Independent Samples t-Test

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Are there mean differences in risk scores (`risk_score`) by whether or not the defendant had a gun in possession for their current case (`gun`)?

Here, we’ll be working from the Defendants2025 data set, to examine mean differences in a defendant’s risk score the miles per gallon of a car (risk_score: measured as an interval-ratio variable), by whether or not they had a gun in their possession (gun: measured no or yes).

What is the Independent Samples t-Test?

The t-test examines the differences in means between two groups, in effort to see if the differences reflect true differences that we could expect to find in the population.

Load the Necessary Stuff

library(MASS)
library(psych)
library(vannstats)

Reading in the Data

data1 <- Defendants2025

Assumptions and Diagnostics for the Independent Samples t-Test

The assumptions for a t-test are…

Independence of Observations
Equal Sample Sizes
Homogeneity of Variance
Normality

1. Independence of Observations (Examine Data Collection Strategy)

Groups are not related or dependent upon each other. Case can’t be in more than one group. No ties between observations. Examine data collection strategy to see if there are linkages between observations.
- Given that the Defendants2025 data have been randomly-sampled, we have met the assumption of independence of observations.

2. Equal Sample Sizes (Examine N for each group)

The number of cases in each group should be relatively similar. (If not, use pooled variance/unequal variances asssume t-test formula)

3. Homogeneity of Variance (Examine SD² for each group)

Both groups have approximately equal variances (SD²). The distributions (or spread) for the groups are approximately equal. Keppel & Zedeck (1989) suggest that variance comparison should not exceed 10:1 ratio (or… alternatively, the SDs, when compared, should not exceed around a 3:1 ratio). In the past, you may have been instructed to use the Levene’s test to assess the degree of similarity in variances across groups. This is wrong. Unfortunately, tests such as these are overly-sensitive to trivial deviations from homogeneity of variance. It is a better practice to compare group variances/SDs based on the ratios listed above.

For both of the above (2 and 3) assumptions, we can examine the univariate data table, broken out by group:

describeBy(data1$risk_score, data1$gun)

## 
##  Descriptive statistics by group 
## group: no
##    vars    n mean   sd median trimmed  mad  min  max range skew kurtosis   se
## X1    1 1015 4.24 2.66   4.01    4.13 3.25 0.01 9.98  9.97 0.27    -0.96 0.08
## ------------------------------------------------------------------------------ 
## group: yes
##    vars   n mean   sd median trimmed  mad  min max range skew kurtosis   se
## X1    1 723 4.88 2.91   4.77    4.85 3.66 0.04  10  9.96 0.07    -1.21 0.11

Given that the group sizes are similar, we have met the assumption of equal sample sizes. Further, given that the standard deviations for both groups do not exceed a 3:1 ratio, we have met the assumption of homogeneity of variance.

4. Normality (Examine Plots: Histogram, Q-Q Normality Plots, Box-and-Whiskers Plots)

Distribution must be relatively normal. (If violated, use “unequal variances assumed” formula, otherwise, use “equal variances assumed”). In the past, you may have been instructed to use the Shapiro-Wilk test to assess normality. This is wrong. Unfortunately, tests such as these are overly-sensitive to trivial deviations from normality, and may result in you believing you must correct for normality by transforming your data. Please do not do this. The good thing is the t-test is super-robust – robust enough to provide results even in the presence of data that are not fully normally-distributed.

4a. Histogram

Plot the histogram for risk score (Y variable) broken out by whether or not the defendant was in possession of a gun (levels of the X variable)…

hst(data1, risk_score, gun)

We can see from the histograms that the distributions of the outcome variable (risk_score) by the predictor/grouping/independent variable (gun), are relatively normal.

4b. Boxplots (Box-and-Whisker Plots)

Boxplots also provide a visual representation of the normality of a distribution. The boxplot has a box, a line through the box, two whiskers on either end of the box, and sometimes dots/points outside the whiskers. Below, we get a sense of what each part of the boxplot represents…

Bottom (or left end) of the whisker represents the minimum score for that variable’s distribution
Bottom (or left end) of the box represents the first quartile (the 25th percentile case)
Middle line (or dot) inside the box represents the median, also known as the second quartile (the 50th percentile case)
Top (or right end) of the box represents the third quartile (the 75th percentile case)
Top (or right end) of the whisker represents the maximum score for that variable’s distribution
Outside dots represent outliers - extreme high or extreme low values for that variable.

To tell if a variable is normally-distrubted using the box-and-whisker plot, generally, we want to see that there is some distance between the box and the end of the whiskers, that the box isn’t pushed too close to either whisker, that the median line (dot) is near the center of the box, and that there aren’t many outliers (dots) on the outside of the whiskers.

To plot a boxplot, broken out by gun possession, we can do the following…

box(data1, risk_score, gun)

We can see from the boxplots that the data for both groups tend to be normally-distributed: The medians generally fall in the center of the interquartile range and that interquartile range is generally centered between the whiskers. Interestingly, the interquartile range is slightly larger for those with gun possession. However, the data seem normal enough. It is safe to assume that these data are close enough to normal, since they aren’t drastically different from normal, and therefore safe to proceed with the statistical test.

4c. Normal Q-Q (Quantile-Quantile) Plots

The quantile-quantile plot is a visual tool to help us figure out if the empirical distribution of our variable fits (or rather, comes from) a theoretical normal distribution.

We assess normality an break this plot out by a grouping variable.

qq(data1, risk_score, gun)

We can see from the Q-Q plot that group distributions of the outcome variable (risk_score), the data are somewhat normal, however, it is clear that the data tend to curl away from the normality line at the tails of each distribution. This indicates some deviation from normality. Therefore, it is safe to proceed with the statistical test.

Across all three plots of risk_score broken out by gun, the variables do not seem to drastically deviate from normality. Therefore, we can assume normality.

The Independent Samples t-Test Calculation

The calculation for the t-Test is:

\(t = \frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{SD_1^2}{n_1}+\frac{SD_2^2}{n_2}}}\)

where…

\(\bar{x}_1\) is the mean for group 1
\(\bar{x}_2\) is the mean for group 2
\(SD_1^2\) is the variance (\(SD^2\)) for group 1
\(SD_2^2\) is the variance (\(SD^2\)) for group 2
\(n_1\) is the number of observations (\(N\)) for group 1
\(n_2\) is the number of observations (\(N\)) for group 2

In addition, the degrees of freedom (\(df\)) for the test is…
\(df = n_1 + n_2 -2\) (aka \(df = N-2\))

Running the Independent Samples t-Test in R

To run the independent samples t-test in R, we use the traditional t.test function. But, in the vannstats package, we can use the is.t.

Within the is.t function, the data frame is listed first, followed by the (interval-ratio level) dependent variable, followed by the (discrete/categorical) independent variable is listed second.

If you meet the assumptions of the independent samples t-test, you can assume equal variances, which is assumed by default in the function (using the call var.equal=TRUE). If you violate this assumption, you must add the following call to the function: var.equal=FALSE.

#t.test(data1$risk_score ~ data1$gun, var.equal=TRUE)
#ttest <- is.t(data1, risk_score, gun, var.equal=TRUE) #this is same as below
ttest <- is.t(data1, risk_score, gun)
summary(ttest)

## Call:
## is.t(df = data1, var1 = risk_score, var2 = gun)
## 
## Independent Samples (Two Sample) t-test: 
## 
##       𝑡 Critical 𝑡   df   p-value    
##  -4.714      1.961 1736 2.623e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Group Means:
##    x̅: no   x̅: yes 
## 4.241941 4.876763

In the output above, we see the t-obtained value (-4.714, or rather, \(\pm\) 4.714), the degrees of freedom (1736), and the p-value (.000002623, which is less than our set alpha level of .05).

To interpret the findings, we report the following information:

The test used
If you reject or fail to reject the null hypothesis
The variables used in the analysis
The degrees of freedom, calculated value of the test (\(t_{obtained}\)), and \(p-value\)
- \(t(df) = t_{obtained}\), \(p-value\)

“Using an independent samples t-test, I reject/fail to reject the null hypothesis that there is no mean difference between group 1 and group 2, in the population, \(t(?) = ?, p ? .05\)”

“Using an independent samples t-test, I reject the null hypothesis that there is no difference between the mean risk score for those with and without a gun charge, in the population, \(t(1736) = \pm 4.714, p \lt .05\)”