Question

The drug ancrod was tested in a double-blinded clinical trial in which subjects who had strokes were randomly assigned to get either ancrod or a placebo. One outcome of interest 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 in the placebo group.

a. Compute a 90% confidence interval for the difference of the proportions of patients experiencing hemorrhaging in the two groups.

• Point estimate

• Standard error

• Multiplier

• Confidence interval

• Check for large sample size

b. Conduct an appropriate test to determine whether the proportions of patients experiencing hemorrhaging in the two groups are different. Use a significance level of 0.01.

• Pooled sample proportion

• Test statistic

• Rejection region

• Decision

• Check for large sample size

Answer 1

first sample size, ancrod n1=   248          
number of successes, sample 1 =     x1=   13          
proportion success of sample 1 , p̂1=   x1/n1=   0.0524          
                     
second sample size, placebo n2 =    252          
number of successes, sample 2 =     x2 =    6          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.024          
                  
Point estimate =difference in sample proportions, p̂1 - p̂2 =     0.0524   -   0.0238   =   0.0286

level of significance, α =   0.10              
Z critical value =   Z α/2 =    1.645   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0171      

   
multiplier =margin of error , E = Z*SE =    1.645   *   0.0171   =   0.0281
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    0.029   -   0.0281   =   0.0005
upper limit = (p̂1 - p̂2) + E =    0.029   +   0.0281   =   0.0567
                  
so, confidence interval is (   0.0005   < p1 - p2 <   0.0567   )  

b)

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 ╪   0          
                  
sample #1   ----->   experimental          
first sample size,     n1=   248          
number of successes, sample 1 =     x1=   13          
proportion success of sample 1 , p̂1=   x1/n1=   0.0524          
                  
sample #2   ----->   standard          
second sample size,     n2 =    252          
number of successes, sample 2 =     x2 =    6          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.024          
                  
difference in sample proportions, p̂1 - p̂2 =     0.0524   -   0.0238   =   0.0286
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.0380         
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0171          
Z-statistic = (p̂1 - p̂2)/SE = (   0.029   /   0.0171   ) =   1.6729
                  

p-value =        0.0943   [excel formula =2*NORMSDIST(z)]      
decision :    p-value>α,Don't reject null hypothesis

THANKS

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Conclusion:   There is not enough evidence to