 Principles of Systematics, Module 132, Paired T-Test Paired T-Test, we apply at that time when we have to compare the means of two samples and we have the same entities i.e. the sample data we collected earlier we collected the same sample data later and we want to compare the results we had earlier and the results later there is a statistically significant difference or not if we look at this slide we have two samples one is x-sample and one is y-sample i.e. when we recorded the data we named it as x and later when we collected the same sample data we named it as y the first data we had was 9, 12, 13, 20 and 5 after that when we collected the same data later we got the answers 12, 14, 19, 25 and 10 we can say that the score of the students before the training and the score of the students after the training now we want to compare this so we represented the difference in both the samples so here we have differences minus 3, minus 2, minus 6, minus 5 and minus 5 after that the difference we got we took the scale of this difference after taking the scale we summed all of these summation d-square we summed all of these now for the parity test for the parity test the formula we have is d-bar this is basically the difference of both the samples is called the mean d-bar and after that this is the variance and here is the number now when we took the variance we got the variance 20.34 this is the formula of the variance when we put the variance the answer we got is 20.34 after that d-bar we got 4.2 now for the variance and d-bar and n we put all the values of the parity test in the formula and then we got the value of t i minus 2.08 this is the value we calculated because we have calculated now for the parity test in the parity test because there are period observations because of this all the period observations we get minus 1 so we get the degree of freedom now because there were 5 period observations n minus 1 4 so this means the degree of freedom is 4 the independent test the degree of freedom of both the samples is removed then it is added whereas in this case because there are period observations because of this only one degree of freedom comes now the calculated value is minus 2.08 whereas the table value is 2.77 now here we ignored the sign our table value is bigger and our answer the calculated value is less this means the samples we compared they there is no statistically significant difference so in this way through the parity test we found out before training and after training the score there is a statistical difference or not so the parity test we apply at that time when we have to compare the means of two samples and we have the period observations apart from that the first score or the score later we get the difference we get the difference after that variance and d bar d bar n and variance we put the values in the t test formula and we get the calculated value later on we compare the calculated value with the standard value and we find out that there is a statistical difference or not so in this case both the values in the first and later there is no statistical difference in the mean