Running a T-test
Gaby Kovac
2026-04-01
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/projectdatagaby.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 the perceived stress scale by people’s level of ownership of pets, between dog and no pets
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': 681 obs. of 7 variables:
## $ X : int 520 2814 3146 3295 717 6056 4753 5365 2044 1965 ...
## $ relationship_status: chr "Single, never married" "Single, never married" "Prefer not to say" "Single, never married" ...
## $ pet : chr "cat" "other" "no pets" "no pets" ...
## $ big5_ext : num 2 2.67 4.67 4.33 1.67 ...
## $ pas_covid : num 3 3.44 4.67 2.44 1.56 ...
## $ pss : num 2.75 2.25 3 2 1.75 2 1 1.25 3 1.25 ...
## $ isolation_c : num 1 1 1 1 1 1 1 1 1 1 ...# 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$pet <- as.factor(d$pet)
str(d)## 'data.frame': 681 obs. of 7 variables:
## $ X : int 520 2814 3146 3295 717 6056 4753 5365 2044 1965 ...
## $ relationship_status: chr "Single, never married" "Single, never married" "Prefer not to say" "Single, never married" ...
## $ pet : Factor w/ 8 levels "bird","cat","cat and dog",..: 2 8 7 7 7 7 7 2 2 7 ...
## $ big5_ext : num 2 2.67 4.67 4.33 1.67 ...
## $ pas_covid : num 3 3.44 4.67 2.44 1.56 ...
## $ pss : num 2.75 2.25 3 2 1.75 2 1 1.25 3 1.25 ...
## $ isolation_c : num 1 1 1 1 1 1 1 1 1 1 ...table(d$pet, useNA = "always")##
## bird cat cat and dog
## 3 121 80
## dog fish multiple types of pet
## 112 19 62
## no pets other <NA>
## 244 40 0## Checking the Continuous variable (DV)
# you can use the describe() command on an entire dataframe (d) or just on a single variable within your dataframe -- which we will do here
describe(d$pss)## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 681 3.13 0.95 3.25 3.15 1.11 1 5 4 -0.13 -0.74 0.04# also use a histogram to visualize your continuous variable
hist(d$pss)
# use the describeBy() command to view the means and standard deviations by group
# it's very similar to the describe() command but splits the dataframe according to the 'group' variable
describeBy(d$pss, group=d$pet)##
## Descriptive statistics by group
## group: bird
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 3 2.42 1.18 2 2.42 0.74 1.5 3.75 2.25 0.31 -2.33 0.68
## ------------------------------------------------------------
## group: cat
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 121 3.27 0.92 3.25 3.3 1.11 1 5 4 -0.31 -0.58 0.08
## ------------------------------------------------------------
## group: cat and dog
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 80 3.19 0.86 3.25 3.21 0.74 1.25 5 3.75 -0.2 -0.71 0.1
## ------------------------------------------------------------
## group: dog
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 112 3.15 1 3.25 3.17 1.11 1 5 4 -0.14 -0.74 0.09
## ------------------------------------------------------------
## group: fish
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 19 3.12 0.89 3 3.13 1.11 1.5 4.5 3 -0.07 -1.18 0.2
## ------------------------------------------------------------
## group: multiple types of pet
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 62 3.16 0.91 3.25 3.17 1.11 1.25 4.75 3.5 -0.13 -1.14 0.12
## ------------------------------------------------------------
## group: no pets
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 244 3.05 0.97 3 3.05 1.11 1 5 4 0 -0.7 0.06
## ------------------------------------------------------------
## group: other
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 40 3.08 1 3.12 3.09 1.3 1.25 4.75 3.5 -0.13 -1.12 0.16# lastly, use a boxplot to examine your chosen continuous and categorical variables together
boxplot(d$pss~d$pet)
5 Check Your Assumptions
5.1 T-test Assumptions
- IV must have two levels.
- Data values must be independent (independent t-test only)
- Data obtained via a random sample
- Dependent variable must be normally distributed.
- Variances of the two groups are approx. equal.
# If the IV has more than 2 levels, you must DROP any additional levels in order to meet the first assumption of a t-test.
## NOTE: This is a FOUR STEP process!
d <- subset(d, pet != "bird") # use subset() to remove all participants from the additional level
d <- subset(d, pet != "cat and dog") # use subset() to remove all participants from the additional level
table(d$pet, useNA = "always") # verify that now there are ZERO participants in the additional level##
## bird cat cat and dog
## 0 121 0
## dog fish multiple types of pet
## 112 19 62
## no pets other <NA>
## 244 40 0d$pet <- droplevels(d$pet) # use droplevels(bird) to drop the empty factor
table(d$pet, useNA = "always") # verify that now there are ZERO participants in the additional level##
## cat dog fish
## 121 112 19
## multiple types of pet no pets other
## 62 244 40
## <NA>
## 0## Repeat ALL THE STEPS ABOVE if your IV has more levels that need to be DROPPED. Copy the 4 lines of code, and replace the level name in the subset() command.
d <- subset(d, pet != "cat") # use subset() to remove all participants from the additional level
table(d$pet, useNA = "always") # verify that now there are ZERO participants in the additional level##
## cat dog fish
## 0 112 19
## multiple types of pet no pets other
## 62 244 40
## <NA>
## 0d$pet <- droplevels(d$pet) # use droplevels(bird) to drop the empty factor
table(d$pet, useNA = "always") # verify that now there are ZERO participants in the additional level##
## dog fish multiple types of pet
## 112 19 62
## no pets other <NA>
## 244 40 0d <- subset(d, pet != "multiple types of pet") # use subset() to remove all participants from the additional level
table(d$pet, useNA = "always") # verify that now there are ZERO participants in the additional level##
## dog fish multiple types of pet
## 112 19 0
## no pets other <NA>
## 244 40 0d$pet <- droplevels(d$pet) # use droplevels(bird) to drop the empty factor
table(d$pet, useNA = "always") # verify that now there are ZERO participants in the additional level##
## dog fish no pets other <NA>
## 112 19 244 40 0d <- subset(d, pet != "other") # use subset() to remove all participants from the additional level
table(d$pet, useNA = "always") # verify that now there are ZERO participants in the additional level##
## dog fish no pets other <NA>
## 112 19 244 0 0d$pet <- droplevels(d$pet) # use droplevels(bird) to drop the empty factor
table(d$pet, useNA = "always") # verify that now there are ZERO participants in the additional level##
## dog fish no pets <NA>
## 112 19 244 0d <- subset(d, pet != "multiple types of pet") # use subset() to remove all participants from the additional level
table(d$pet, useNA = "always") # verify that now there are ZERO participants in the additional level##
## dog fish no pets <NA>
## 112 19 244 0d$pet <- droplevels(d$pet) # use droplevels(bird) to drop the empty factor
table(d$pet, useNA = "always") # verify that now there are ZERO participants in the additional level##
## dog fish no pets <NA>
## 112 19 244 0d <- subset(d, pet != "fish") # use subset() to remove all participants from the additional level
table(d$pet, useNA = "always") # verify that now there are ZERO participants in the additional level##
## dog fish no pets <NA>
## 112 0 244 0d$pet <- droplevels(d$pet) # use droplevels(bird) to drop the empty factor
table(d$pet, useNA = "always") # verify that now there are ZERO participants in the additional level##
## dog no pets <NA>
## 112 244 05.2 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(pss~pet, data = d)## Levene's Test for Homogeneity of Variance (center = median)
## Df F value Pr(>F)
## group 1 0.0766 0.7821
## 354Levene’s test revealed that our data has does not show heterogeneity in variances between the two comparison groups, pet and no pet, on their perceived levels of stress.
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 a variety of scale items eg., cat and dog, fish, multiple types of pet, bird, other, and cat 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 does not have an issue regarding homogeneity of variance, as Levene’s test was insignificant. 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$pss~d$pet) # t_output will now show in your Global Environment7 View Test Output
t_output##
## Welch Two Sample t-test
##
## data: d$pss by d$pet
## t = 0.92228, df = 209.02, p-value = 0.3574
## alternative hypothesis: true difference in means between group dog and group no pets is not equal to 0
## 95 percent confidence interval:
## -0.1190447 0.3283538
## sample estimates:
## mean in group dog mean in group no pets
## 3.151786 3.0471318 Calculate Cohen’s d - Effect Size
# once again, we use the same formula, y~x, to calculate cohen's d
# We **only** calculate effect size if the test is SIG!
d_output <- cohen.d(d$pss~d$pet) # d_output will now show in your Global Environment9 View Effect Size
d_output##
## Cohen's d
##
## d estimate: 0.1066108 (negligible)
## 95 percent confidence interval:
## lower upper
## -0.1179957 0.3312174## 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 participants in our sample would report significantly higher levels of stress if they do not have a pet, we used an independent-samples t-test. This required us to drop a variety of scale items 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 (p < .001). This suggests that there is 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 not having a pet (M = 3.13, SD = 0.95) reported significantly higher levels of stress (M = 3.05, SD = 0.97); t(1283.50) = -16.05, p < .001 (see Figure 1). The effect size was calculated using Cohen’s d, with a value of 0.75 (medium effect; Cohen, 1988).
[Revise the above statements for you HW assignment.]
References
Cohen J. (1988). Statistical Power Analysis for the Behavioral Sciences. New York, NY: Routledge Academic.