Question

Pharmaceutical companies promote their prescription drugs using television advertising. In a survey of 76 randomly sampled television viewers, 19 indicated that they asked their physician about using a prescription drug they saw advertised on TV.

Develop a 99% confidence interval for the proportion of viewers who discussed a drug seen on TV with their physician. (Round your answers to 3 decimal places.)

Answer 1

We have given,              
              
x=19          
n=76          
Estimate for sample proportion=   0.25  
Level of significance is =1-0.99=0.01      
Z critical value(using Z table)=2.576      
              
Confidence interval formula is              


=(0.122,0.378)