5 undergraduate degree
## 282 0
## 6 graduate degree or higher prefer not to say
## 93 62
## <NA>
## 0d$education <- droplevels(d$education) # use droplevels() to drop the empty factor
table(d$education, useNA = "always") # verify that now the entire factor level is removed##
## 2 equivalent to high school completion 6 graduate degree or higher
## 282 93
## prefer not to say <NA>
## 62 0d <- subset(d, education != "prefer not to say") # use subset() to remove all participants from the additional level
table(d$education, useNA = "always") # verify that now there are ZERO participants in the additional level##
## 2 equivalent to high school completion 6 graduate degree or higher
## 282 93
## prefer not to say <NA>
## 0 0d$education <- droplevels(d$education) # use droplevels() to drop the empty factor
table(d$education, useNA = "always") # verify that now the entire factor level is removed##
## 2 equivalent to high school completion 6 graduate degree or higher
## 282 93
## <NA>
## 0## 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(phq~education, data =d)## Levene's Test for Homogeneity of Variance (center = median)
## Df F value Pr(>F)
## group 1 32.498 2.428e-08 ***
## 373
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Levene’s test revealed that our data has significantly different variances between the two comparison groups, two levels of education on depression.
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 more than two levels. To proceed with this analysis, I will drop the “equivalent to not completing high school,” “equivalent to vocational/technical program completion,” “equivalent to AP/IB completion,” “undergraduate degree,” and “prefer not to say” participants from my sample. I will make a note to discuss this issue in my Methods section write-up and in my Discussion section as a limitation of my study.
My data also has an issue regarding homogeneity of variance, as Levene’s test was significant. To accommodate for this heterogeneity of variance, I will use Welch’s t-test instead of Student’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$phq~d$education) # t_output will now show in your Global Environment7 View Test Output
t_output##
## Welch Two Sample t-test
##
## data: d$phq by d$education
## t = 10.15, df = 242.45, p-value < 2.2e-16
## alternative hypothesis: true difference in means between group 2 equivalent to high school completion and group 6 graduate degree or higher is not equal to 0
## 95 percent confidence interval:
## 0.6180841 0.9157373
## sample estimates:
## mean in group 2 equivalent to high school completion
## 2.269898
## mean in group 6 graduate degree or higher
## 1.5029878 Calculate Cohen’s d - Effect Size
# once again, we use the same formula, y~x, to calculate cohen's d
# We **only** calculate effect size if the test is SIG!
d_output <- cohen.d(d$phq~d$education) # d_output will now show in your Global Environment9 View Effect Size
d_output##
## Cohen's d
##
## d estimate: 0.9870205 (large)
## 95 percent confidence interval:
## lower upper
## 0.741442 1.232599## Remember to always take the ABSOLAUTE VALUE of the effect size value (i.e., it will never be negative)10 Write Up Results
To test our hypothesis that individuals with a graduate degree in our sample would report significantly lower levels of depression than individuals that only completed high school, we used an Independent Samples t-test. This required us to drop our “equivalent to not completing high school,” “equivalent to vocational/technical program completion,” “equivalent to AP/IB completion,” “undergraduate degree,” and “prefer not to say” participants 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 heterogeneity of variance (p < .001). This suggests that there is an increased chance of Type 1 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.
As predicted, we found that individuals with a graduate degree (M = 1.50, SD = 0.55 ) reported significantly lower levels of depression than individuals that only completed high school (M = 2.27, SD =0.84); t(242.45) =10.15 , p < .001 (see Figure 1). The effect size was calculated using Cohen’s d, with a value of 0.99 (large effect; Cohen, 1988).
References
Cohen J. (1988). Statistical Power Analysis for the Behavioral Sciences. New York, NY: Routledge Academic.