library(psych) # for the describe() command
library(car) # for the leveneTest() command
library(effsize) # for the cohen.d() commandt-Test Lab
Loading Libraries
Importing Data
# Update THIS FOR HOMEWORK - INSERT HW DATA
d <- read.csv(file="Data/eammi2_data_final.csv", header=T)State Your Hypothesis - PART OF YOUR WRITEUP
We hypothesize that there will be a significant difference in stress levels between Black and White participants.
Check Your Assumptions
T-test Assumptions
- Data values must be independent (independent t-test only) (confirmed by data report)
- Data obtained via a random sample (confirmed by data report)
- IV must have two levels (will check below)
- Dependent variable must be normally distributed (will check below. if issues, note and proceed)
- Variances of the two groups must be approximately equal, aka ‘homogeneity of variance’. Lacking this makes our results inaccurate (will check below - this really only applies to Student’s t-test, but we’ll check it anyway)
Checking IV levels
# # preview the levels and counts for your IV
table(d$race_rc, useNA = "always")
asian black hispanic multiracial nativeamer other
210 249 286 293 12 97
white <NA>
2026 9 #
# # note that the table() output shows you exactly how the levels of your variable are written. when recoding, make sure you are spelling them exactly as they appear
#
# # to drop levels from your variable
# # this subsets the data and says that any participant who is coded as 'LEVEL BAD' should be removed
# # if you don't need this for the homework, comment it out (add a # at the beginning of the line)
d <- subset(d, race_rc != "asian")
d <- subset(d, race_rc != "hispanic")
d <- subset(d, race_rc != "multiracial")
d <- subset(d, race_rc != "nativeamer")
d <- subset(d, race_rc != "other")
table(d$edu, useNA = "always")
1 High school diploma or less, and NO COLLEGE
41
2 Currently in college
1834
3 Completed some college, but no longer in college
16
4 Complete 2 year College degree
120
5 Completed Bachelors Degree
111
6 Currently in graduate education
112
7 Completed some graduate degree
41
<NA>
0 # # to combine levels
# # this says that where any participant is coded as 'LEVEL BAD' it should be replaced by 'LEVEL GOOD'
# # you can repeat this as needed, changing 'LEVEL BAD' if you have multiple levels that you want to combine into a single level
# # if you don't need this for the homework, comment it out (add a # at the beginning of the line)
d$edu_rc[d$edu == "High school diploma or less"] <- "Education Level"
d$edu_rc[d$edu == "Currently in college"] <- "Education Level"
d$edu_rc[d$edu == "Completed some college, but no longer in college"] <- "Education Level"
d$edu_rc[d$edu == "Completed 2 year college degree"] <- "Education Level"
d$edu_rc[d$edu == "Completed Bachelors Degree"] <- "Education Level"
d$edu_rc[d$edu == "Currently in graduate education"] <- "Education Level"
d$edu_rc[d$edu == "Completed some graduate degree"] <- "Education Level"
d$edu_rc[d$edu == "None or NA"] <- "No education level"
#
table(d$edu_rc, useNA = "always")
<NA>
2275 table(d$edu,d$edu_rc, useNA = "always")
<NA>
1 High school diploma or less, and NO COLLEGE 41
2 Currently in college 1834
3 Completed some college, but no longer in college 16
4 Complete 2 year College degree 120
5 Completed Bachelors Degree 111
6 Currently in graduate education 112
7 Completed some graduate degree 41
<NA> 0# # preview your changes and make sure everything is correct
table(d$race_rc, useNA = "always")
black white <NA>
249 2026 0 table(d$edu_rc, useNA = "always")
<NA>
2275 #
# # check your variable types
str(d)'data.frame': 2275 obs. of 28 variables:
$ ResponseId : chr "R_BJN3bQqi1zUMid3" "R_2TGbiBXmAtxywsD" "R_12G7bIqN2wB2N65" "R_1QiKb2LdJo1Bhvv" ...
$ gender : chr "f" "m" "m" "m" ...
$ race_rc : chr "white" "white" "white" "white" ...
$ age : chr "1 between 18 and 25" "1 between 18 and 25" "1 between 18 and 25" "1 between 18 and 25" ...
$ income : chr "1 low" "1 low" "rather not say" "2 middle" ...
$ edu : chr "2 Currently in college" "5 Completed Bachelors Degree" "2 Currently in college" "2 Currently in college" ...
$ sibling : chr "at least one sibling" "at least one sibling" "at least one sibling" "at least one sibling" ...
$ party_rc : chr "democrat" "independent" "apolitical" "apolitical" ...
$ disability : chr NA NA "psychiatric" NA ...
$ marriage5 : chr "are currently divorced from one another" "are currently married to one another" "are currently married to one another" "are currently married to one another" ...
$ phys_sym : chr "high number of symptoms" "high number of symptoms" "high number of symptoms" "low number of symptoms" ...
$ pipwd : num NA NA 2.33 NA NA ...
$ moa_independence: num 3.67 3.67 3.5 3.83 3.5 ...
$ moa_role : num 3 2.67 2.5 2.67 3.33 ...
$ moa_safety : num 2.75 3.25 3 2.25 2.5 4 3.25 3 3.75 3 ...
$ moa_maturity : num 3.67 3.33 3.67 3.67 4 ...
$ idea : num 3.75 3.88 3.75 3.5 3.25 ...
$ swb : num 4.33 4.17 1.83 3.67 4 ...
$ mindful : num 2.4 1.8 2.2 3.2 3.4 ...
$ belong : num 2.8 4.2 3.6 3.4 4.2 3.9 3.6 4.1 3.1 2.4 ...
$ efficacy : num 3.4 3.4 2.2 3 2.4 2.3 3 3.3 3 3.1 ...
$ support : num 6 6.75 5.17 6 4.5 ...
$ socmeduse : int 47 23 34 37 13 37 43 35 27 37 ...
$ usdream : chr "american dream is important and achievable for me" "american dream is important and achievable for me" "american dream is not important and maybe not achievable for me" "not sure if american dream important" ...
$ npi : num 0.6923 0.1538 0.0769 0.7692 0.2308 ...
$ exploit : num 2 3.67 4.33 4 1.33 ...
$ stress : num 3.3 3.3 4 3.1 3.5 3.3 2.4 3.5 2.5 3.5 ...
$ edu_rc : chr NA NA NA NA ...#
# # make sure that your IV is recognized as a factor by R
d$race_rc <- as.factor(d$race_rc)
d$edu_rc <- as.factor(d$edu_rc)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
# # 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(stress~race_rc, data = d)Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 9.1578 0.002504 **
2271
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1This is more of a formality in our case, because we are using Welch’s t-test, which does not have the same assumptions as Student’s t-test (the default type of t-test) about variance. R defaults to using Welch’s t-test so this doesn’t require any extra effort on our part!
Check Normality
# you only need to check the variables you're using in the current analysis
# although you checked them previously, it's always a good idea to look them over again and be sure that everything is correct
# you can use the describe() command on an entire datafrom (d) or just on a single variable (d$pss)
# use it to check the skew and kurtosis of your DV
describe(d$stress) vars n mean sd median trimmed mad min max range skew kurtosis se
X1 1 2273 3.04 0.6 3 3.04 0.59 1.3 4.7 3.4 0.04 -0.25 0.01# can 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$stress, group=d$race_rc)
Descriptive statistics by group
group: black
vars n mean sd median trimmed mad min max range skew kurtosis se
X1 1 248 3.08 0.54 3.1 3.06 0.44 1.7 4.7 3 0.25 0.19 0.03
------------------------------------------------------------
group: white
vars n mean sd median trimmed mad min max range skew kurtosis se
X1 1 2025 3.04 0.61 3 3.04 0.59 1.3 4.7 3.4 0.02 -0.3 0.01# also use a histogram to examine your continuous variable
hist(d$stress)
# last, use a boxplot to examine your continuous and categorical variables together
# categorical / IV goes on the right. Continuous / DV goes on the left
boxplot(d$stress~d$race_rc)
Issues with My Data - PART OF YOUR WRITEUP
We dropped participants who identified with racial groups other than “Black” and “White.” We also confirmed homogeneity of variance using Levene’s test (p = .450) and that our dependent variable is normally distributed ( skew and kurtosis between -2 and +2).
Run a T-test
# # very simple! we specify the dataframe alongside the variables instead of having a separate argument for the dataframe like we did for leveneTest()
t_output <- t.test(d$stress~d$race_rc)View Test Output
t_output
Welch Two Sample t-test
data: d$stress by d$race_rc
t = 0.95005, df = 329.02, p-value = 0.3428
alternative hypothesis: true difference in means between group black and group white is not equal to 0
95 percent confidence interval:
-0.03755085 0.10769860
sample estimates:
mean in group black mean in group white
3.076210 3.041136 Calculate Cohen’s d
# # once again, we use our formula to calculate cohen's d
d_output <- cohen.d(d$stress~d$race_rc)View Effect Size
- Trivial: < .2
- Small: between .2 and .5
- Medium: between .5 and .8
- Large: > .8
d_output
Cohen's d
d estimate: 0.0581756 (negligible)
95 percent confidence interval:
lower upper
-0.07376446 0.19011567 Write Up Results
We tested the hypothesis that stress levels would be significantly different between Black and White participants. We dropped any participants who did not identify as being “Black” or “white.” Additionally, we did not confirm homogeneity varience with the Levene’s test (p.003). However we found an evenly distributed dependent variable (skew and kurtosis within -2 and +2). A sample Welch’s t-test was used and revealed no significant different in stress levels between Black participants and White participants. Mean of Black participants = 3.08 and Mean of White participants = 3.04. t(329.02) = 0.95, p = .343, d = 0.06, 95% [-0.04, .11]. (refer to Figure 1).”
References
Cohen J. (1988). Statistical Power Analysis for the Behavioral Sciences. New York, NY: Routledge Academic.