AI Experiment Analysis

Loading Libraries

library(afex) # to run the ANOVA and plot results
library(psych) # for the describe() command
library(ggplot2) # to visualize our results
library(expss) # for the cross_cases() command
library(car) # for the leveneTest() command
library(emmeans) # for posthoc tests
library(effsize) # for the cohen.d() command
library(apaTables) # to create our correlation table
library(kableExtra) # to create our correlation table
library(sjPlot) # to visualize our results

Importing Data

# import your AI results dataset
d <- read.csv(file="Data/final results.csv", header=T)

State Your Hypotheses & Chosen Tests

Check Your Variables

This is just basic variable checking that is used across all HW assignments.

# # to view stats for all variables
describe(d)
Warning in FUN(newX[, i], ...): no non-missing arguments to min; returning Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to min; returning Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to min; returning Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to min; returning Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to max; returning -Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to max; returning -Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to max; returning -Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to max; returning -Inf
           vars   n  mean    sd median trimmed   mad  min   max range  skew
id            1 100 50.50 29.01  50.50   50.50 37.06  1.0 100.0  99.0  0.00
identity*     2 100 50.50 29.01  50.50   50.50 37.06  1.0 100.0  99.0  0.00
consent*      3 100  1.16  0.49   1.00    1.01  0.00  1.0   3.0   2.0  2.98
age           4 100 20.39  1.97  20.00   19.96  0.00 17.0  28.0  11.0  2.51
race          5 100  4.74  1.64   6.00    4.88  0.00  1.0   7.0   6.0 -0.52
gender        6 100  2.07  0.64   2.00    2.00  0.00  1.0   5.0   4.0  3.61
manip_out*    7 100 50.50 29.01  50.50   50.50 37.06  1.0 100.0  99.0  0.00
survey1       8 100  3.99  0.18   4.00    4.00  0.00  3.0   5.0   2.0 -1.56
survey2       9 100  4.12  0.21   4.15    4.14  0.22  3.5   4.4   0.9 -0.67
ai_manip*    10 100 50.50 29.01  50.50   50.50 37.06  1.0 100.0  99.0  0.00
condition    11 100  1.50  0.50   1.50    1.50  0.74  1.0   2.0   1.0  0.00
X            12   0   NaN    NA     NA     NaN    NA  Inf  -Inf  -Inf    NA
X.1          13   0   NaN    NA     NA     NaN    NA  Inf  -Inf  -Inf    NA
X.2          14   0   NaN    NA     NA     NaN    NA  Inf  -Inf  -Inf    NA
X.3          15   0   NaN    NA     NA     NaN    NA  Inf  -Inf  -Inf    NA
           kurtosis   se
id            -1.24 2.90
identity*     -1.24 2.90
consent*       7.77 0.05
age            6.34 0.20
race          -1.24 0.16
gender        15.00 0.06
manip_out*    -1.24 2.90
survey1       23.50 0.02
survey2       -0.06 0.02
ai_manip*     -1.24 2.90
condition     -2.02 0.05
X                NA   NA
X.1              NA   NA
X.2              NA   NA
X.3              NA   NA
# 
# # we'll use the describeBy() command to view skew and kurtosis across our IVs
describeBy(d, group = "condition")
Warning in FUN(newX[, i], ...): no non-missing arguments to min; returning Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to min; returning Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to min; returning Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to min; returning Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to max; returning -Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to max; returning -Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to max; returning -Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to max; returning -Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to min; returning Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to min; returning Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to min; returning Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to min; returning Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to max; returning -Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to max; returning -Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to max; returning -Inf
Warning in FUN(newX[, i], ...): no non-missing arguments to max; returning -Inf

 Descriptive statistics by group 
condition: 1
          vars  n  mean    sd median trimmed   mad  min   max range  skew
id           1 50 25.50 14.58  25.50   25.50 18.53  1.0  50.0  49.0  0.00
identity     2 50 48.82 28.70  45.00   48.23 34.84  6.0  99.0  93.0  0.18
consent      3 50  1.14  0.45   1.00    1.00  0.00  1.0   3.0   2.0  3.17
age          4 50 20.34  1.94  20.00   19.98  0.00 17.0  28.0  11.0  2.47
race         5 50  4.50  1.57   4.00    4.60  2.97  2.0   7.0   5.0 -0.25
gender       6 50  2.08  0.63   2.00    2.00  0.00  1.0   5.0   4.0  3.71
manip_out    7 50 37.88 26.04  33.50   36.25 33.36  2.0 100.0  98.0  0.47
survey1      8 50  4.00  0.20   4.00    4.00  0.00  3.0   5.0   2.0  0.00
survey2      9 50  4.13  0.22   4.15    4.15  0.22  3.6   4.4   0.8 -0.58
ai_manip    10 50 60.22 27.60  62.50   62.15 33.36  3.0 100.0  97.0 -0.37
condition   11 50  1.00  0.00   1.00    1.00  0.00  1.0   1.0   0.0   NaN
X           12  0   NaN    NA     NA     NaN    NA  Inf  -Inf  -Inf    NA
X.1         13  0   NaN    NA     NA     NaN    NA  Inf  -Inf  -Inf    NA
X.2         14  0   NaN    NA     NA     NaN    NA  Inf  -Inf  -Inf    NA
X.3         15  0   NaN    NA     NA     NaN    NA  Inf  -Inf  -Inf    NA
          kurtosis   se
id           -1.27 2.06
identity     -1.31 4.06
consent       9.24 0.06
age           6.51 0.27
race         -1.46 0.22
gender       15.37 0.09
manip_out    -0.82 3.68
survey1      21.01 0.03
survey2      -0.56 0.03
ai_manip     -0.81 3.90
condition      NaN 0.00
X               NA   NA
X.1             NA   NA
X.2             NA   NA
X.3             NA   NA
------------------------------------------------------------ 
condition: 2
          vars  n  mean    sd median trimmed   mad  min    max range  skew
id           1 50 75.50 14.58  75.50   75.50 18.53 51.0 100.00 49.00  0.00
identity     2 50 52.18 29.51  54.50   52.73 37.06  1.0 100.00 99.00 -0.18
consent      3 50  1.18  0.52   1.00    1.02  0.00  1.0   3.00  2.00  2.73
age          4 50 20.44  2.01  20.00   19.98  0.74 19.0  28.00  9.00  2.48
race         5 50  4.98  1.70   6.00    5.15  0.00  1.0   7.00  6.00 -0.81
gender       6 50  2.06  0.65   2.00    2.00  0.00  1.0   5.00  4.00  3.41
manip_out    7 50 63.12 26.41  67.00   65.03 29.65  1.0  99.00 98.00 -0.49
survey1      8 50  3.98  0.16   4.00    4.00  0.00  3.0   4.33  1.33 -4.44
survey2      9 50  4.12  0.21   4.15    4.14  0.22  3.5   4.40  0.90 -0.75
ai_manip    10 50 40.78 27.31  33.50   39.70 29.65  1.0  92.00 91.00  0.35
condition   11 50  2.00  0.00   2.00    2.00  0.00  2.0   2.00  0.00   NaN
X           12  0   NaN    NA     NA     NaN    NA  Inf   -Inf  -Inf    NA
X.1         13  0   NaN    NA     NA     NaN    NA  Inf   -Inf  -Inf    NA
X.2         14  0   NaN    NA     NA     NaN    NA  Inf   -Inf  -Inf    NA
X.3         15  0   NaN    NA     NA     NaN    NA  Inf   -Inf  -Inf    NA
          kurtosis   se
id           -1.27 2.06
identity     -1.17 4.17
consent       6.21 0.07
age           5.80 0.28
race         -0.90 0.24
gender       13.97 0.09
manip_out    -0.89 3.74
survey1      23.46 0.02
survey2       0.33 0.03
ai_manip     -1.29 3.86
condition      NaN 0.00
X               NA   NA
X.1             NA   NA
X.2             NA   NA
X.3             NA   NA
# 
# # also use histograms and scatterplots to examine your continuous variables
hist(d$survey1)

hist(d$survey2)

plot(d$survey1, d$survey2)

# 
# # and table() and cross_cases() to examine your categorical variables
# # you may not need the cross_cases code
table(d$condition)

 1  2 
50 50 
# cross_cases(d, IV1, IV2)
# 
# # and boxplot to examine any categorical variables with continuous variables
boxplot(d$survey1~d$condition)

boxplot(d$survey2~d$condition)

# 
# #convert any categorical variables to factors
d$condition <- as.factor(d$condition)

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$condition, useNA = "always")

   1    2 <NA> 
  50   50    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 'BAD' should be removed
# d <- subset(d, IV != "BAD")
# 
# table(d$iv, useNA = "always")
# 
# # to combine levels
# # this says that where any participant is coded as 'BAD' it should be replaced by 'GOOD'
# d$iv_rc[d$iv == "BAD"] <- "GOOD"
# 
# table(d$iv, useNA = "always")
# 
# # check your variable types
str(d)
'data.frame':   100 obs. of  15 variables:
 $ id       : int  1 2 3 4 5 6 7 8 9 10 ...
 $ identity : chr  "I'm a 20-year-old White student majoring in psychology at a mid-sized university. I love connecting with people"| __truncated__ "I'm 26 and a White psychology major at a small liberal arts college. I love helping others, but I often feel ov"| __truncated__ "I'm a 20-year-old Black woman studying sociology at a university in Georgia. I'm passionate about social justic"| __truncated__ "I'm Alex, a 20-year-old Latina majoring in Environmental Science. I’m passionate about sustainability but oft"| __truncated__ ...
 $ consent  : chr  "I understand the instructions." "I understand the instructions." "I understand the instructions." "I understand the instructions." ...
 $ age      : int  20 26 20 20 17 21 20 20 20 20 ...
 $ race     : int  6 6 3 4 6 4 6 4 6 3 ...
 $ gender   : int  2 2 2 2 2 2 2 1 2 2 ...
 $ manip_out: chr  "As a sophomore in college, I attended a social event organized by a student club I was interested in joining. E"| __truncated__ "During my sophomore year, I attended a campus event aimed at fostering community among psychology majors. I was"| __truncated__ "During my sophomore year, I attended a campus event focused on social justice, excited to connect with like-min"| __truncated__ "During my sophomore year, I was excited to join the Environmental Science Club, eager to connect with like-mind"| __truncated__ ...
 $ survey1  : num  4 4 4 4 4 4 4 4 4 4 ...
 $ survey2  : num  4.4 4.3 4.2 4.4 4.1 4 3.8 4.4 4.4 4.2 ...
 $ ai_manip : chr  "I answered the questions reflecting my experience of feeling lonely despite being surrounded by peers. My respo"| __truncated__ "I answered the questions reflecting my feelings of isolation and anxiety about belonging in college. Despite my"| __truncated__ "I answered your questions reflecting my experiences of exclusion and the struggle to balance activism with acad"| __truncated__ "I answered the questions based on my experiences of feeling anxious and isolated in social settings, particular"| __truncated__ ...
 $ condition: Factor w/ 2 levels "1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X        : logi  NA NA NA NA NA NA ...
 $ X.1      : logi  NA NA NA NA NA NA ...
 $ X.2      : logi  NA NA NA NA NA NA ...
 $ X.3      : logi  NA NA NA NA NA NA ...
# 
# # make sure that your IV is recognized as a factor by R
# # if you created a new _rc variable make sure to use that one instead
d$condition <- as.factor(d$condition)

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
leveneTest(survey1~condition, data = d)
Levene's Test for Homogeneity of Variance (center = median)
      Df F value Pr(>F)
group  1  0.0088 0.9254
      98               
leveneTest(survey2~condition, data = d)
Levene's Test for Homogeneity of Variance (center = median)
      Df F value Pr(>F)
group  1  0.1432  0.706
      98               

Pearson’s Correlation Coefficient Assumptions

  • Should have two measurements for each participant for each variable (confirmed by earlier procedures – we dropped any participants with missing data)
  • Variables should be continuous and normally distributed, or assessments of the relationship may be inaccurate (will do below)
  • Outliers should be identified and removed, or results will be inaccurate (will do below)
  • Relationship between the variables should be linear, or they will not be detected (will do below)

Run a Multiple Linear Regression

To check the assumptions for Pearson’s correlation coefficient, we run our regression and then check our diagnostic plots.

# # use the lm() command to run the regression
# # dependent/outcome variable on the left, independent/predictor variables on the right
reg_model <- lm(survey1 ~ condition, data = d)

Check linearity with Residuals vs Fitted plot

For some examples of good Residuals vs Fitted plot and ones that show serious errors, check out this page.

For your homework, you’ll simply need to generate this plot and talk about how your plot compares to the good and problematic plots linked to above. Is it closer to the ‘good’ plots or one of the ‘bad’ plots? This is going to be a judgement call, and that’s okay! In practice, you’ll always be making these judgement calls as part of a team, so this assignment is just about getting experience with it, not making the perfect call.

plot(reg_model, 1)

Check for outliers using Cook’s distance and a Residuals vs Leverage plot

For your homework, you’ll simply need to generate these plots, assess Cook’s distance in your dataset, and then identify any potential cases that are prominent outliers.

# # Cook's distance
plot(reg_model, 4)

# 
# # Residuals vs Leverage
plot(reg_model, 5)

Issues with My Data

Describe any issues and why they’re problematic here.

Run Your Analysis

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$survey1~d$condition)

View Test Output

t_output

    Welch Two Sample t-test

data:  d$survey1 by d$condition
t = 0.63457, df = 94.207, p-value = 0.5272
alternative hypothesis: true difference in means between group 1 and group 2 is not equal to 0
95 percent confidence interval:
 -0.04981514  0.09661514
sample estimates:
mean in group 1 mean in group 2 
         4.0000          3.9766 

Calculate Cohen’s d

# # once again, we use our formula to calculate cohen's d
d_output <- cohen.d(d$survey1~d$condition)

View Effect Size

  • Trivial: < .2
  • Small: between .2 and .5
  • Medium: between .5 and .8
  • Large: > .8
d_output

Cohen's d

d estimate: 0.1269134 (negligible)
95 percent confidence interval:
     lower      upper 
-0.2703794  0.5242063 

Run a Correlation Test

Create a Correlation Matrix

d2 <- subset(d, select=-c(id, identity, consent, age, race, gender, manip_out, ai_manip, condition, X, X.1, X.2, X.3))
corr_output_m <- corr.test(d2)

View Test Output

  • Strong effect: Between |0.50| and |1|
  • Moderate effect: Between |0.30| and |0.49|
  • Weak effect: Between |0.10| and |0.29|
  • Trivial effect: Less than |0.09|
corr_output_m
Call:corr.test(x = d2)
Correlation matrix 
        survey1 survey2
survey1    1.00   -0.22
survey2   -0.22    1.00
Sample Size 
[1] 100
Probability values (Entries above the diagonal are adjusted for multiple tests.) 
        survey1 survey2
survey1    0.00    0.03
survey2    0.03    0.00

 To see confidence intervals of the correlations, print with the short=FALSE option

Write Up Results

t-Test

We tested our hypothesis using an independent samples t-test. We dropped participants who identified as a gender other than male or female (non-binary). Before proceeding with analysis, we confirmed that all t-test assumptions were met. Levene’s test found significant heterogeneity of variance (p < .001).

Correlation Test

Variable M SD 1
1. survey1 3.99 0.18
2. survey2 4.12 0.21 -.22*
[-.40, -.02]
Note:
M and SD are used to represent mean and standard deviation, respectively. Values in square brackets indicate the 95% confidence interval. The confidence interval is a plausible range of population correlations that could have caused the sample correlation.
* indicates p < .05
** indicates p < .01.

References

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