Question

A clinical trial is conducted to determine if a certain type of inoculation has an effect on the incidence of a certain disease. A sample of 1000 rats was kept in a controlled environment for a period of 1 year and 500 of the rats were given the inoculation. Of the group not given the drug, there were 120 incidences of the disease, while 98 of the inoculated group contracted it. If we call p1 the probability of incidence of the disease in uninoculated rats and p2 the probability of incidence after receiving the drug, compute a 90% confidence interval for p1 – p2.

Answer 1

Solution

Confidence Interval for Diffrence of Proportion
CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
No. of chances( X1 )=500
No.Of Observed (n1)=1000
P1= X1/n1=0.5
Proportion 2
No. of chances(X2)=98
No.Of Observed (n2)=120
P2= X2/n2=0.817
C.I = (0.5-0.817) ±Z a/2 * Sqrt( (0.5*0.5/1000) + (0.817*0.183/120) )
      =(0.5-0.817) ± 1.64* Sqrt(0.001)
      =-0.317-0.063,-0.317+0.063
     =[-0.38,-0.253]