# 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':
##
## logit
library(effsize) # for the cohen.d() command
##
## Attaching package: 'effsize'
## The following object is masked from 'package:psych':
##
## cohen.d
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 semester
We predict that individuals with divorced/separated parents will report significantly different levels of perceived stress than individuals with married parents.
# you **only** need to check the variables you're using in the current analysis
## Checking the Categorical variable (IV)
str(d)
## 'data.frame': 3141 obs. of 7 variables:
## $ ResponseID: chr "R_BJN3bQqi1zUMid3" "R_2TGbiBXmAtxywsD" "R_12G7bIqN2wB2N65" "R_39pldNoon8CePfP" ...
## $ 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" ...
## $ income : chr "1 low" "1 low" "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 ...
## $ 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 ...
## $ swb : num 4.33 4.17 1.83 5.17 3.67 ...
# 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$marriage5 <- as.factor(d$marriage5)
str(d)
## 'data.frame': 3141 obs. of 7 variables:
## $ ResponseID: chr "R_BJN3bQqi1zUMid3" "R_2TGbiBXmAtxywsD" "R_12G7bIqN2wB2N65" "R_39pldNoon8CePfP" ...
## $ marriage5 : Factor w/ 4 levels "are currently divorced from one another",..: 1 2 2 2 2 2 2 2 2 3 ...
## $ income : chr "1 low" "1 low" "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 ...
## $ 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 ...
## $ swb : num 4.33 4.17 1.83 5.17 3.67 ...
table(d$marriage5, useNA = "always")
##
## are currently divorced from one another
## 733
## are currently married to one another
## 2119
## never married each other and are not together
## 243
## never married each other but are currently together
## 46
## <NA>
## 0
## Checking the Continuous variable (DV)
# you can use the describe() command on an entire dataframe (d) or just on a single variable within your dataframe -- which we will do here
describe(d$stress)
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 3141 3.05 0.6 3 3.05 0.59 1.3 4.7 3.4 0.03 -0.16 0.01
# also use a histogram to visualize your continuous variable
hist(d$stress)
# use the describeBy() command to view the means and standard deviations by group
# it's very similar to the describe() command but splits the dataframe according to the 'group' variable
describeBy(d$stress, group=d$marriage5)
##
## Descriptive statistics by group
## group: are currently divorced from one another
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 733 3.08 0.6 3.1 3.08 0.59 1.3 4.6 3.3 0.08 -0.26 0.02
## ------------------------------------------------------------
## group: are currently married to one another
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 2119 3.04 0.59 3 3.03 0.59 1.4 4.7 3.3 0.02 -0.16 0.01
## ------------------------------------------------------------
## group: never married each other and are not together
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 243 3.09 0.63 3.1 3.09 0.59 1.3 4.7 3.4 -0.07 -0.11 0.04
## ------------------------------------------------------------
## group: never married each other but are currently together
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 46 2.96 0.71 3 2.95 0.59 1.4 4.6 3.2 0.17 -0.13 0.11
# lastly, use a boxplot to examine your chosen continuous and categorical variables together
boxplot(d$stress~d$marriage5)
# If the IV has more than 2 levels, you must DROP any additional levels in order to meet the first assumption of a t-test.
## NOTE: This is a FOUR STEP process!
d <- subset(d, marriage5 != "never married each other and are not together") # use subset() to remove all participants from the additional level
table(d$marriage5, useNA = "always") # verify that now there are ZERO participants in the additional level
##
## are currently divorced from one another
## 733
## are currently married to one another
## 2119
## never married each other and are not together
## 0
## never married each other but are currently together
## 46
## <NA>
## 0
d$marriage5 <- droplevels(d$marriage5) # use droplevels() to drop the empty factor
table(d$marriage5, useNA = "always") # verify that now the entire factor level is removed
##
## are currently divorced from one another
## 733
## are currently married to one another
## 2119
## never married each other but are currently together
## 46
## <NA>
## 0
d <- subset(d, marriage5 != "never married each other but are currently together") # use subset() to remove all participants from the additional level
table(d$marriage5, useNA = "always") # verify that now there are ZERO participants in the additional level
##
## are currently divorced from one another
## 733
## are currently married to one another
## 2119
## never married each other but are currently together
## 0
## <NA>
## 0
d$marriage5 <- droplevels(d$marriage5) # use droplevels() to drop the empty factor
table(d$marriage5, useNA = "always") # verify that now the entire factor level is removed
##
## are currently divorced from one another are currently married to one another
## 733 2119
## <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.
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(stress~marriage5, data =d )
## Levene's Test for Homogeneity of Variance (center = median)
## Df F value Pr(>F)
## group 1 0.9152 0.3388
## 2850
Levene’s test revealed that our data has no significant variances between the two comparison groups, individuals with married parents and divorced parents, on their 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.
My independent variable has more than two levels. To proceed with this analysis, I will drop the never married each other and are not together, and never married each other but are currently together participants 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 does not have an issue regarding homogeneity of variance, as Levene’s test was not significant. However, I will still use Welch’s t-test instead of Student’s t-test in my analysis.
# 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$stress~d$marriage5) # t_output will now show in your Global Environment
t_output
##
## Welch Two Sample t-test
##
## data: d$stress by d$marriage5
## t = 1.8699, df = 1245.6, p-value = 0.06173
## alternative hypothesis: true difference in means between group are currently divorced from one another and group are currently married to one another is not equal to 0
## 95 percent confidence interval:
## -0.002368441 0.098692288
## sample estimates:
## mean in group are currently divorced from one another
## 3.084311
## mean in group are currently married to one another
## 3.036149
# once again, we use the same formula, y~x, to calculate cohen's d
# We **only** calculate effect size if the test is SIG!
#d_output <- cohen.d(d$stress~d$marriage5) # d_output will now show in your Global Environment
#d_output
## Remember to always take the ABSOLAUTE VALUE of the effect size value (i.e., it will never be negative)
To test our hypothesis that individuals with divorced parents in our sample would report significantly different levels of perceived stress than individuals with married parents, we used an independent samples t-test. This required us to drop participants whose parents had never married but were currently together and participants whose parents had never married and were not currently together 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 no signs of heterogeneity (p = 0.34). This indicates that the assumption of equal variances was met. However, we still 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, individuals with divorced parents (M = 3.08, SD = 0.6) reported slightly higher levels of perceived stress than individuals with married parents (M = 3.04, SD = 0.59); however, this difference was not statistically significant, t(1245.6) = 1.87, p = 0.06 (see Figure 1). The effect size was not calculated for this analysis.
References
Cohen J. (1988). Statistical Power Analysis for the Behavioral Sciences. New York, NY: Routledge Academic.