Running a T-test HW
Ananya Adur
2026-10-29
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
We predict that people with one or more siblings will report significantly higher levels of social support than people without siblings.
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': 3156 obs. of 7 variables:
## $ ResponseID: chr "R_BJN3bQqi1zUMid3" "R_2TGbiBXmAtxywsD" "R_12G7bIqN2wB2N65" "R_39pldNoon8CePfP" ...
## $ sibling : chr "at least one sibling" "at least one sibling" "at least one sibling" "at least one sibling" ...
## $ race_rc : chr "white" "white" "white" "other" ...
## $ belong : num 2.8 4.2 3.6 4 3.4 4.2 3.9 3.6 2.9 2.5 ...
## $ support : num 6 6.75 5.17 5.58 6 ...
## $ npi : num 0.6923 0.1538 0.0769 0.0769 0.7692 ...
## $ efficacy : num 3.4 3.4 2.2 2.8 3 2.4 2.3 3 3 3.7 ...# 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$sibling <- as.factor(d$sibling)
str(d)## 'data.frame': 3156 obs. of 7 variables:
## $ ResponseID: chr "R_BJN3bQqi1zUMid3" "R_2TGbiBXmAtxywsD" "R_12G7bIqN2wB2N65" "R_39pldNoon8CePfP" ...
## $ sibling : Factor w/ 2 levels "at least one sibling",..: 1 1 1 1 1 1 1 2 1 1 ...
## $ race_rc : chr "white" "white" "white" "other" ...
## $ belong : num 2.8 4.2 3.6 4 3.4 4.2 3.9 3.6 2.9 2.5 ...
## $ support : num 6 6.75 5.17 5.58 6 ...
## $ npi : num 0.6923 0.1538 0.0769 0.0769 0.7692 ...
## $ efficacy : num 3.4 3.4 2.2 2.8 3 2.4 2.3 3 3 3.7 ...table(d$sibling, useNA = "always")##
## at least one sibling only child <NA>
## 2854 302 0## Checking the Continuous variable (DV)
# you can use the describe() command on an entire dataframe (d) or just on a single variable within your dataframe -- which we will do here
describe(d$support)## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 3156 5.54 1.13 5.75 5.66 0.99 0 7 7 -1.1 1.43 0.02# also use a histogram to visualize your continuous variable
hist(d$support)
# use the describeBy() command to view the means and standard deviations by group
# it's very similar to the describe() command but splits the dataframe according to the 'group' variable
describeBy(d$support, group=d$sibling)##
## Descriptive statistics by group
## group: at least one sibling
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 2854 5.52 1.13 5.75 5.64 0.99 0 7 7 -1.08 1.34 0.02
## ------------------------------------------------------------
## group: only child
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 302 5.66 1.11 5.92 5.79 0.99 1 7 6 -1.34 2.46 0.06# lastly, use a boxplot to examine your chosen continuous and categorical variables together
boxplot(d$support~d$sibling)
5 Check Your Assumptions
5.1 T-test Assumptions
- IV must have two levels.
- Data values must be independent (independent t-test only)
- Data obtained via a random sample
- Dependent variable must be normally distributed.
- Variances of the two groups are approx equal.
# If the IV has more than 2 levels, you must DROP any additional levels in order to meet the first assumption of a t-test.
## NOTE: This is a FOUR STEP process!
#d <- subset(d, gender != "nb") # use subset() to remove all participants from the additional level
#table(d$gender, useNA = "always") # verify that now there are ZERO participants in the additional level
#d$gender <- droplevels(d$gender) # use droplevels() to drop the empty factor
#table(d$gender, useNA = "always") # verify that now the entire factor level is removed
## 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(support~sibling, data =d)## Levene's Test for Homogeneity of Variance (center = median)
## Df F value Pr(>F)
## group 1 0.6004 0.4385
## 3154Levene’s test revealed that our data does not have significantly different variances between the two comparison groups, people with no siblings and people with at least one sibling, on their levels of perceived social support
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 only has two levels. To proceed with this analysis, I will not be dropping any participants from my sample. However, the kurtosis in the only child group of my study was above 2, so the data did not meet all the requirements for running a t-test.
My data did not have an issue regarding homogeneity of variance, as Levene’s test was not significant. Regardless of the homogeneity 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$support~d$sibling) # t_output will now show in your Global Environment7 View Test Output
t_output##
## Welch Two Sample t-test
##
## data: d$support by d$sibling
## t = -1.9679, df = 370.48, p-value = 0.04983
## alternative hypothesis: true difference in means between group at least one sibling and group only child is not equal to 0
## 95 percent confidence interval:
## -0.2645562463 -0.0001007705
## sample estimates:
## mean in group at least one sibling mean in group only child
## 5.523301 5.6556298 Calculate Cohen’s d - Effect Size
# once again, we use the same formula, y~x, to calculate cohen's d
# We **only** calculate effect size if the test is SIG!
#d_output <- cohen.d(d$support~d$sibling) # d_output will now show in your Global Environment9 View Effect Size
#d_output
## Remember to always take the ABSOLUTE VALUE of the effect size value (i.e., it will never be negative)10 Write Up Results
To test our hypothesis that people with siblings in our sample would report significantly higher levels of perceived social support than those without siblings, we used an independent samples t-test. We tested the homogeneity of variance with Levene’s test and found signs of homogeneity (p = 0.46). However, we used Welch’s t-test, which does not assume homogeneity of variance. Our data did not meet all other assumptions of an independent samples t-test, since the kurtosis for the only children group was above 2.
Contrary to our prediction, we found that there was no significant difference between only children (M = 5.51, SD = 1.15) and those with at least one sibling (M = 5.64, SD = 1.13); t(373.09) = -1.89, p = .0596 (see Figure 1).
References
Cohen J. (1988). Statistical Power Analysis for the Behavioral Sciences. New York, NY: Routledge Academic.