Question

In a research to study the gender differences in receiving Lidocaine therapy in Acure Myocardial infarction (AMI) patients older than 75 years who have a history of hypertension and strokes, there were 81 patients in total (30 males and 51 females). The following table provides the outcomes.

	Therapy
Gender	Receive	Not receive	Total
Male	8	22	30
Female	16	35	51
Total	24	57	81

1. According to the hypothesis that there is no association between receive Lidocaine therapy and gender, what does the minimum expected cell count in this table?Based on the answer to part a, which test should you use to test the hypothesis of no association between receive Lidocaine therapy and gender?Test the hypothesis whether the receiving Lidocaine therapy is independent of their gender at .05 significance level. Find the critical value and p-value, respectively. If the numbers in each cell are changed to above table, what test should be used to test the hypothesis whether the receiving Lidocaine therapy is independent of their gender?

2. 	Therapy
   Gender	Receive	Not receive	Total
   Male	2	28	30
   Female	5	46	51
   Total	7	74	81

Answer 1

Solution

Final answers in the stipulated format are given below. Back-up Theory and Details of calculations follow at the end.

Part (a)

Sub-part (i)

According to the hypothesis that there is no association between receive Lidocaine therapy and gender, the minimum expected cell count in this table is 8.89 Answer 1

Sub-part (ii)

Based on the answer to part a, which test should you use to test the hypothesis of no association between receive Lidocaine therapy and gender:

Chi-square test for independence, also known as Contingency Chi-square Test. Answer 2

Sub-part (iii)

The hypothesis that the receiving Lidocaine therapy is independent of their gender is accepted at .05 significance level. Answer 3

Critical value = 3.8415 Answer 4

p-value = 0.6542 Answer 5

Part (b)

Here, the expected frequencies in two cells fall below 5. Hence, Chi-square test cannot be applied.

The test to be used is:

p1 = p2, where p1 and p2 are respectively the proportion for male and female under ‘receive therapy’ category. Answer 5

Back-up Theory and Details of calculations

Suppose we have a contingency table with r rows representing r levels/grades of one attribute and c columns representing c levels/grades of another attribute.

The Chi-square Test of Independence is designed to test if the two attributes are associated.

Hypotheses

Null H₀: The two attributes are independent Vs

Alternative H₁: The two attributes are not independent

Test Statistic

χ² = ∑(i = 1 to r, j = 1 to c){(Oij - Eij)²/Eij}, where Oij and Eij are respectively, the observed and the expected frequencies of the ijth cell of the contingency table.

Under H₀, Eij = (Oi. x O.j)/O.., where Oi.,O.j, and O.. are respectively the ith row total, jth column total and grand total.

Calculations

DF	1	α	0.05	χ2(crit)	3.8415
χ2(cal)	0.2006	p-value	0.6542	Accept H0
Oij
i	j = 1	j = 2	Oi.
1	8	22	30
2	16	35	51
O.j	24	57	81
Eij
i	j = 1	j = 2	Ei.
1	8.8889	21.1111	30
2	15.1111	35.8889	51
E.j	24	57	81
χ2ij
i	j = 1	j = 2	Total
1	0.0889	0.0374	0.1263
2	0.0523	0.0220	0.0743
Total	0.1412	0.0594	0.2006

Distribution

Under H₀, χ² ~ χ²_n, where n = {(r - 1)(s - 1)}

Critical Value

Given level of significance as α, critical value is the upper α% point of χ²_n.

p-value

P(χ²_n > χ²_cal)

Critical value and p-value obtained using Excel Function: Statistical CHIINV and CHIDIST are given in the above table.

Decision

Since χ²_cal < χ²_crit, or equivalently, p-value > α, H₀is accepted.

Conclusion

There is enough evidence to suggest that the two attributes, are independent.

DONE