### Paired sample t-test in R

### Introduction

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 include case-control studies or repeated measures.

The paired sample t-test has the following assumptions:

- The dependent variable must be numeric and continuous
- The observations are independent of one another
- The dependent variable should be approximately normally distributed
- The dependent variable should not contain any outliers

Let’s carry out paired sample t-test in R studio. Suppose you are interested in evaluating the effectiveness of a company training program. One approach you might consider would be to measure the performance of a sample of employees before and after completing the program, and analyze the differences using a *paired sample t-test*.

Suppose a training program was conducted to improve the participants’ knowledge on ICT. Data were collected from a selected sample of \(10\) individuals before and after ICT training program. Test the hypothesis that the training is effective to improve the participants’ knowledge on ICT at \(5\) percent level of significance.

### 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.

\(H_0\): there is no difference in participants’ knowledge before and after the ICT training

\(H_1\): 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:

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 data matrix using `data.frame()`

function. In this function you can define the variables. First variable is subject or individuals. As sample size is \(10\) so we shall repeat \(1\) to \(10\) twice. Second variable is time interval containing two levels. Each level will be repeated \(10\) times. 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 14.6
```

Use `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.

```
# 'data.frame': 20 obs. of 3 variables:
# $ subject: int 1 2 3 4 5 6 7 8 9 10 ...
# $ time : Factor w/ 2 levels "after","before": 2 2 2 2 2 2 2 2 2 2 ...
# $ score : num 12.2 14.6 13.4 11.2 12.7 10.4 15.8 13.9 9.5 14.2 ...
```

### 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.

```
# subject time score
# Min. : 1.0 after :10 Min. : 8.70
# 1st Qu.: 3.0 before:10 1st Qu.:11.97
# Median : 5.5 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 summary of each time intervals 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.

```
# data[, "time"]: after
# subject time score
# Min. : 1.00 after :10 Min. : 8.70
# 1st Qu.: 3.25 before: 0 1st Qu.:12.95
# Median : 5.50 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 after : 0 Min. : 9.50
# 1st Qu.: 3.25 before:10 1st Qu.:11.45
# Median : 5.50 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 of scores before and after ICT training. Subtract the ICT score before training from the one after training to get the score differences. Again using summary function for `diff`

object will result in summary statistics for the differences.

```
# 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 time variable in `x`

and `y`

arguments. In method argument we shall use **Pearson** which is most commonly used method. Let’s test this relationship at 0.95 confidence level.

```
#
# 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 shows that there is significantly strong relationship in before and after training scores. The correlation coefficient is \(0.94\) which is very close to one. It reflects strong positive relationship or association in before and after ICT training score.

#### Visualizing samples

Its good practice to visualize 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

#### Homogeneity in variances

Before proceeding for test statistics, first determine the homogeneity of variances of time variable for ICT training scores. You can use `bartlett.test()`

where **score** is separated by **time** variable.

```
#
# Bartlett test of homogeneity of variances
#
# data: score by time
# Bartlett's K-squared = 0.050609, df = 1, p-value = 0.822
```

The probability value is \(0.82\) which is higher than \(0.05\). This indicated that there is no significant difference in variances. It also means variances are homogeneous.

### Paired sample t-test

Now let’s apply the paired sample t-test on data set by using `t.test()`

function. Set the value for `formula`

argument as score is separated by time. In `alternative`

argument set the value of alternative hypothesis as **“two.sided”** in this case for two tailed hypothesis. The `mu`

argument indicate the true value of difference in means for a two sample test. You can set the value of this argument according to the null hypothesis as in this case null hypothesis is the hypothesis of no difference (\(H_0: \mu_1 - \mu_2=0\)). Set *TRUE* for `paired`

argument as this is paired sample data and each subject is measured twice before and after the ICT training. The variances are homogeneous as indicated by the test of homogeneity computed earlier so set *TRUE* for `var.equal`

argument. Keep confidence level at \(0.95\) in `conf.level`

argument. This corresponds to \(95\) percent chance of obtaining a result like the one that was observed if the null hypothesis was false.

```
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 difference in means is greater than 0
# 95 percent confidence interval:
# 0.1043169 Inf
# sample estimates:
# mean of the differences
# 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.

`Please comment below if you have any questions.`

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