Introduction Inferential statistics is a type of statistics that extends beyond the data set to conclude. Inferential statistics utilizes sample of data that are randomly selected from a population to explain and infer. They aid in drawing inferences from a sample of the population and may be used to determine if there are differences between groups or if links between variables have been established. They may be used to compare group means and determine if there are substantial variations across groups (Guetterman, 2019). Inferential statistics are frequently used to contrast treatment groups and draw conclusions about the greater population of interest. A one-way ANOVA is often used when there are three or more independent groups, although it may also be employed when there are just two groups. The paper argues that descriptive statistical analyses are used in conjunction with inferential statistics to derive inferences from a population sample utilizing practical tests such as t-tests and ANOVA statistics. The Average, Mode, and Median of Discharges Determined per State In 2014 In 2014, frequency data for 31 states were compiled to establish the mode, mean, and median of discharges. To analyze the data for the variable discharges, the central tendency approach was calculated. Dispersion measures were estimated to understand better the variation of discharges across all states for the 2014 set. The study yielded the following results: N = 31, the average is 604175, the median is 393002, the model is 49564, SD is 609284. The research deduced from the substantial standard deviation that the number of discharges differed significantly across states in 2014. Application of ANOVA to compare the number of discharges in 2015, 2012, and 2010 across all states In 2010, 2012, and 2015, a one-way ANOVA was done on 92 discharges from all states. The Levene statistic revealed a significant result, p 0.001. As a result, the categories are statistically significantly distinct, and their variance is not homogeneous since F is (32,59) = 1102.4, and P is 0.05. In general, there was a statistically significant variation number of discharges across the states. Post hoc analysis of the proportion of discharges was not possible because at least one state had less than two data points. As a result, the Welch t-test was developed since the statistic could not be computed. However, the difference between the groups is greater than the difference within the groups, and the number of discharges was not substantially different among the groups. Comparison Between 2010 and 2015, of the Number of Discharges in All States Between 2010 and 2015, an independent sample–test was used to compare the mean discharges for all states. Discharges in 2015 (M = 593000, SD = 613848) and 2010 (M = 756629, SD = 860411) were not significantly different. The findings imply that identical variances between 2010 and 2015 are assumed since p = 0.387 and t (61) = 0.871. Using Levine’s test to carry out the assumption, the null hypothesis tests the variance of the population, showing homogeneity. Therefore, the null hypothesis of equality of variances is rejected, and it is inferred that the population’s variances vary. ANOVA to Exami