Question

 A newspaper article reported that 350 people in one state were surveyed and 60% were opposed to a recent court decision. The same article reported that a similar survey of 550 people in another state indicated oppo- sition by only 20%. Construct a 95% confidence inter- val of the difference in population proportions based on the data.

Answer 1

% C.I=(p₁-p₂)-z$\cdot \sqrt{\frac{p1\cdot(1-p1)}{n1}+\frac{p2\cdot(1-p2)}{n2}}$,(p₁-p₂)+z $\cdot \sqrt{\frac{p1\cdot(1-p1)}{n1}+\frac{p2\cdot(1-p2)}{n2}}$

p₁=0.6

n₁=350

p₂=0.2

n₂=550

Z=1.96

95% C.I=(0.6-0.2)-1.96 $\cdot \sqrt{(\frac{0.6\cdot(1-0.6)}{350}+\frac{0.2\cdot(1-0.2)}{550}}$ ,(0.6-0.2)+1.96 $\cdot \sqrt{(\frac{0.6\cdot(1-0.6)}{350}+\frac{0.2\cdot(1-0.2)}{550})}$

95% C.I=(0.4-0.0613,0.4+0.0613)

95% C.I=(0.3387,0.4613)