When do you use paired t-test and how to apply the same in a practical situation?

In this article, we will talk about paired t-test analysis calculation based on a mathematical formula and using R software.

## Mathematical Approach

Let us consider two-variable x1 and x2 which are normally distributed. In a practical situation, observations are taken from the same item considered for paired t-test analysis. ie item have pre and post-values.

x1-x2 is the difference denoted as d with mean u and variance sigma1+sigma2-2sigma1sigma2.

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### Null Hypothesis

In this case, the null hypothesis is

Ho µd=0 and H1 is µd≠0

the paired t-test statistic

t statistic should be

t=mean(d)/(Sd/sqrt(n))

Let consider x1 values are 2.265, 2.267, 2.264, 2.267, 2.268, 2.263, 2.264, 2.258 and x2 values are 2.270, 2.268, 2.269, 2.273, 2.270, 2.270, 2.268, 2.268

Test pre and post values are significantly different or not.

mean of difference=0.005, sd=0.0028 and t=5.05. According tabled value n-1=7 and confidence level =0.95, Given tabled value t=2.2365.

The calculated value greater than the tabled value, which means rejecting Ho and accepting H1.

In other words, a significant difference was observed between Pre and Post values at a 95% significance level.

### R Script

Let see how to calculate paired t-test analysis in R software.

The advantage here is with a single line of R script we can execute the analysis.

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Let store values of x1 in

X1<-c( 2.265, 2.267, 2.264, 2.267, 2.268, 2.263, 2.264, 2.258)

and values of x2 in

X2<-c(2.270, 2.268, 2.269, 2.273, 2.270, 2.270, 2.268, 2.268)

#### Null Hypothesis

Ho=No difference was observed between Pre and post values

H1=Difference was observed between Pre and Post values

R Function

The R function is mentioned below

t.test(X1, X2, paired = TRUE, alternative = "two.sided")

**Result**:-

data: X1 and X2 t = -5, df = 7, p-value = 0.001565 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -0.007364624 -0.002635376 sample estimates: mean of the differences -0.005

## Conclusion: –

the p-value is less than 0.05 which means a significant difference was observed between pre and post-values. Using R with a single line of the script can easily execute the analysis.

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