Question

3. A clinical trial examined the effectiveness of aspirin in the treatment of cerebral ischemia(stroke). Patients were randomized into treatment and control groups. The study wasdouble-blind. After six months of treatment, the attending physicians evaluated eachpatient’s progress as either favorable or unfavorable. Of the 78 patients in the aspiringroup, 63 had favorable outcomes; 43 of the 77 control patients had favorable outcomes.(A) The physicians conducting the study had concluded from previous research theaspirin was likely to increase the chance of a favorable outcome. Carry out a significancetest to confirm this conclusion. State the hypotheses, find aP-value, and write a summaryof your results.(B) Estimate the difference between the favorable proportions in the treatment andcontrol groups. Use 95% confidence.

Answer 1

Part A

Here, we have to use z test for difference between population proportions.

Null hypothesis: H₀: The aspirin was not likely to increase the chance of favourable outcomes.

Alternative hypothesis: H_a: The aspirin was likely to increase the chance of favourable outcomes.

H₀: p₁ ≤ p₂ versus H_a: p > p₂

This is a right tailed or an upper tailed test.

We assume level of significance = α = 0.05

The test statistic formula is given as below:

Z = (P1 – P2) / sqrt(P*(1 – P)*((1/N1) + (1/N2)))

Where,

X1 = 63

X2 = 43

N1 = 78

N2 = 77

P = (X1+X2)/(N1+N2) = (63 + 43) / (78 + 77) = 0.6839

P1 = X1/N1 = 63/78 = 0.807692308

P2 = X2/N2 = 43/77 = 0.558441558

P1 - P2 = 0.807692308 - 0.558441558 = 0.249250749

Z = (P1 – P2) / sqrt(P*(1 – P)*((1/N1) + (1/N2)))

Z = 0.249250749 / sqrt(0.6839*(1 – 0.6839)*((1/78) + (1/77)))

Z = 3.3369

P-value = 0.0004

(by using z-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the aspirin was likely to increase the chance of favourable outcomes.

Part B

Here, we have to find the 95% confidence interval for difference between two population proportions.

Confidence interval for difference between two population proportions:

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

P1 - P2 = 0.807692308 - 0.558441558 = 0.249250749

Confidence interval = 0.249250749 ± 1.96*sqrt[(0.807692308*(1 – 0.807692308)/78) + (0.558441558*(1 – 0.558441558)/77)]

Confidence interval = 0.249250749 ± 0.1413

Lower limit = 0.249250749 - 0.1413 = 0.1080

Upper limit = 0.249250749 + 0.1413 = 0.3905

Confidence interval = (0.1080, 0.3905)

Confidence interval = (10.8%, 39.1%)