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Problem 3 [Two sample test for the difference in proportions]: The drug ancrod was tested in a double-blinded clinical trial in which subjects who had a strokes were randomly assigned to get either ancrod or a placebo. One response variable in the study was whether or not a subject experienced intracranial hemorrhaging. In the ancrod group 13 of 248 experienced hemorrhaging compared to 6 patients out of 252 on the placebo group.

(i) Compute a 90% confidence interval for the difference of the proportion of patients experiencing hemorrhaging in the two groups.

(iv) Use a test to compare the two proportions. Step 1: Parameter: H0: ____________ Ha: _______________ Significance level:

Step 2: Verify necessary data conditions and compute an appropriate test statistic:

Step 3: Assuming the H0 is true, define decision rule:

Step 4: Decision [Circle One]: Reject H0 Fail to Reject H0

Step 5: Conclusion:

Explanation / Answer

(i)

For 90% confidence interval critical z=1.645

Phat1=13/248 =0.0524

Phat2=6/252 =0.0238

Sample size n1=248 and n2=252

Standard error of difference in proportion SED=sqrt(phat1*(1-phat1)/n1+ phat2*(1-phat2)/n2)

=sqrt(0.0524*(1-0.0524)/248+0.0238*(1-0.0238)/252)

=0.0171

So 90% confidence interval =(phat1-phat2)±z*SED =(0.0524-0.0238)±1.645*0.0171

=(0.0005               0.0567)

(ii) Step-1: H0:1=2   Ha: 12  

Significance level =0.10

Step-2: n1*phat1=13>5 and n2*phat2=6>5 , so assumption are satisfied.
Test statistic Z=(phat1-phat2)/SED=(0.0524-0.0238)/0.0171 =1.6725

Step-3: We reject the null hypothesis if |Zcalculated|>1.645

Step4: As z=1.6725>1.645 , we reject H0

Step5: We conclude that there is significant difference in the proportion of patients experiencing hemorrhaging in the two groups pf ancrod and placebo.

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