Running a T-test
Zora Parchman
2026-03-25
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 [Positive Identity as a Person with a Disability] by people’s level of [Disability status], between [Chronic health] and [Psychiatry].
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': 788 obs. of 7 variables:
## $ ResponseID: chr "R_12G7bIqN2wB2N65" "R_3lLnoV2mYVYHFvf" "R_1gTNDGWsqikPuEX" "R_2QLjdu3yoxqQ21c" ...
## $ race_rc : chr "white" "white" "white" "other" ...
## $ disability: chr "psychiatric" "other" "learning" "psychiatric" ...
## $ pipwd : num 2.33 2.33 3 2.93 3.33 ...
## $ npi : num 0.0769 0.7692 0 0.0769 0.6923 ...
## $ exploit : num 4.33 7 1.67 2.67 1.33 ...
## $ stress : num 4 3.5 4.4 3.6 3 3 3.7 3.4 2.7 3.8 ...# 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$disability <- as.factor(d$disability)
str(d)## 'data.frame': 788 obs. of 7 variables:
## $ ResponseID: chr "R_12G7bIqN2wB2N65" "R_3lLnoV2mYVYHFvf" "R_1gTNDGWsqikPuEX" "R_2QLjdu3yoxqQ21c" ...
## $ race_rc : chr "white" "white" "white" "other" ...
## $ disability: Factor w/ 6 levels "chronic health",..: 5 3 2 5 3 3 5 1 1 2 ...
## $ pipwd : num 2.33 2.33 3 2.93 3.33 ...
## $ npi : num 0.0769 0.7692 0 0.0769 0.6923 ...
## $ exploit : num 4.33 7 1.67 2.67 1.33 ...
## $ stress : num 4 3.5 4.4 3.6 3 3 3.7 3.4 2.7 3.8 ...table(d$disability, useNA = "always")##
## chronic health learning other physical psychiatric
## 138 113 76 44 354
## sensory <NA>
## 63 0## Checking the Continuous variable (DV)
# you can use the describe() command on an entire dataframe (d) or just on a single variable within your dataframe -- which we will do here
describe(d$pipwd)## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 788 2.92 0.7 3 2.93 0.69 1.13 5 3.87 0 -0.08 0.03# also use a histogram to visualize your continuous variable
hist(d$pipwd)
# use the describeBy() command to view the means and standard deviations by group
# it's very similar to the describe() command but splits the dataframe according to the 'group' variable
describeBy(d$pipwd, group=d$disability)##
## Descriptive statistics by group
## group: chronic health
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 138 2.93 0.77 3 2.93 0.74 1.27 4.8 3.53 0 -0.25 0.07
## ------------------------------------------------------------
## group: learning
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 113 3.16 0.62 3.13 3.17 0.59 1.4 4.73 3.33 -0.31 0.57 0.06
## ------------------------------------------------------------
## group: other
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 76 2.94 0.73 2.93 2.93 0.79 1.13 4.67 3.53 0.11 -0.44 0.08
## ------------------------------------------------------------
## group: physical
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 44 3.23 0.67 3.17 3.24 0.44 1.53 5 3.47 0.04 0.46 0.1
## ------------------------------------------------------------
## group: psychiatric
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 354 2.72 0.65 2.73 2.72 0.69 1.27 4.8 3.53 0.15 0 0.03
## ------------------------------------------------------------
## group: sensory
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 63 3.37 0.49 3.27 3.36 0.49 2.13 5 2.87 0.38 0.68 0.06# lastly, use a boxplot to examine your chosen continuous and categorical variables together
boxplot(d$pipwd~d$disability)
5 Check Your Assumptions
5.1 T-test Assumptions
- IV must have tow levels.
- Data values must be independent (independent t-test only).
- Data obtained via a random sample.
- Dependent variable must be normally distributed.
- Variances of the two groups are approx. equal
# If the IV has more than 2 levels, you must DROP any additional levels in order to meet the first assumption of a t-test.
## NOTE: This is a FOUR STEP process!
d <- subset(d,disability!= "learning") # use subset() to remove all participants from the additional level
table(d$disability, useNA = "always") # verify that now there are ZERO participants in the additional level##
## chronic health learning other physical psychiatric
## 138 0 76 44 354
## sensory <NA>
## 63 0 d$disability<- droplevels(d$disability) # use droplevels() to drop the empty factor
table(d$disability, useNA = "always") # verify that now the entire factor level is removed ##
## chronic health other physical psychiatric sensory
## 138 76 44 354 63
## <NA>
## 0d <- subset(d,disability!= "other") # use subset() to remove all participants from the additional level
table(d$disability, useNA = "always") # verify that now there are ZERO participants in the additional level##
## chronic health other physical psychiatric sensory
## 138 0 44 354 63
## <NA>
## 0 d$disability<- droplevels(d$disability) # use droplevels() to drop the empty factor
table(d$disability, useNA = "always") # verify that now the entire factor level is removed ##
## chronic health physical psychiatric sensory <NA>
## 138 44 354 63 0d <- subset(d,disability!= "physical") # use subset() to remove all participants from the additional level
table(d$disability, useNA = "always") # verify that now there are ZERO participants in the additional level##
## chronic health physical psychiatric sensory <NA>
## 138 0 354 63 0 d$disability<- droplevels(d$disability) # use droplevels() to drop the empty factor
table(d$disability, useNA = "always") # verify that now the entire factor level is removed ##
## chronic health psychiatric sensory <NA>
## 138 354 63 0d <- subset(d,disability!= "sensory") # use subset() to remove all participants from the additional level
table(d$disability, useNA = "always") # verify that now there are ZERO participants in the additional level##
## chronic health psychiatric sensory <NA>
## 138 354 0 0 d$disability<- droplevels(d$disability) # use droplevels() to drop the empty factor
table(d$disability, useNA = "always") # verify that now the entire factor level is removed ##
## chronic health psychiatric <NA>
## 138 354 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(pipwd~disability, data =d)## Levene's Test for Homogeneity of Variance (center = median)
## Df F value Pr(>F)
## group 1 3.3363 0.06837 .
## 490
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Levene’s test revealed that our data has no sig different variances between the two comparison groups, Chronic Health and Psychiatry, on their levels of Postive Identity as a Person with a Disability
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 sensory, learning, other,and physical repsonse options from my sample. I will make a note to discuss this issue in my Methods section write-up and in my Disucssion section as a limitation of my study.
My data has no issue regarding homogeneity of variance , as Levene’s test was not significant but all students have to do the Welch’s test, I will use Welch’s’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$pipwd~d$disability) # t_output will now show in your Global Environment7 View Test Output
t_output##
## Welch Two Sample t-test
##
## data: d$pipwd by d$disability
## t = 2.8509, df = 217.89, p-value = 0.004779
## alternative hypothesis: true difference in means between group chronic health and group psychiatric is not equal to 0
## 95 percent confidence interval:
## 0.06544513 0.35861121
## sample estimates:
## mean in group chronic health mean in group psychiatric
## 2.932367 2.7203398 Calculate Cohen’s d - Effect Size
# once again, we use the same formula, y~x, to calculate cohen's d
# We **only** calculate effect size if the test is SIG!
d_output <- cohen.d(d$pipwd~d$disability) # d_output will now show in your Global Environment9 View Effect Size
#delete if hypothsis is wrong or not sig
## 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 there will be a significant difference in [Positive Identity as a Person with a Disability] by people’s level of [Disability status], between [Chronic health] and [Psychiatry]., we used an Independent-samples T-test . This required us to drop drop the sensory, learning, other,and physical repsonse options from my 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 didn’t find signs of heterogeneity (p = .068). I 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 my hypothesis, we found that people with Chronic Health issues (M = 2.93 , SD =0.77 ) had no significant difference in Positive Identity as a Person With A Disability as a person with Psychiatric issues (M = 2.72 , SD = 0.65); t(2.85) =.0048 , p <.001 (see Figure 1). Effect size was calculated do to the non significance.
[Revise the above statements for you HW assignment.]
References
Cohen J. (1988). Statistical Power Analysis for the Behavioral Sciences. New York, NY: Routledge Academic.