In a sample of 900 female adults 76 of them said that they a

In a sample of 900 female adults, 76% of them said that they are in favour of stricter laws to reduce air pollution. In a sample of 800 male adults, 68% said they are in favour of stricter laws to reduce air pollution.

a) Construct a 98% confidence interval for the difference between the proportions of all female adults and all male adults who are in favour of stricter laws to reduce air pollution.

b) Test at the 2.5% significance level if the proportion of all female adults who are in favour of stricter laws to reduce the air pollution is greater than the proportion of all male adults who are in favour of stricter laws to reduce air pollution.

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
Probability Succuses( X1 )=684
No.Of Observed (n1)=900
P1= X1/n1=0.76
Proportion 2
Probability Succuses(X2)=544
No.Of Observed (n2)=800
P2= X2/n2=0.68
C.I = (0.76-0.68) ±Z a/2 * Sqrt( (0.76*0.24/900) + (0.68*0.32/800) )
=(0.76-0.68) ± 2.33* Sqrt(0.0005)
=0.08-0.0508,0.08+0.0508
=[0.0292,0.1308]

b)
Test For Significance of Difference Of Proportion
Null Hypothesis, female adults is lessr than the proportion of all male adults Ho: p1 < p2
Alternate Hypothesis, female adults is greater than the proportion of all male adults, H1: p1 > p2
Test Statistic
Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2)
P^=0.722
Q^ Value For Proportion= 1-P^=0.278
we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2))
Zo =(0.76-0.68)/Sqrt((0.722*0.278(1/900+1/800))
Zo =3.676
| Zo | =3.676
Critical Value
The Value of |Z | at LOS 0.025% is 1.96
We got |Zo| =3.676 & | Z | =1.96
Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho
P-Value: Right Tail -Ha : ( P > 3.6763 ) = 0.00012
Hence Value of P0.025 > 0.00012,Here we Reject Ho

female adults is greater than the proportion of all male adults