# 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
## Warning: package 'psych' was built under R version 4.5.2
library(car) # for the leveneTest() command
## Warning: package 'car' was built under R version 4.5.2
## 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
There will be a significant difference in self-efficacy by people’s level of age, between 18 and 25 and 26 and 35.
# you **only** need to check the variables you're using in the current analysis
## Checking the Categorical variable (IV)
str(d)
## 'data.frame': 2065 obs. of 7 variables:
## $ ResponseID : chr "R_BJN3bQqi1zUMid3" "R_2TGbiBXmAtxywsD" "R_12G7bIqN2wB2N65" "R_39pldNoon8CePfP" ...
## $ 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" "rather not say" ...
## $ moa_independence: num 3.67 3.67 3.5 3 3.83 ...
## $ moa_safety : num 2.75 3.25 3 1.25 2.25 2.5 4 3.25 2.75 3.5 ...
## $ efficacy : num 3.4 3.4 2.2 2.8 3 2.4 2.3 3 3 3.7 ...
## $ stress : num 3.3 3.3 4 3.2 3.1 3.5 3.3 2.4 2.9 2.7 ...
# 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$age <- as.factor(d$age)
str(d)
## 'data.frame': 2065 obs. of 7 variables:
## $ ResponseID : chr "R_BJN3bQqi1zUMid3" "R_2TGbiBXmAtxywsD" "R_12G7bIqN2wB2N65" "R_39pldNoon8CePfP" ...
## $ age : Factor w/ 4 levels "1 between 18 and 25",..: 1 1 1 1 1 1 1 1 1 1 ...
## $ income : chr "1 low" "1 low" "rather not say" "rather not say" ...
## $ moa_independence: num 3.67 3.67 3.5 3 3.83 ...
## $ moa_safety : num 2.75 3.25 3 1.25 2.25 2.5 4 3.25 2.75 3.5 ...
## $ efficacy : num 3.4 3.4 2.2 2.8 3 2.4 2.3 3 3 3.7 ...
## $ stress : num 3.3 3.3 4 3.2 3.1 3.5 3.3 2.4 2.9 2.7 ...
table(d$age, useNA = "always")
##
## 1 between 18 and 25 2 between 26 and 35 3 between 36 and 45 4 over 45
## 1901 110 37 17
## <NA>
## 0
## Checking the Continuous variable (DV)
# you can use the describe() command on an entire dataframe (d) or just on a single variable within your dataframe -- which we will do here
describe(d$efficacy)
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 2065 3.11 0.44 3.1 3.12 0.44 1.2 4 2.8 -0.21 0.4 0.01
# also use a histogram to visualize your continuous variable
hist(d$efficacy)
# use the describeBy() command to view the means and standard deviations by group
# it's very similar to the describe() command but splits the dataframe according to the 'group' variable
describeBy(d$efficacy, group=d$age)
##
## Descriptive statistics by group
## group: 1 between 18 and 25
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 1901 3.09 0.44 3 3.1 0.44 1.2 4 2.8 -0.21 0.46 0.01
## ------------------------------------------------------------
## group: 2 between 26 and 35
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 110 3.26 0.43 3.2 3.27 0.44 1.8 4 2.2 -0.29 -0.01 0.04
## ------------------------------------------------------------
## group: 3 between 36 and 45
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 37 3.43 0.41 3.3 3.45 0.3 2.5 4 1.5 -0.12 -0.88 0.07
## ------------------------------------------------------------
## group: 4 over 45
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 17 3.44 0.55 3.5 3.51 0.44 1.8 4 2.2 -1.49 2.1 0.13
# lastly, use a boxplot to examine your chosen continuous and categorical variables together
boxplot(d$efficacy~d$age)
# If the IV has more than 2 levels, you must DROP any additional levels in order to meet the first assumption of a t-test.
## NOTE: This is a FOUR STEP process!
d <- subset(d, age != "3 between 36 and 45") # use subset() to remove all participants from the additional level
table(d$age, useNA = "always") # verify that now there are ZERO participants in the additional level
##
## 1 between 18 and 25 2 between 26 and 35 3 between 36 and 45 4 over 45
## 1901 110 0 17
## <NA>
## 0
d$age <- droplevels(d$age) # use droplevels() to drop the empty factor
table(d$age, useNA = "always") # verify that now the entire factor level is removed
##
## 1 between 18 and 25 2 between 26 and 35 4 over 45 <NA>
## 1901 110 17 0
d <- subset(d, age != "4 over 45") # use subset() to remove all participants from the additional level
table(d$age, useNA = "always") # verify that now there are ZERO participants in the additional level
##
## 1 between 18 and 25 2 between 26 and 35 4 over 45 <NA>
## 1901 110 0 0
d$age <- droplevels(d$age) # use droplevels() to drop the empty factor
table(d$age, useNA = "always")
##
## 1 between 18 and 25 2 between 26 and 35 <NA>
## 1901 110 0
# verify that now the entire factor level is removed
## 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(efficacy~age, data =d)
## Levene's Test for Homogeneity of Variance (center = median)
## Df F value Pr(>F)
## group 1 0.1168 0.7325
## 2009
Levene’s test revealed that our data does not have significantly different variances between the two comparison groups, between ages 18 and 25 and between ages 26 and 35, on their levels of self-efficacy.
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 between 36 and 45 and over 45 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 did not have an issue regarding homogeneity of variance, as Levene’s test was not significant . Even though I do not need to accommodate for this heterogeneity of variance, I will 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$efficacy~d$age) # t_output will now show in your Global Environment
t_output
##
## Welch Two Sample t-test
##
## data: d$efficacy by d$age
## t = -3.8271, df = 122.36, p-value = 0.0002057
## alternative hypothesis: true difference in means between group 1 between 18 and 25 and group 2 between 26 and 35 is not equal to 0
## 95 percent confidence interval:
## -0.24636366 -0.07838695
## sample estimates:
## mean in group 1 between 18 and 25 mean in group 2 between 26 and 35
## 3.094897 3.257273
# once again, we use the same formula, y~x, to calculate cohen's d
# We **only** calculate effect size if the test is SIG!
d_output <- cohen.d(d$efficacy~d$age) # d_output will now show in your Global Environment
d_output
##
## Cohen's d
##
## d estimate: -0.3703093 (small)
## 95 percent confidence interval:
## lower upper
## -0.5629717 -0.1776469
## Remember to always take the ABSOLAUTE VALUE of the effect size value (i.e., it will never be negative)
To test our hypothesis that there will be a significant difference in self-efficacy by people’s level of age, between 18 and 25 and 26 and 35, we used an independent samples t-test . This required us to drop our between 36 and 45 and over 45 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 no signs of heterogeneity (p = .733). Even though we did not have issues of homogeneity, 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 ages between 18 and 25 (M = 3.09 , SD = 0.44) reported significantly different levels of self-efficacy with ages between 26 and 35 (M = 3.26, SD = 0.43); t(122.36) = -3.83, p < .001 (see Figure 1). The effect size was calculated using Cohen’s d, with a value of 0.37 (small 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.