First Step:
Import the data.
dat <- read.csv("uav.csv")dat$Training <- factor(dat$Training, labels = c("Standard Training", "New Training"))
datsummary(dat) ## Training Maneuver Sensor Knowledge
## Standard Training:6 Min. :14.00 Min. :25.00 Min. :14.00
## New Training :5 1st Qu.:18.50 1st Qu.:30.50 1st Qu.:17.50
## Median :20.00 Median :41.00 Median :20.00
## Mean :21.64 Mean :39.91 Mean :20.82
## 3rd Qu.:25.50 3rd Qu.:50.00 3rd Qu.:24.00
## Max. :29.00 Max. :54.00 Max. :29.00levels(dat$Training)## [1] "Standard Training" "New Training"
To operate a Raven B (a small unmanned aerial vehicle), 11 special forces soldiers completed either Standard training (0) or New training
(1). Upon completion of training, the soldiers had scores on various criteria including Maneuvering Efficacy, Sensor Operation Proficiency, and Declarative Knowledge. (Higher scores are better.)
library(car)## Loading required package: carDataleveneTest(Maneuver ~ Training, dat, center = mean)According to Levene’s Test for Equality of Variances F(1)=.1779, p=.683, it is not statistically significant and the homogeneity of variance assumption is not violated. We will accept equal variances assumed.
Student’s T-Test
if(!require(lsr)){install.packages("lsr")}## Loading required package: lsrindependentSamplesTTest(Maneuver ~ Training, data = dat, var.equal = T)##
## Student's independent samples t-test
##
## Outcome variable: Maneuver
## Grouping variable: Training
##
## Descriptive statistics:
## Standard Training New Training
## mean 18.333 25.600
## std dev. 2.733 3.507
##
## Hypotheses:
## null: population means equal for both groups
## alternative: different population means in each group
##
## Test results:
## t-statistic: -3.87
## degrees of freedom: 9
## p-value: 0.004
##
## Other information:
## two-sided 95% confidence interval: [-11.514, -3.019]
## estimated effect size (Cohen's d): 2.343The scores t(9)=-.87, p=.004 are statistically significant to support different population means in each group. The null hypothesis is rejected. Participants in the new training scored higher on maneuvering efficacy (M=25.6, SD=3.51) compared to standard training participants (M=18.33, SD=2.73). The effect size was d=2.34.
leveneTest(Sensor ~ Training, dat, center = mean)According to Levene’s Test for Equality of Variances F(1)=1.27, p=.29, it is not statistically significant and the homogeneity of variance assumption is not violated. We will accept equal variances assumed.
Student’s T-Test
independentSamplesTTest(Sensor ~ Training, data = dat, var.equal = T)##
## Student's independent samples t-test
##
## Outcome variable: Sensor
## Grouping variable: Training
##
## Descriptive statistics:
## Standard Training New Training
## mean 34.000 47.000
## std dev. 10.392 6.671
##
## Hypotheses:
## null: population means equal for both groups
## alternative: different population means in each group
##
## Test results:
## t-statistic: -2.404
## degrees of freedom: 9
## p-value: 0.04
##
## Other information:
## two-sided 95% confidence interval: [-25.235, -0.765]
## estimated effect size (Cohen's d): 1.455The scores t(9)=-2.404, p=.04 are statistically significant to support different population means in each group. The null hypothesis is rejected. Participants in the new training score higher on sensor operation proficiency (M=47, SD=6.67) compared to standard training participants (M=34, SD=10.39).The effect size was d=1.46.
leveneTest(Knowledge ~ Training, dat, center = mean)According to Levene’s Test for Equality of Variances F(1)=.9 p=.37, it is not statistically significant and the homogeneity of variance assumption is not violated. We will accept equal variances assumed.
Student’s T-Test
independentSamplesTTest(Knowledge ~ Training, data = dat, var.equal = T)##
## Student's independent samples t-test
##
## Outcome variable: Knowledge
## Grouping variable: Training
##
## Descriptive statistics:
## Standard Training New Training
## mean 18.333 23.800
## std dev. 3.502 4.817
##
## Hypotheses:
## null: population means equal for both groups
## alternative: different population means in each group
##
## Test results:
## t-statistic: -2.182
## degrees of freedom: 9
## p-value: 0.057
##
## Other information:
## two-sided 95% confidence interval: [-11.135, 0.202]
## estimated effect size (Cohen's d): 1.321The scores t(9)=-2.182, p=.06 are not statistically significant to support different population means in each group. The researchers will fail to reject the null hypothesis. There was no significant difference in knowledge between new training (M=23.8, SD=4.82) compared to standard training participants (M=18.3, SD=3.5).The effect size was d=1.32.
The special forces soldiers that completed the new training had a significantly higher performance on maneuvering efficacy and sensor operation proficiency. The new training did not have a significant impact on declarative knowledge.
2.) For the two independent samples t-test, to ensure the validity of the test statistic, an important statistical assumption is homoscedasticity or homogeneity of variance. (a) Conceptually, what does this assumption mean? (b) Explain why this assumption was never mentioned in the case of a one sample test on a mean.
Homogeneneity of variance is the concept of the variances between the two indepent sample sizes are equal. Conceptually, the variance does not differ between the two datasets and the Student’s T-test will be used instead of Welch’s T. The assumption was never mentioned with one sample because two samples are required to measure variance homogeneity.