In statistical research, clearly articulated hypotheses that guide the vector of further data analysis are of paramount importance. In works that are based on quantitative paradigms, hypotheses can be directional or non-directional, depending on what question the researcher is asking. Once the null and alternative hypotheses have been formulated, the researcher must select an appropriate statistical test based on the nature of the variables, the end goal, and the resources available. Examples of such tests are correlation and regression analysis, as well as t-tests, ANOVAs, posterior tests, and other methods designed specifically for a narrow problem. The results of the statistical analysis performed are used to compare the previously indicated level of significance (alpha) with the calculated level; when this p-value is above the level of significance, the null hypothesis is rejected and accepted otherwise.
Alternative and Null Hypotheses
The formulation of the null and alternative hypothesis must strictly follow the researcher’s question. In the current assignment, the researcher is examining the potential effectiveness of a treatment to lower cholesterol levels in 114 patients. Excessive cholesterol concentrations are known to stress normal cardiovascular function and may be predictors of pathology (Vourakis et al., 2021). Accordingly, the null hypothesis would postulate that treatment has no significant effect on reducing cholesterol levels, whereas the alternative hypothesis reports that treatment is effective in reducing cholesterol in patients.
Test Statistics
Both the data collected by the researcher and the nature of the research question point to the fact that nonparametric tests should be addressed in this case. Nonparametric analysis, unlike classical parametric analysis, does not need to evaluate the assumption of normality of the distribution, that is, it does not impose restrictions on population parameters (Turney, 2022). In this case, the chi-square test of independence, a traditional nonparametric test assessing differences between two or more categorical variables, can be useful for use. It is worth noting that the nonparametric nature does not mean that the conclusions of the analysis cannot be extrapolated to the population. On the contrary, since the chi-square test of independence is a logical method, sample test results are also applicable to the population if the significance is confirmed. The rationale for this choice is simple: the investigator’s data are dichotomous categorical and include the fact of treatment (Yes/No) and the effect on cholesterol concentration (Decreases/No Decreases). Accordingly, this test is designed to test whether both variables are related or whether their relationship is not statistically confirmed (Turney, 2022). If their relationship is confirmed, it can be concluded that a change in one of the variables affects a change in the other — in which case, the treatment affects the change in cholesterol, as postulated by the alternative hypothesis.
The mechanism of this test statistic comes down to comparing the observed frequencies, that is, those that have already been obtained by the researcher, with the expected frequencies in the case that the variables would have been unrelated. When performing the test manually, expected values can be calculated using the formula ((row total × column total)/N), but when conducting the test using computer programs (SPSS, MS Excel), expected frequencies are calculated automatically (Turney, 2022). The chi-square criterion is calculated as the sum of the square of the difference between the expected and observed values divided by the expected value. A p-value is calculated for this parameter, taking into account the number of degrees of freedom, and comparing it to the chosen level of significance.
Calculations
Since the assignment requires manual chi-square calculations for the 2×2 data table below, this section shows and comments on all steps in detail; by the end of this block, the calculated p-value.
Table 1. Observed and expected frequencies for crossings of variables.
Now that the observed and expected frequencies for each of the intersections are known, it is possible to determine the chi-square:
And also, the number of degrees of freedom:
Using standard p-values tables for chi-tests, taking into account the calculated criterion and the number of degrees of significance, the final p-value for this test was.07921.
Discussion and Interpretation
In the present work, the researcher sought to determine if the treatment program had an effect on cholesterol reduction in 114 patients. To answer this question, the necessary data were collected and organized in a 2×2 table and then used to perform a chi-square test for independence. The results showed that χ2(1, N = 114) = 3.081, p =.07921 – when comparing the calculated p-value with the alpha of significance (.05), it can be seen that.07921 >.05. It follows that there is not enough evidence to reject the null hypothesis and therefore it should be accepted. That is, the results are not significant, and therefore the treatment proposed by the researcher has no effect on reducing cholesterol concentrations in the body.
References
Turney, S. (2022). Chi-square test of independence | formula, guide & examples. Scribbr.
Vourakis, M., Mayer, G., & Rousseau, G. (2021). The role of gut microbiota on cholesterol metabolism in atherosclerosis. International Journal of Molecular Sciences, 22(15), 1-8.