Question

1. The following data was collected in a clinical trial evaluating a new compound designed to improve wound healing in trauma patients. The new compound was compared against a placebo. After treatment for 5 days with the new compound or placebo, the extent of wound healing was measured and the data are shown below. (10 points)

	Percent Wound Healing
Treatment	0-25%	26-50%	51-75%	76-100%
New Compound (n=125)	15	37	32	41
Placebo (n=125)	36	45	34	10

Is there a difference in the extent of wound healing by treatment? (Hint: Are treatment and the percent wound healing independent?) Run the appropriate test at a 5% level of significance.

2. Use the data in Problem #1 and pool the data across the treatments into one sample of size n=250. Use the pooled data to test whether the distribution of the percent wound healing is approximately normal. Specifically, use the following distribution: 30%, 40%, 20% and 10% and a=0.05 to run the appropriate test. (10 points)

	Percent Wound Healing
Treatment	0-25%	26-50%	51-75%	76-100%	Total
Number of Patients	51	82	66	51	250

Answer 1

Chi-square test of independence

The Chi-square test of independence is performed to test the association between the extent of wound healing and the type of treatment

Hypothesis:

H0​: The two variables are independent

Ha: The two variables are dependent

Chi-Square statistic:

The observed values are,

	Percent Wound Healing
Treatment	0-25%	26-50%	51-75%	76-100%	Total
New Compound (n=125)	15	37	32	41	125
Placebo (n=125)	36	45	34	10	125
Total	51	82	66	51	250

The expected values are obtained using the following formula,

	Percent Wound Healing
Treatment	0-25%	26-50%	51-75%	76-100%	Total
New Compound (n=125)	125*51/250=25.5	125*82/250=41	125*66/250=33	125*51/250=25.5	125
Placebo (n=125)	125*51/250=25.5	125*82/250=41	125*66/250=33	125*51/250=25.5	125
Total	51	82	66	51	250

The Chi-Square value is obtained using the formula

Observed, Oi	Expected, Ei
15	25.5	4.32
37	41	0.39
32	33	0.03
41	25.5	9.42
36	25.5	4.32
45	41	0.39
34	33	0.03
10	25.5	9.42
	Sum	28.33

Critical value:

The critical value for the chi-square statistic is obtained from the chi-square distribution table for,

degree of freedom = (r-1)(c-1) = (2-1)(4-1) = 3

Where r is the number of rows and c is the number of columns in the table.

For significance level = 0.05 and degree of freedom = 3, the critical value is,

Conclusion:

Since the chi-square value is greater than the critical value, It can be concluded that the null hypothesis is rejected at a 5% significance level.

Hence the extent of wound healing and the type of treatment are dependent.

Chi-Square Goodness of fit test

The Chi-Square Goodness of fit test is used here to test whether the samples are from normal distribution within a given proportion.

Hypothesis:

Null hypothesis: There is no significant difference between the observed frequency and the expected frequency.

Chi-Square statistic:

The observed values are,

Treatment	Observed
0-25%	51
26-50%	82
51-75%	66
76-100%	51
Total	250

The expected values are obtained by multiplying the total population by the given proportion,

Treatment	Expected
0-25%	250*0.30=75
26-50%	250*0.40=100
51-75%	250*0.20=50
76-100%	250*0.10=25

The Chi-Square statistic is obtained using the formula,

Treatment	Observed	Expected
0-25%	51	75	7.68
26-50%	82	100	3.24
51-75%	66	50	5.12
76-100%	51	25	27.04
Total			43.08

Critical value:

The critical value for the chi-square statistic is obtained from the chi-square distribution table for,

degree of freedom = 4-1 = 3 and significance level = 0.05

Conclusion:

Since the chi-square statistic is greater than the critical value at a 5% significance level. it can be concluded that the null hypothesis is rejected. Hence we can conclude that the distribution of percent wound healing doesn't follow a normal distribution.