Question

Independent random samples were selected from two binomial populations, with sample sizes and the number of successes given below. Population 1 2 500 500 120 147 Sample Size Number of Successes Construct a 95% confidence interval for the difference in the population proportions. (Use P, - Pg. Round your answers to four decimal places.) - 1087 to

Answer 1

We have given,              
              
x1=120          
n1=500          
              
x2=147          
n2=500          
              
Level of significance =   0.05          
Z critical value (by using Z table)=1.96
Estimate for sample proportion 1$=\hat{p}_{1}=0.24$
              
Estimate for sample proportion 2 $= \hat{p}_{2}=0.294$
      
Confidence interval formula is              
$\left ( \hat{p}_{1}-\hat{p}_{2} \right )\pm Z*\sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}}+\frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}}}$
$=\left ( 0.24-0.294\right )\pm 1.96*\sqrt{\frac{0.24(1-0.24)}{500}+\frac{0.294(1-0.294)}{500}}$
=(-0.1087,0.0007)          
              
              
Lower limit for confidence interval=-0.1087
              
Upper limit for confidence interval is=0.0007