Construct a 90 confidence interval for the difference betwee

Construct a 90% confidence interval for the difference between the population proportions. (Negative values should be indicated by a minus sign. Round intermediate calculations and final answer to 4 decimal places.)

Is there a statistical difference between the population proportions at the 10% significance level?

Solution

a)
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 Observed (n1)=400
P1= X1/n1=0.85
Proportion 2
No.Of Observed (n2)=350
P2= X2/n2=0.9
C.I = (0.85-0.9) ±Z a/2 * Sqrt( (0.85*0.15/400) + (0.9*0.1/350) )
=(0.85-0.9) ± 1.64* Sqrt(0.0006)
=-0.05-0.0394,-0.05+0.0394
=[-0.0894,-0.0106]

b)
Yes, since the confidence interval does not include the value 0