Running a T-test
Joseph Piontek
2026-11-02
1 Loading Libraries
# install any packages you have not previously used, then comment them back out.
#install.packages("psych")
#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 resilience by people’s level of hours of sleep, between less than 5 hours and 7-8 hours.
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': 292 obs. of 7 variables:
## $ X : int 7888 7365 8747 7357 8760 8654 8272 8738 7911 8463 ...
## $ gender : chr "male" "female" "female" "female" ...
## $ sleep_hours: chr "3 7-8 hours" "2 5-6 hours" "4 8-10 hours" "4 8-10 hours" ...
## $ mfq_state : num 3.75 4 5.88 4.75 5.5 ...
## $ phq : num 2 1.78 1 1.11 1.11 ...
## $ gad : num 4 1.43 1.14 1 1 ...
## $ brs : num 2 3.83 3.83 4 4.67 ...# 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$gender <- as.factor(d$gender)
d$sleep_hours <- as.factor(d$sleep_hours)
str(d)## 'data.frame': 292 obs. of 7 variables:
## $ X : int 7888 7365 8747 7357 8760 8654 8272 8738 7911 8463 ...
## $ gender : Factor w/ 4 levels "female","I use another term",..: 3 1 1 1 3 3 3 1 1 1 ...
## $ sleep_hours: Factor w/ 5 levels "1 < 5 hours",..: 3 2 4 4 3 3 3 3 4 2 ...
## $ mfq_state : num 3.75 4 5.88 4.75 5.5 ...
## $ phq : num 2 1.78 1 1.11 1.11 ...
## $ gad : num 4 1.43 1.14 1 1 ...
## $ brs : num 2 3.83 3.83 4 4.67 ...table(d$gender, useNA = "always")##
## female I use another term male Prefer not to say
## 228 17 41 6
## <NA>
## 0## Checking the Continuous variable (DV)
# you can use the describe() command on an entire dataframe (d) or just on a single variable within your dataframe -- which we will do here
describe(d$brs)## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 292 2.65 0.85 2.58 2.65 0.86 1 5 4 0.16 -0.58 0.05# also use a histogram to visualize your continuous variable
hist(d$brs)
# use the describeBy() command to view the means and standard deviations by group
# it's very similar to the describe() command but splits the dataframe according to the 'group' variable
describeBy(d$brs, group=d$sleep_hours)##
## Descriptive statistics by group
## group: 1 < 5 hours
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 36 2.35 0.77 2.33 2.34 0.99 1 3.83 2.83 0.09 -0.97 0.13
## ------------------------------------------------------------
## group: 2 5-6 hours
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 104 2.54 0.82 2.5 2.53 0.86 1 4.33 3.33 0.07 -0.8 0.08
## ------------------------------------------------------------
## group: 3 7-8 hours
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 97 2.77 0.89 2.83 2.76 0.99 1 4.67 3.67 0.15 -0.72 0.09
## ------------------------------------------------------------
## group: 4 8-10 hours
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 40 3.02 0.73 3 3.03 0.99 1.67 4.5 2.83 0.01 -0.99 0.11
## ------------------------------------------------------------
## group: 5 > 10 hours
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 15 2.44 0.97 2.17 2.36 0.74 1 5 4 0.99 0.78 0.25# lastly, use a boxplot to examine your chosen continuous and categorical variables together
boxplot(d$brs~d$sleep_hours)
5 Check Your Assumptions
5.1 T-test Assumptions
- IV must have two levels
- Data values must be independent (independent t-test only)
- Data obtained via a random sample
- Dependent variable must be equally distributed
- Variances of the two groups are approx. equal
# If the IV has more than 2 levels, you must DROP any additional levels in order to meet the first assumption of a t-test.
## NOTE: This is a FOUR STEP process!
d <- subset(d, sleep_hours != "2 5-6 hours") # use subset() to remove all participants from the additional level
table(d$sleep_hours, useNA = "always") # verify that now there are ZERO participants in the additional level##
## 1 < 5 hours 2 5-6 hours 3 7-8 hours 4 8-10 hours 5 > 10 hours <NA>
## 36 0 97 40 15 0d$sleep_hours <- droplevels(d$sleep_hours) # use droplevels() to drop the empty factor
table(d$sleep_hours, useNA = "always") # verify that now the entire factor level is removed##
## 1 < 5 hours 3 7-8 hours 4 8-10 hours 5 > 10 hours <NA>
## 36 97 40 15 0d <- subset(d, sleep_hours != "4 8-10 hours") # use subset() to remove all participants from the additional level
table(d$sleep_hours, useNA = "always") # verify that now there are ZERO participants in the additional level##
## 1 < 5 hours 3 7-8 hours 4 8-10 hours 5 > 10 hours <NA>
## 36 97 0 15 0d$sleep_hours <- droplevels(d$sleep_hours) # use droplevels() to drop the empty factor
table(d$sleep_hours, useNA = "always") # verify that now the entire factor level is removed##
## 1 < 5 hours 3 7-8 hours 5 > 10 hours <NA>
## 36 97 15 0d <- subset(d, sleep_hours != "5 > 10 hours") # use subset() to remove all participants from the additional level
table(d$sleep_hours, useNA = "always") # verify that now there are ZERO participants in the additional level##
## 1 < 5 hours 3 7-8 hours 5 > 10 hours <NA>
## 36 97 0 0d$sleep_hours <- droplevels(d$sleep_hours) # use droplevels() to drop the empty factor
table(d$sleep_hours, useNA = "always") # verify that now the entire factor level is removed##
## 1 < 5 hours 3 7-8 hours <NA>
## 36 97 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(brs~sleep_hours, data = d)## Levene's Test for Homogeneity of Variance (center = median)
## Df F value Pr(>F)
## group 1 1.53 0.2183
## 131Levene’s test revealed that our data does not have significantly different variances between the two comparison groups, less than 5 hours and 7-8 hours, on their levels of resilience
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 participants who recorded 5-6 hours, 8-10 hours, and more than 10 hours 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 did not have any issues regarding homogeneity of variance, as Levene’s test was not 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$brs~d$sleep_hours)7 View Test Output
t_output##
## Welch Two Sample t-test
##
## data: d$brs by d$sleep_hours
## t = -2.659, df = 71.795, p-value = 0.009656
## alternative hypothesis: true difference in means between group 1 < 5 hours and group 3 7-8 hours is not equal to 0
## 95 percent confidence interval:
## -0.7312246 -0.1045906
## sample estimates:
## mean in group 1 < 5 hours mean in group 3 7-8 hours
## 2.351852 2.7697598 Calculate Cohen’s d - Effect Size
# once again, we use the same formula, y~x, to calculate cohen's d
# We **only** calculate effect size if the test is SIG!
d_output <- cohen.d(d$brs~d$sleep_hours) # d_output will now show in your Global Environment9 View Effect Size
d_output##
## Cohen's d
##
## d estimate: -0.4857658 (small)
## 95 percent confidence interval:
## lower upper
## -0.87630715 -0.09522447## 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 7-8 hours in our sample would report significantly higher levels of resilience than less than 5 hours, we used an independent samples t-test . This required us to drop our 5-6 hours, 8-10 hours, and more than 10 hours 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 did not find signs of heterogeneity (p < .001). This suggests that there is not an increased chance of type I 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 7-8 hours (M = 2.77, SD = .89) reported significantly higher levels of resilience than less than 5 hours (M = 2.35, SD = .77); t(71.80) = -2.66, p = .009 (see Figure 1). The effect size was calculated using Cohen’s d, with a value of .49(small effect; Cohen, 1988).
References
Cohen J. (1988). Statistical Power Analysis for the Behavioral Sciences. New York, NY: Routledge Academic.