Question

How is the chi-square independence test similar to the goodness-of-fit test? How is it different?

What is the difference between McMemar’s Test and the Chi-squared test for 2 by 2 table.

Question 2

A clinic administers two drugs to two groups of randomly assigned patients to cure the same disease: 70 patients received Drug 1 and 80 patients received Drug 2. The following table gives the information about the number of patients cured and the once not cured by each of these two drugs

	Cured	Not cured
Drug 1	59	11
Drug 2	61	19

A. For this contingency table, what is the number of degrees of freedom?

1. 4
2. 3
3. 1
4. 2

B. For this contingency table, what are the observed frequencies for the first row?

1. 56 and 14
2. 59 and 11
3. 59 and 61
4. 56 and 64

C. For this contingency table, what are the expected frequencies for the first row?

1. 59 and 11
2. 59 and 61
3. 56 and 64
4. 56 and 14

D. For this contingency table, what is the critical value of chi square, use 10% significance level?

1. 5.991
2. 4.605
3. 3.841
4. 2.706

E. What is the test statistics for this contingency table?

1. 1.51
2. 2.15
3. 1.61
4. 2.15

F. Based on chi-squared test, we can say that

1. The drug is effective
2. The drug is not effective

For this question, please answer in the following format:

A	B	C	D	E	F

Answer 1

A	B	C	D	E	F
1	59 and 11	56 and 14	2.706	1.51	The drug is not effective

below are the details":

A)

degree of freedom(df) =(rows-1)*(columns-1)=

1

B)

observed frequencies for the first row 59 and 11

c) Ei=row total*column total/grand total

from above

expected frequencies for the first row =56 and 14

d)

for 1 df and 0.1 level , critical value       χ2=

2.706

e)

Applying chi square test of independence:

chi square    χ2	=(Oi-Ei)2/Ei	c	d	Total
	a	0.161	0.643	0.8036
	b	0.141	0.5625	0.7031
	total	0.3013	1.2054	1.507
test statistic X2 =		1.51

F)

since test statistic <critical value

The drug is not effective