Question

One month before the election, a poll of 634 randomly selected voters showed 54% planning to vote for a certain candidate. A week later it became known that he had had an extramarital affair, and a new poll showed only 51% of 1010 voters supporting him. Do these results indicate a decrease in voter support for his candidacy?

a. What is the value of x1?

b. Calculate the value of the one tail test for 2 proportion(s). Round your answer to 2 decimal places.

c. Find the lower bound of 95% CI for the difference in two proportions.

d. Find the upper bound of 95% CI for the difference in two proportions.

Answer 1

Part a)

X1 = 54% * 634 = 342.36

X2 = 51% * 1010 = 515.1

To Test :-

H0 :- P1 = P2
H1 :- P1 < P2

Test Statistic :-
Z = ( p̂1 - p̂2 ) / √(p̂ * q̂ * (1/n1 + 1/n2) ) )
p̂ is the pooled estimate of the proportion P
p̂ = ( x1 + x2) / ( n1 + n2)
p̂ = ( 342.36 + 515.1 ) / ( 634 + 1010 )
p̂ = 0.5216
q̂ = 1 - p̂ = 0.4784
Z = ( 0.54 - 0.51) / √( 0.5216 * 0.4784 * (1/634 + 1/1010) )
Z = 1.1853 ≈ 1.19

Test Criteria :-
Reject null hypothesis if Z < -Z(α)
Z(α) = Z(0.05) = 1.645
Z > -Z(α) = 1.1853 > -1.645, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

Decision based on P value
P value = P ( Z < 1.1853 ) = 0.882 ( From Z table )
Reject null hypothesis if P value < α = 0.05
Since P value = 0.882 > 0.05, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

There is insufficient evidence to suport the claim that there is decrease in voter support.

(p̂1 - p̂2) ± Z(α/2) * √( ((p̂1 * q̂1)/ n1) + ((p̂2 * q̂2)/ n2) )
Z(α/2) = Z(0.05 /2) = 1.96 ( From Z table )
Lower Limit = ( 0.54 - 0.51 )- Z(0.05/2) * √(((0.54 * 0.46 )/ 634 ) + ((0.51 * 0.49 )/ 1010 ) = -0.0196
upper Limit = ( 0.54 - 0.51 )+ Z(0.05/2) * √(((0.54 * 0.46 )/ 634 ) + ((0.51 * 0.49 )/ 1010 )) = 0.0796
95% Confidence interval is ( -0.0196 , 0.0796 )
( -0.0196 < ( P1 - P2 ) < 0.0796 )

Part c)

Lower Limit = ( 0.54 - 0.51 )- Z(0.05/2) * √(((0.54 * 0.46 )/ 634 ) + ((0.51 * 0.49 )/ 1010 ) = -0.0196

Part d)

Upper Limit = ( 0.54 - 0.51 )+ Z(0.05/2) * √(((0.54 * 0.46 )/ 634 ) + ((0.51 * 0.49 )/ 1010 )) = 0.0796