A Washington Post Poll March 18 2013 and a Pew Poll March 17

A Washington Post Poll (March 18, 2013) and a Pew Poll (March 17, 2013) both claimed to ask a random sample of adults in the United States whether they supported or opposed gay marriage. In the Washington Post Poll 581 supported and 360 opposed gay marriage. In the Pew Poll, 735 supported and 660 opposed gay marriage. Find the percentages supporting gay marriage in these two polls and compare them. Test the hypothesis that the population proportions are not equal at the 0.05 significance level Using methods learned in Chapter 7, find a 95% confidence interval for the difference between the two percentages, and interpret it. Does it capture 0? What does that show?

Solution

Null, There Is No Significance between them Ho: p1 = p2
Alternate, There Is Significance between them H1: p1 != p2
Test Statistic
Sample 1 : X1 =581, n1 =941, P1= X1/n1=0.617
Sample 2 : X2 =735, n2 =1395, P2= X2/n2=0.527
Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2)
P^=0.563
Q^ Value For Proportion= 1-P^=0.437
we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2))
Zo =(0.617-0.527)/Sqrt((0.563*0.437(1/941+1/1395))
Zo =4.328
| Zo | =4.328
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =4.328 & | Z | =1.96
Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho
P-Value: Two Tailed ( double the one tail ) -Ha : ( P != 4.3278 ) = 0
Hence Value of P0.05 > 0,Here we Reject Ho

ANS:
a. P1= X1/n1=0.617,P2= X2/n2=0.527
b. We have evidence that Proportions are not equal at 0.05 LOS

c.
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 chances( X1 )=581
No.Of Observed (n1)=941
P1= X1/n1=0.617
Proportion 2
No. of chances(X2)=735
No.Of Observed (n2)=1395
P2= X2/n2=0.527
C.I = (0.617-0.527) ±Z a/2 * Sqrt( (0.617*0.383/941) + (0.527*0.473/1395) )
=(0.617-0.527) ± 1.96* Sqrt(0)
=0.091-0.041,0.091+0.041
=[0.05,0.131]

It does n\'t contain the value 0