Exercise 3
Winelyn Pelias
2022-09-28
Paired sample t-test using RICT training data
Paired sample T-test is also called the dependent sample T-test. It compares differences between two dependent mean scores. It is used to determine whether the mean difference between two sets of observations is zero. In a paired sample t-test each subject or entity is measured twice. This results in pairs of observations. The common application of paired sample t-test includes case-control studies or repeated measures.
Formulation of hypothesis
Like many statistical procedures, the paired sample t-test has two competing hypotheses, the null hypothesis and the alternative hypothesis. The null hypothesis assumes that the true mean difference between the paired samples is zero. Under this model, all observable differences are explained by random variation. Conversely, the alternative hypothesis assumes that the true mean difference between the paired samples is not equal to zero.
- H0H0: there is no difference in participants’ knowledge before and after the ICT training
- H1H1: ICT training affected the participant’s knowledge
The alternative hypothesis can take one of several forms depending on the expected outcome. If the direction of the difference does not matter, a two-tailed hypothesis is used. Otherwise, an upper-tailed or lower-tailed hypothesis can be used to increase the power of the test.
Note: It is important to remember that hypotheses are never about data, they are about the processes which produce the data.Example data set
We shall test this hypothesis against the alternative hypothesis. The data is shown below which is the ICT training data
Let’s create this data set in Rstudio. First, create before and after as objects containing the scores of ICT training.
before <- c(12.2, 14.6, 13.4, 11.2, 12.7, 10.4, 15.8, 13.9, 9.5, 14.2)
after <- c(13.5, 15.2, 13.6, 12.8, 13.7, 11.3, 16.5, 13.4, 8.7, 14.6)Now create a data matrix using data.frame() function. In this function, you can define the variables. The first variable is the subject or individuals. As the sample size is 1010 so we shall repeat 11 to 1010 twice. The second variable is time interval containing two levels. Each level will be repeated 1010 times. The third variable is ICT score representing response variable values. Use objects before and after as created earlier to represent scores. Let’s print the data to see how it looks like.
data <- data.frame(subject = rep(c(1:10), 2),
time = rep(c("before", "after"), each = 10),
score = c(before, after))
print(data) subject time score
1 1 before 12.2
2 2 before 14.6
3 3 before 13.4
4 4 before 11.2
5 5 before 12.7
6 6 before 10.4
7 7 before 15.8
8 8 before 13.9
9 9 before 9.5
10 10 before 14.2
11 1 after 13.5
12 2 after 15.2
13 3 after 13.6
14 4 after 12.8
15 5 after 13.7
16 6 after 11.3
17 7 after 16.5
18 8 after 13.4
19 9 after 8.7
20 10 after 14.6Use str() function to see the variable structure. The structure of the data variables is fine. The function attach() gives direct access to the variables of a data frame by typing the name of a variable as it is written on the first line of the file.
str(data)'data.frame': 20 obs. of 3 variables:
$ subject: int 1 2 3 4 5 6 7 8 9 10 ...
$ time : chr "before" "before" "before" "before" ...
$ score : num 12.2 14.6 13.4 11.2 12.7 10.4 15.8 13.9 9.5 14.2 ...attach(data)Descriptive analysis
Before proceeding for t-test let’s see the summary statistics of the data. Use summary() function to get the summary of scores. The summary() is a generic function used to produce result summaries of the results of various model fitting functions.
summary(data) subject time score
Min. : 1.0 Length:20 Min. : 8.70
1st Qu.: 3.0 Class :character 1st Qu.:11.97
Median : 5.5 Mode :character Median :13.45
Mean : 5.5 Mean :13.06
3rd Qu.: 8.0 3rd Qu.:14.30
Max. :10.0 Max. :16.50 You can also get a summary of each time interval using by() function. In data argument type the name of the data file which was created as object earlier. In INDICES argument specify the second variable in quotation marks within square brackets. This will specify the time variable to split it into its components. Putting summary in FUN argument will apply summary to each component.
by(data = data,
INDICES = data[,"time"],
FUN = summary)data[, "time"]: after
subject time score
Min. : 1.00 Length:10 Min. : 8.70
1st Qu.: 3.25 Class :character 1st Qu.:12.95
Median : 5.50 Mode :character Median :13.55
Mean : 5.50 Mean :13.33
3rd Qu.: 7.75 3rd Qu.:14.38
Max. :10.00 Max. :16.50
------------------------------------------------------------
data[, "time"]: before
subject time score
Min. : 1.00 Length:10 Min. : 9.50
1st Qu.: 3.25 Class :character 1st Qu.:11.45
Median : 5.50 Mode :character Median :13.05
Mean : 5.50 Mean :12.79
3rd Qu.: 7.75 3rd Qu.:14.12
Max. :10.00 Max. :15.80 You may be interested in the summary statistics of the difference in scores before and after ICT training. Subtract the ICT score before training from the one after training to get the score differences. Again using the summary function for diff object will result in summary statistics for the differences.
diff = after - before
summary(diff) Min. 1st Qu. Median Mean 3rd Qu. Max.
-0.800 0.250 0.650 0.540 0.975 1.600 Association or correlation test
Now let’s see the association or correlation between the paired samples. Use cor.test() function to test this association. Type the components of the time variable in x and y arguments. In the method argument, we shall use Pearson which is the most commonly used method. Let’s test this relationship at 0.95 confidence level.
library(stats)
cor.test(x = before, y = after,
method = c("pearson"),
conf.level = 0.95)
Pearson's product-moment correlation
data: before and after
t = 7.5468, df = 8, p-value = 6.628e-05
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
0.7474506 0.9851801
sample estimates:
cor
0.9363955 The results show that there is a significantly strong relationship between before and after training scores. The correlation coefficient is 0.940.94 which is very close to one. It reflects a strong positive relationship or association in before and after ICT training scores.
Visualizing samples
Its good practice to visualize the data set before applying test statistics. To set query of graphical parameters use par() function. To specify square plotting region of the plot set “s” for pty argument. To produce a box plot use boxplot() function where the argument formula is set to score ~ time. You can add additional arguments like col to set colors, main to add title and xlab and ylab arguments to add labels to X and Y-axis labels.
par(pty = "s")
boxplot(score ~ time)
boxplot(score ~ time,
col = c("#003C67FF", "#EFC000FF"),
main = "ICT training score improves knowlege",
xlab = "Time", ylab = "Score")
In a box plot we can get the information of distributional characteristics:
- Upper and lower whisker represents the scores outside the middle 50 percent
- Upper and lower quartiles
- 75 percent of the scores fall below the upper quartile and 25% of the scores fall below the lower quartile
- The middle box represents the middle 50 percent of the score (Interquartile range)
- The median or middle quartile represent the midpoint of the data
Characteristics of Boxplot
Homogeneity in variances
Before proceeding with test statistics, first determine the homogeneity of variances of time variables for ICT training scores. You can use bartlett.test() where the score is separated by time variable.
bartlett.test(score ~ time)
Bartlett test of homogeneity of variances
data: score by time
Bartlett's K-squared = 0.050609, df = 1, p-value = 0.822The probability value is 0.820.82 which is higher than 0.050.05. This indicated that there is no significant difference in variances. It also means variances are homogeneous.
Paired sample t-test
t.test(formula = score ~ time,
alternative = "greater",
mu = 0,
paired = TRUE,
var.equal = TRUE,
conf.level = 0.95)
Paired t-test
data: score by time
t = 2.272, df = 9, p-value = 0.0246
alternative hypothesis: true mean difference is greater than 0
95 percent confidence interval:
0.1043169 Inf
sample estimates:
mean difference
0.54 Interpretation
Statistical significance is determined by looking at the p-value. The p-value gives the probability of observing the test results under the null hypothesis. The lower the p-value, the lower the probability of obtaining a result like the one that was observed if the null hypothesis was true. Thus, a low p-value indicates decreased support for the null hypothesis.
The results showed that the probability value is lower than 0.05. Lower the P-value, lower the evidence we have to support the null hypothesis. Based on this result, we shall reject the null hypothesis of no difference. It means ICT training significantly improved the participants’ knowledge.