That is the significant differences in the means of two independently sampled populations can be evaluated. The two assumptions are the independence of population and within each population, the variable of interest is normally distributed with equal variances. The mathematical derivation of this test statistic is as follows:where n1 and n2 are the numbers of observations in each of the two groups. x1 and x2 are the means of the two groups. is the estimate of the standard deviation of (ẋ1 – ẋ2) and is calculated using the formula:Consider the situation when we would like to determine whether there was a significant difference in the infant birth weight between mothers who smoked during pregnancy and those who did not. The mean birth weight measured among ten infants whose mothers smoked was 5 lbs., while the mean birth weight measured among the same number of infants whose mothers did not smoke was 8 lbs. Based on the weight measurements of the 20 infants, the pooled sample variance was obtained as 3 lbs. Using the above formula, the calculated t-test statistic was approximately 3.9 with 18 degrees of freedom. The two-sided p-value associated with this test is approximately 0.0006. In other words, even if there was no connection between birth weight and maternal smoking, it could be said that there is a 6 out of 10,000 chance of observing a difference at least as large as 3 lbs. by chance alone. Hence we would conclude that the observed mean difference of 3 lbs. was statistically significant as it could not be explained as to be by chance.The chi-square test is used to compare the observed data with data expected to obtain according to a specific hypothesis. The data involves categorical variables only. It requires the observed data in a tabular format containing the categories in rows and columns. Thus chi-square test statistic is designed to test the null hypothesis that there is no association between the rows and columns.