1 Loading Libraries

library(psych) # for the describe() command
library(car) # for the leveneTest() command
## Warning: package 'car' was built under R version 4.3.3
## Loading required package: carData
## Warning: package 'carData' was built under R version 4.3.3
## 
## Attaching package: 'car'
## The following object is masked from 'package:psych':
## 
##     logit
library(effsize) # for the cohen.d() command
## Warning: package 'effsize' was built under R version 4.3.3
## 
## Attaching package: 'effsize'
## The following object is masked from 'package:psych':
## 
##     cohen.d

2 Importing Data

# for the homework: 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/mydata.csv", header=T)

3 State Your Hypothesis

We predict that younger individuals will report significantly more anxiety than older individuals, as measured by general anxiety disorder (gad).

(Note: The above hypothesis is for the lab. You will need to come up with your own hypothesis, specific to your chosen variables, for the homework.) (need IV with 2 levels like gender and a continuous DV)

4 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':    912 obs. of  7 variables:
##  $ X        : int  20 30 31 33 57 58 81 104 113 117 ...
##  $ age      : chr  "1 under 18" "1 under 18" "4 between 36 and 45" "4 between 36 and 45" ...
##  $ mhealth  : chr  "anxiety disorder" "none or NA" "none or NA" "none or NA" ...
##  $ pas_covid: num  4.56 3.33 4.22 3.22 4.56 ...
##  $ phq      : num  3.33 1 2.33 1.11 2.33 ...
##  $ gad      : num  3.86 1.14 2 1.43 2.86 ...
##  $ swemws   : num  2.29 4.29 3.29 4 3.29 ...
d$age <- as.factor(d$age)

table(d$age, useNA = "always")
## 
##          1 under 18 2 between 18 and 25 3 between 26 and 35 4 between 36 and 45 
##                 585                  54                   6                  87 
##           5 over 45                <NA> 
##                 180                   0
# you can use the describe() command on an entire datafrom (d) or just on a single variable (d$pss)
describe(d$gad)
##    vars   n mean  sd median trimmed  mad min max range skew kurtosis   se
## X1    1 912 1.98 0.9   1.71    1.88 0.85   1   4     3 0.75    -0.61 0.03
# also use a histogram to examine your continuous variable
hist(d$gad)

# 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$gad, group=d$age)
## 
##  Descriptive statistics by group 
## group: 1 under 18
##    vars   n mean   sd median trimmed  mad min max range skew kurtosis   se
## X1    1 585 2.16 0.93      2    2.09 1.06   1   4     3 0.47    -1.04 0.04
## ------------------------------------------------------------ 
## group: 2 between 18 and 25
##    vars  n mean   sd median trimmed  mad min max range skew kurtosis   se
## X1    1 54 2.43 0.89    2.5    2.41 0.95   1   4     3  0.1    -1.05 0.12
## ------------------------------------------------------------ 
## group: 3 between 26 and 35
##    vars n mean   sd median trimmed  mad  min  max range skew kurtosis   se
## X1    1 6 1.95 0.67   1.71    1.95 0.21 1.57 3.29  1.71 1.23     -0.3 0.27
## ------------------------------------------------------------ 
## group: 4 between 36 and 45
##    vars  n mean   sd median trimmed  mad min max range skew kurtosis   se
## X1    1 87 1.68 0.73   1.43    1.56 0.64   1   4     3 1.49     1.87 0.08
## ------------------------------------------------------------ 
## group: 5 over 45
##    vars   n mean   sd median trimmed  mad min  max range skew kurtosis   se
## X1    1 180 1.42 0.49   1.29    1.33 0.42   1 3.29  2.29 1.54     2.11 0.04
# last, use a boxplot to examine your continuous and categorical variables together
boxplot(d$gad~d$age)

#variable is DV and group is IV (gender) must be in order for boxplot 

5 Check Your Assumptions

5.1 T-test Assumptions

  • IV must be a categorical variable with two levels
  • Data values must be independent (in this case, that the collection of one variable isn’t based on the response to another variable)
  • Data obtained via a random sample
  • Dependent variable must be normally distributed
  • Variances of the two groups are approximately equal

Some of these we can check in R, while others are down to our research design. These assumptions are confirmed by our research design, so we don’t have to do anything now:

  • Data values must be independent (in this case, that the collection of one variable isn’t based on the response to another variable) – confirmed by the data report that details how the data was collected
  • Data obtained via a random sample – confirmed by the data report that details how the data was collected

This assumption is not met:

  • IV must be a categorical variable with two levels – participants could answer M, F, or NB for their gender, so there are three levels, not two

This assumption was confirmed in the section above:

  • Dependent variable must be normally distributed – we checked skew and kurtosis above. For our class, even if your variables don’t meet the assumption we’ll proceed anyway, just make a note of it.

So we only have one assumption to test:

  • Variances of the two groups are approximately equal – AKA homogeneity of variance. Tested below!

5.2 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!

# subetting to drop the nb group so that our IV only has two levels
d <- subset(d, age == c("1 under 18", "5 over 45"))
d$age <- droplevels(d$age) # using droplevels() to drop the empty factor

# 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(d$gad~d$age, data = d)
## Levene's Test for Homogeneity of Variance (center = median)
##        Df F value    Pr(>F)    
## group   1  74.325 < 2.2e-16 ***
##       373                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

As you can see, our data is statistically 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 checking the homogeneity of our variance, even if we already have a solution for any potential problems.

5.3 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. My Levene’s test was statistically significant. To accommodate any potential heterogeneity of variance, I will use Welch’s t-test instead of Student’s t-test.

6 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$gad~d$age)

7 View Test Output

t_output
## 
##  Welch Two Sample t-test
## 
## data:  d$gad by d$age
## t = 11.515, df = 310.93, p-value < 2.2e-16
## alternative hypothesis: true difference in means between group 1 under 18 and group 5 over 45 is not equal to 0
## 95 percent confidence interval:
##  0.7120941 1.0055966
## sample estimates:
## mean in group 1 under 18  mean in group 5 over 45 
##                 2.243461                 1.384615

8 Calculate Cohen’s d

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

9 View Effect Size

d_output
## 
## Cohen's d
## 
## d estimate: 1.003087 (large)
## 95 percent confidence interval:
##     lower     upper 
## 0.7555174 1.2506571

10 Write Up Results

To test our hypothesis that younger individuals in our sample would report significantly more anxiety than older individuals, we used an two-sample or independent t-test. This required us to drop our non-binary and other 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 = <.001). 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. Our data met all other assumptions of a t-test.

As predicted, we found that younger individuals (M = 2.16, SD = .93) reported significantly higher anxiety than older individuals (M = 1.42, SD = .49); t(310.93) = 11.515, p < .001 (see Figure 1). The effect size was calculated using Cohen’s d, with a value of .75 (small effect; Cohen, 1988).

look at describe by command for M and SD

References

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