In the context of outbreak of COVID – 19, a sample of 200 patients are selected. Out of them, 100 were given a drug and the others were not given any drug. The results are as follows:Number of patients Drug No Drug Total Cured 6555120 Not cured 354580 Total 100100200 Test whether the drug is effective or not. (The Chi-square value as 5% level of significance for 1 degree of freedom is 3.841)

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In the context of outbreak of COVID – 19, a sample of 200 patients are selected. Out of them, 100 were given a drug and the others were not given any drug. The results are as follows:Number of patients Drug  No Drug  Total  Cured 6555120 Not cured 354580 Total 100100200 Test whether the drug is effective or not. (The Chi-square value as 5% level of significance for 1 degree of freedom is 3.841)

Davon Friedman · Accepted Answer

The chi-square test is applied when:1. Data is in the form of frequencies and when it is discrete. However, continuous data can also be used which is reducible to categories in such a way that it becomes discrete data.2. There is quantitative data, which means numerical data, usually discrete quantitative data.3. The observations are independent.4. The sample size is large enough, at least 50.Assume the null and alternate hypothesis is:H0: Drug is effectiveH1:Drug is not effectiveThe formula of chi-square test statistic is:C=∑i∑j[(Oij−Eij)2Eij]Where Oij and Eij denote observed and expected frequencies corresponding to ith row and jth column.And, i=(Cured, Not Cured), j=(Drug, No Drug)The expected frequencies are obtained by the formula:Eij=∑i∑j[(Ri−Cj)2G]Where Ri,Cj, and G denote row total corresponding to ith row, column total corresponding to jth column, and grand total across all rows and columns respectively.Therefore, the value of expected frequencies is:ECured Drug=120×100200=60ECured No Drug=120×100200=60ENot Cured Drug=80×100200=40ENot Cured No Drug=80×100200=40The observed frequencies are given as:OCured Drug=65OCured No Drug=65ONot Cured Drug=35ONot Cured No Drug=35Hence, the value of chi-square test statistic is obtained as:C=(65−60)260+(55−60)260+(35−40)240+(45−40)240=2.083The chi-square critical value at 5% level of significance for 1 degree of freedom is given as 3.841.Since, the value of chi-square test statistic is less than chi-square critical value at 5% level of significance for 1 degree of freedom, therefore, the null hypothesis is failed to reject at 5% level of significance and hence it is reasonable to conclude that the drug is effective.

In the context of outbreak of COVID – 19, a sample of 200 patients are selected. Out of them, 100 we

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