library(psych) # for the describe() command
library(car) # for the leveneTest() command
library(effsize) # for the cohen.d() command
t-Test HW
Loading Libraries
Importing Data
# UPDATE THIS FOR HW!
<- read.csv(file="Data/mydata.csv", header=T) d
State Your Hypothesis - PART OF YOUR WRITEUP
Females will report higher levels of worry than males.
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$gender, useNA = "always")
female I use another term male Prefer not to say
1028 28 199 17
<NA>
0
# # note that the table() output shows you exactly how the levels of your variable are written. when recoding, make sure you are spelling them exactly as they appear
#
# # to drop levels from your variable
# # this subsets the data and says that any participant who is coded as 'LEVEL BAD' should be removed
# # if you don't need this for the homework, comment it out (add a # at the beginning of the line)
<- subset(d, gender != "I use another term")
d <- subset(d, gender != "Prefer not to say")
d
table(d$mhealth, useNA = "always")
<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$mhealth_rc[d$mhealth == "anxiety disorder"] <- "mental health diagnosis"
# d$mhealth_rc[d$mhealth == "bipolar"] <- "mental health diagnosis"
# d$mhealth_rc[d$mhealth == "depression"] <- "mental health diagnosis"
# d$mhealth_rc[d$mhealth == "eating disorders"] <- "mental health diagnosis"
# d$mhealth_rc[d$mhealth == "obsessive compulsive disorder"] <- "mental health diagnosis"
# d$mhealth_rc[d$mhealth == "other"] <- "mental health diagnosis"
# d$mhealth_rc[d$mhealth == "ptsd"] <- "mental health diagnosis"
# d$mhealth_rc[d$mhealth == "none or NA"] <- "no mental health diagnosis"
#table(d$mhealth_rc, useNA = "always")
#table(d$mhealth, d$mhealth_rc, useNA = "always")
# # preview your changes and make sure everything is correct
table(d$gender, useNA = "always")
female male <NA>
1028 199 0
#table(d$mhealth_rc, useNA = "always")
#
# # check your variable types
str(d)
'data.frame': 1227 obs. of 6 variables:
$ gender : chr "male" "female" "female" "female" ...
$ age : chr "1 under 18" "1 under 18" "4 between 36 and 45" "4 between 36 and 45" ...
$ big5_ext : num 1.67 6 5 4.33 5.67 ...
$ pswq : num 0.851 -1.1235 1.1627 1.8077 0.0159 ...
$ covid_pos: int 0 0 0 0 0 0 0 0 0 0 ...
$ covid_neg: int 0 0 0 0 0 0 0 0 0 0 ...
# # make sure that your IV is recognized as a factor by R
$gender <- as.factor(d$gender)
d#d$mhealth_rc <- as.factor(d$mhealth_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!
This is the code needed in lab
# # 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(pswq~gender, data = d)
Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 0.2593 0.6107
1225
This 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$pss)
# use it to check the skew and kurtosis of your DV
describe(d$pswq)
vars n mean sd median trimmed mad min max range skew kurtosis se
X1 1 1227 -0.04 1 0 -0.04 1.15 -2.25 2.38 4.63 -0.05 -0.92 0.03
# 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$pswq, group=d$gender)
Descriptive statistics by group
group: female
vars n mean sd median trimmed mad min max range skew kurtosis se
X1 1 1028 0.06 0.97 0.1 0.08 1.12 -2.16 2.38 4.54 -0.13 -0.85 0.03
------------------------------------------------------------
group: male
vars n mean sd median trimmed mad min max range skew kurtosis se
X1 1 199 -0.6 0.96 -0.77 -0.66 1.06 -2.25 1.7 3.95 0.49 -0.65 0.07
# also use a histogram to examine your continuous variable
hist(d$pswq)
# last, use a boxplot to examine your continuous and categorical variables together
# categorical/IV goes on the right, continuous/DV goes on the left
boxplot(d$pswq~d$gender)
Issues with My Data - PART OF YOUR WRITEUP
“We dropped participants who had genders other than female or male (I use another term, prefer not to say). We also confirmed homogeneity of variance using Levene’s test (p = .610) and that our dependent variable is normally distributed (skew and kurtosis between -2 and +2).”
Run a T-test
# # very simple! we specify the dataframe alongside the variables instead of having a separate argument for the dataframe like we did for leveneTest()
<- t.test(d$pswq~d$gender) t_output
View Test Output
t_output
Welch Two Sample t-test
data: d$pswq by d$gender
t = 8.8576, df = 281.2, p-value < 2.2e-16
alternative hypothesis: true difference in means between group female and group male is not equal to 0
95 percent confidence interval:
0.5149021 0.8091462
sample estimates:
mean in group female mean in group male
0.06428784 -0.59773629
Calculate Cohen’s d
# # once again, we use our formula to calculate cohen's d
<- cohen.d(d$pswq~d$gender) d_output
View Effect Size
- Trivial: < .2
- Small: between .2 and .5
- Medium: between .5 and .8
- Large: > .8
d_output
Cohen's d
d estimate: 0.6829534 (medium)
95 percent confidence interval:
lower upper
0.5286231 0.8372836
Write Up Results
We tested our hypothesis that females would report significantly more worry than males using an independent samples t-test. Our data met all of the assumptions of a t-test, we did find a significant difference, t(281.2) = 8.86, p<.001, d = .68, 95% [0.51, 0.80] (refer to Figure 1.
Our effect size was large according to Cohen (1988)
References
Cohen J. (1988). Statistical Power Analysis for the Behavioral Sciences. New York, NY: Routledge Academic.