An experiment was conducted to test the efficacy of chloromycetin in checking typhoid. In a certain
hospital chloromycetin was given to 285 out of the 392 patients suffering from typhoid. The number of
typhoid cases were as follows: Typhoid No Typhoid Total
Chloromycetin 35 250 285
No chloromycetin 50 57 107
Total 85 307 392
With the help of χ2 , test the effectiveness of chloromycetin in checking typhoid
Answers
It is a example of Chi square.
it is statistic test that measures how a model compares to actual observed data
Explanation:
From given, the the typhoid casesthe amount of Chloromycetin and no Chloromycetin and the total patients.
The observed data does not have any relation with the no of typhoid cases hence there will be some assumptions one should follow:
the χ2 assumes that the data for the study is obtained through random selection, i.e. they are randomly picked from the population
like here it is effectiveness of chloromycetin in checking typhoid
Answer:
Chi-Square Value that has been calculated is 54.33 and degree of freedom for the calculated result is (c-1)(r-1) which is (2-1)(2-1) = 1 and for 1 degree of freedom for 5% level of significance; the tabulated Chi-Square value is 3.841 (already given in the question) so the calculated chi-square value > tabulated chi-square value. i.e. 54.33 > 3.841, so, the null hypothesis is rejected and alternative hypothesis is accepted. If it is less than or equal to the tabulated chi-square value then null hypothesis would be accepted.
Explanation: