1 Loading Libraries

library(expss) # for the cross_cases() command
## Loading required package: maditr
## 
## To select rows from data: rows(mtcars, am==0)
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':
## 
##     logit
## The following object is masked from 'package:expss':
## 
##     recode
library(effsize) # for the cohen.d() command
## 
## Attaching package: 'effsize'
## The following object is masked from 'package:psych':
## 
##     cohen.d

2 Importing Data

# import the dataset you cleaned previously
# this will be the dataset you'll use throughout the rest of the semester
d <- read.csv(file="data/eammi2_clean.csv", header=T)

3 Chi-square: State Your Hypothesis

There will be no gender differences in participation across the income categories (in other words, men (M), women (F), and non-binary (nb) participants will be evenly distributed across the income categories).

4 Chi-square: Check Your Variables

# 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
str(d)
## 'data.frame':    3143 obs. of  7 variables:
##  $ ResponseId: chr  "R_BJN3bQqi1zUMid3" "R_2TGbiBXmAtxywsD" "R_12G7bIqN2wB2N65" "R_39pldNoon8CePfP" ...
##  $ gender    : chr  "f" "m" "m" "f" ...
##  $ income_rc : chr  "20,000 - 39,999" "20,000 - 39,999" "Rather not say" "Rather not say" ...
##  $ stress    : num  3.3 3.3 4 3.2 3.1 3.5 3.3 2.4 2.9 2.7 ...
##  $ swb       : num  4.33 4.17 1.83 5.17 3.67 ...
##  $ socmeduse : int  47 23 34 35 37 13 37 43 37 29 ...
##  $ belong    : num  2.8 4.2 3.6 4 3.4 4.2 3.9 3.6 2.9 2.5 ...
# we can see in the str() command that our categorical variables are being read as character or string variables
# to correct this, we'll use the as.factor() command
d$gender <- as.factor(d$gender)
d$income_rc <- as.factor(d$income_rc)

table(d$gender, useNA = "always")
## 
##    f    m   nb <NA> 
## 2303  786   54    0
table(d$income_rc, useNA = "always")
## 
##   100,000 - 199,999     20,000 - 39,999 200,000 - 1 million     40,000 - 59,999 
##                 388                 360                 139                 343 
##     60,000 - 79,999     80,000 - 99,999      Over 1 million      Rather not say 
##                 301                 236                   7                 852 
##        Under 20,000                <NA> 
##                 517                   0
cross_cases(d, gender , income_rc)
 income_rc 
 100,000 - 199,999   20,000 - 39,999   200,000 - 1 million   40,000 - 59,999   60,000 - 79,999   80,000 - 99,999   Over 1 million   Rather not say   Under 20,000 
 gender 
   f  270 256 99 260 233 170 5 624 386
   m  117 100 37 78 65 61 2 208 118
   nb  1 4 3 5 3 5 20 13
   #Total cases  388 360 139 343 301 236 7 852 517

5 Chi-square: Check Your Assumptions

5.1 Chi-square: Test Assumptions

  • Data should be frequencies or counts
  • Variables and levels should be independent
  • There are two variables
  • At least 5 or more participants per cell

5.2 Issues with My Data

While my data meets the first three assumptions, I don’t have at least 5 participants in all cells. The number of non-binary participants is pretty small, and for one of the income groups it is less than five. The number of participants whose income is over 1 million is also small, with only 5 females and 2 males.

To proceed with this analysis, I will drop the non-binary participants from my sample and I will also drop the participants whose income is over 1 million. Dropping participants is always a difficult choice, and has the potential to further marginalize already minoritized groups, but it’s a necessary compromise for my analysis. I will make a note to discuss this issue in my Method write-up and in my Discussion as a limitation of my study.

# we'll use the subset command to drop our non-binary participants
d <- subset(d, gender != "nb") #using the '!=' sign here tells R to filter out the indicated criteria
# once we've dropped a level from our factor, we need to use the droplevels() command to remove it, or it will still show as 0
d$gender <- droplevels(d$gender)

table(d$gender, useNA = "always")
## 
##    f    m <NA> 
## 2303  786    0
d <- subset(d, income_rc != "Over 1 million")
d$income_rc <- droplevels(d$income_rc)
table(d$income_rc, useNA = "always")
## 
##   100,000 - 199,999     20,000 - 39,999 200,000 - 1 million     40,000 - 59,999 
##                 387                 356                 136                 338 
##     60,000 - 79,999     80,000 - 99,999      Rather not say        Under 20,000 
##                 298                 231                 832                 504 
##                <NA> 
##                   0
# since I made changes to my variables, I am going to re-run the cross_cases() command
cross_cases(d, gender, income_rc)
 income_rc 
 100,000 - 199,999   20,000 - 39,999   200,000 - 1 million   40,000 - 59,999   60,000 - 79,999   80,000 - 99,999   Rather not say   Under 20,000 
 gender 
   f  270 256 99 260 233 170 624 386
   m  117 100 37 78 65 61 208 118
   #Total cases  387 356 136 338 298 231 832 504

6 Chi-square: Run a Chi-square Test

# we use the chisq.test() command to run our chi-square test
# the only arguments we need to specify are the variables we're using for the chi-square test
# we are saving the output from our chi-square test to the chi_output object so we can view it again later
chi_output <- chisq.test(d$gender , d$income_rc)

7 Chi-square: View Test Output

# to view the results of our chi-square test, we just have to call up the output we saved
chi_output
## 
##  Pearson's Chi-squared test
## 
## data:  d$gender and d$income_rc
## X-squared = 10.582, df = 7, p-value = 0.1579

8 Chi-square: View Standardized Residuals

# to view the standardized residuals, we use the $ operator to access the stdres element of the chi_output file that we created
chi_output$stdres
##         d$income_rc
## d$gender 100,000 - 199,999 20,000 - 39,999 200,000 - 1 million 40,000 - 59,999
##        f        -2.3159977      -1.2215989          -0.4841898       1.0563245
##        m         2.3159977       1.2215989           0.4841898      -1.0563245
##         d$income_rc
## d$gender 60,000 - 79,999 80,000 - 99,999 Rather not say Under 20,000
##        f       1.5122056      -0.3515620      0.3395373    1.1415246
##        m      -1.5122056       0.3515620     -0.3395373   -1.1415246

9 Chi-square: Write Up Results

To test my hypothesis that there would be no gender differences in participation across the income categories, I ran a Chi-square test of independence. My variables met most of the criteria for running a chi-square test of analysis (it used frequences, the variables were independent, and there were two variables). However, I had a low number of non-binary participants and the above 1 million category did not meet the criteria for at least five participants per cell. To proceed with this analysis, I dropped the non-binary and above 1 million category from our sample. The final sample for my analysis can be seen in Table 1:

 income_rc 
 100,000 - 199,999   20,000 - 39,999   200,000 - 1 million   40,000 - 59,999   60,000 - 79,999   80,000 - 99,999   Rather not say   Under 20,000 
 gender 
   f  270 256 99 260 233 170 624 386
   m  117 100 37 78 65 61 208 118

As predicted, we did not find a gender difference in participation across the income categories, χ2(7, N = 3082) = 3.38, p = .1579.

–alternative–

As predicted, we did not find a gender difference in participation across the income categories, X^2(7, N = 3082) = 3.38, p = .1579.

10 T-test: State Your Hypothesis

I predict that women will report significantly more feelings of needing to belong than men, as measured by the need to belong scale.

11 T-test: Check Your Variables

str(d)
## 'data.frame':    3082 obs. of  7 variables:
##  $ ResponseId: chr  "R_BJN3bQqi1zUMid3" "R_2TGbiBXmAtxywsD" "R_12G7bIqN2wB2N65" "R_39pldNoon8CePfP" ...
##  $ gender    : Factor w/ 2 levels "f","m": 1 2 2 1 2 1 1 1 1 1 ...
##  $ income_rc : Factor w/ 8 levels "100,000 - 199,999",..: 2 2 7 7 6 7 8 2 1 7 ...
##  $ stress    : num  3.3 3.3 4 3.2 3.1 3.5 3.3 2.4 2.9 2.7 ...
##  $ swb       : num  4.33 4.17 1.83 5.17 3.67 ...
##  $ socmeduse : int  47 23 34 35 37 13 37 43 37 29 ...
##  $ belong    : num  2.8 4.2 3.6 4 3.4 4.2 3.9 3.6 2.9 2.5 ...
d$gender <- as.factor(d$gender)

table(d$gender, useNA = "always")
## 
##    f    m <NA> 
## 2298  784    0
# you can use the describe() command on an entire dataframe (d) or just on a single variable (d$pss)
describe(d$belong)
##    vars    n mean  sd median trimmed  mad min max range  skew kurtosis   se
## X1    1 3082 3.23 0.6    3.3    3.25 0.59 1.3   5   3.7 -0.26    -0.13 0.01
# also use a histogram to examine your continuous variable
hist(d$belong)

# 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$belong, group=d$gender)
## 
##  Descriptive statistics by group 
## group: f
##    vars    n mean   sd median trimmed  mad min max range  skew kurtosis   se
## X1    1 2298 3.28 0.59    3.3     3.3 0.59 1.3   5   3.7 -0.26    -0.17 0.01
## ------------------------------------------------------------ 
## group: m
##    vars   n mean   sd median trimmed  mad min max range  skew kurtosis   se
## X1    1 784 3.08 0.63    3.1     3.1 0.59 1.3 4.9   3.6 -0.16    -0.07 0.02
# last, use a boxplot to examine your continuous and categorical variables together
boxplot(d$belong~d$gender)

12 T-test: Check Your Assumptions

12.1 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 approximately equal

12.2 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
# 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

As you can see, our data is very close to significant. When running a t-test, we can account for heterogeneity in our variance by 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 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 just using Levene’s test here to get into the habit of changing the homogeneity of our variance, even if we already have a solution for any potential problems.

12.3 Issues with My Data

My independent variable has more than two levels. To proceed with this analysis, I will drop the non-binary participants from my sample. I will make a note to discuss this issue in my Method write-up and in my Discussion as a limitation of my study.

My data also has some potential issues regarding homogeneity of variance. Although Levene’s test was not significant, it was close to the significance threshold. To accommodate any potential heterogeneity of variance, I will use Welch’s t-test instead of Student’s t-test.

leveneTest(belong~gender, data = d)
## Levene's Test for Homogeneity of Variance (center = median)
##         Df F value  Pr(>F)  
## group    1   2.747 0.09754 .
##       3080                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

13 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$belong~d$gender)

14 View Test Output

t_output
## 
##  Welch Two Sample t-test
## 
## data:  d$belong by d$gender
## t = 7.8245, df = 1283.6, p-value = 1.058e-14
## alternative hypothesis: true difference in means between group f and group m is not equal to 0
## 95 percent confidence interval:
##  0.1498013 0.2500558
## sample estimates:
## mean in group f mean in group m 
##        3.282071        3.082143

15 Calculate Cohen’s d

# once again, we use our formula to calculate cohen's d
d_output <- cohen.d(d$belong~d$gender)

16 View Effect Size

d_output
## 
## Cohen's d
## 
## d estimate: 0.334042 (small)
## 95 percent confidence interval:
##     lower     upper 
## 0.2525175 0.4155664

17 Write Up Results

To test my hypothesis that women in our sample would report significantly more feeling of needing to belong than men, I used an two-sample or independent t-test. This required me to drop our non-binary gender participants 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 some signs of heterogeneity (p = .097). This suggests that there is an increased chance of Type I error. To correct for this possible issue, we use Welch’s t-test, which does not assume homogeneity of variance. My data met all other assumptions of a t-test.

As predicted, we found that women (M = 3.28, SD = .59) reported significantly higher stress than men (M = 3.08, SD = .63); t(1283.6) = 7.82, p < .001 (see Figure 1). The effect size was calculated using Cohen’s d, with a value of .33 (small effect; Cohen, 1988).

References

Cohen J. (1988). Statistical Power Analysis for the Behavioral Sciences. New York, NY: Routledge Academic.