In randomized doubleblind clinical trials of a new vaccine m

In randomized, double-blind clinical trials of a new vaccine, monkeys were randomly divided into two groups. Subjects in group 1 received the new vaccine while subjects in group 2 received a control vaccine. After the first dose, 101 of 719 subjects in the experimental group (group 1) experienced vomiting as a side effect. After the first dose, 61 of 632 of the subjects in the control group (group 2) experienced vomiting as a side effect. Construct a 99% confidence interval for the difference between the two population proportions, p_1 - p_2. Use x_1 = 101, n_1 = 719, x_2 = 61, and n_2 = 632. The 99% confidence interval for p_1 - p_2 is . (Use ascending order. Round to three decimal places as needed.)

Solution

          
Getting p1^ and p2^,          
          
p1^ = x1/n1 =    0.140472879      
p2 = x2/n2 =    0.096518987      
          
Also, the standard error of the difference is          
          
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] =    0.017490214      
          
For the   99%   confidence level, then  
          
alpha/2 = (1 - confidence level)/2 =    0.005      

z(alpha/2) =    2.575829304      

Margin of error = z(alpha/2)*sd =    0.045051805      
lower bound = p1^ - p2^ - z(alpha/2) * sd =    -0.001097913      

upper bound = p1^ - p2^ + z(alpha/2) * sd =    0.089005696      
          
Thus, the confidence interval is          
          
(   -0.001 ,   0.089 ) [ANSWER]