sThis arises because in science it normally wants to study the relationship of two variables when others are being controlled to verify if the one variable is changes compared to a control group is there any changes in the dependent variables observed and measured appropriately. In these situations two independent samples are selected randomly and keeping other variables constant and observe and measure a variable under study compared to the control sample.

In this article I will discuss using the t-test method to test a null hypotheses whether the differences in the mean value of an observed variable measured in interval scale is due to chance or due to the effect of changes in the independent variable which is different to the control sample at given level of significance accepted by the researcher and relevant to the field of study and phenomenon.

The t-test process to test two different sample mean from two independent samples:

To conduct t-test the researcher must know the basic assumptions of the test. He must ensure the sampling distribution is normal probability distribution. The samples are independent and the participants of the samples are different and not dependent as well the measures are or can be capable of being measured in interval scale. Most importantly they must be randomly chosen from a normally distributed population.

The t-test procedure can be illustrated by a practical example where a researcher wants to know whether different music type may have an influence on the level of concentration of some student population if other things are being constant. Say he chooses two random samples of say a size of 15 and measure the concentration when the two music are played and the two samples have different students but have the same in terms of age, knowledge, interest and other variables which may contribute to the difference in concentration. Say the statistical measures for the sample are as follows:

Mean of the 1st sample = 23.13. The sum of squared deviates SS1 = 119.73

Mean of the 2nd sample = 20.87. The sum of squared deviates SS2 = 175.73

M1- M2 = 2.26

The t- test in the above example is to test whether the difference in due to chance or due to the influence of the music of 1 or 2 at a particular significance level.

Directional and Non-Directional Hypothesis Testing

In the directional hypothesis the researcher will determine in advance the direction of the difference between two means. That is in the previous example the researcher will determine in advance whether the type 1 music is better than type 2 music in terms of concentration. This is a directional hypothesis. If he wants whether there is difference in either direction or wants only to test the difference between the means + or – then it is s non-directional hypothesis.

The t-test procedure for this example

The logic is that if the samples are coming from a normally distributed population then the sample mean difference is equal to the sample mean difference of the population. The null hypothesis is that the difference of the mean is 0. That is there is no difference in mean of the population mean or the sample mean.

Then calculate the t test or Z test co-efficient by dividing the difference in mean by the standard deviation of the population or estimated standard deviation calculated from the sum of squared deviates.

In this example the standard deviation of the population is estimated from the sample sum of squared deviation. This is the method applied in actual practice because the standard deviation of the population is mostly unknown. In mathematical form t = M1-M2/ estimated standard deviation. Estimated standard deviation is equal to square root of variance of population/sample size 1 + variance of population / sample size. However the estimated population variance = sum of squared deviates of sample 1 + sum of square deviated of sample 2/ (sample size 1 + sample size 2). The square root of the above algebraic function gives the estimated standard deviation of the population at the degrees of freedom (sample size 1 + sample size2). In the previous example the estimated variance = 119.73+ 175.73/14+14 = 10.55. There fore applying the above formula for estimated standard deviation of the difference in mean of the population = [10.55/15 +10.55/15] = + or – 1.19. There for t = 23.13-20.87/1.19 = +1.9. This with the degrees of freedom = (15-1) + (15-1) = 28. Then depending on the directional or non-directional hypothesis testing chosen by the researcher use the t-test tables for the accepted significance level and degrees of freedom to determine the t critical value. If the t critical value for the directional or non directional hypothesis is greater than the calculated t- value then the null hypothesis is correct. That is the difference of mean is not significant at the accepted level of confidence interval.  In this example the t critical value at 0.05 levels for directional hypothesis is 1.70 at 28 degrees of freedom and 2.05 for non directional hypothesis. That is the calculated t which is 1.9 is greater than 1.70 for the directional hypothesis which is 1.70. There fore the difference in mean is significant at 0.05 level of significance level. However, for non directional hypothesis the t critical value is more than the calculated t of 1.9. That is if the researcher is only interested in either direction of any differences then the difference is not significant because the t critical value is more than the actual calculated t value.

The t-test process in general

The first step is to verify the assumptions of the test and take due care for the method of selecting the sample.

The second step is to determine whether to test directional null hypothesis.

The third step is to calculate the sum of squared deviates of the two samples and the mean of the samples and the difference.

The fourth step is to calculate from the sum of squared deviates from step 3 to calculate the estimated variance of population and calculate the standard deviation of the mean differences at the sample size-1 and sample size 2-1 degrees of freedom.

The fifth step is to calculate the t- value using the differences in mean and the estimated population standard deviation.

The sixth step is to use the t table and applying the degrees of freedom, significance level and the type of directional or non directional hypothesis testing to identify the t critical value.

The seven steps are to compare the t critical value with the calculated t value. If the t critical value is greater than the t calculated value the null hypothesis is accepted. If not the null hypothesis is rejected at the given significance level.

Conclusion

As explained above it is evident of the practical use of t test in scientific and in social research to test whether the differences exist due to chance or due to the influences of the independent variable other things being equal at a particular level of significance.  As well the usefulness of t-test in comparing the differences in mean of two independent samples where the sample size is quite small. It is also important as explained above its application is relevant only if the variable can be measured in interval scale and its relevance is dependent of the normal distribution assumptions. If these assumptions are not applicable then the t-test must not be conducted and other statistical non parametric tests have to be considered. The above description also highlights the importance of the logic to be understood than the mechanical process of the t-test to understand the statistical principles applicable to the t-test and the sampling methods.