Question

also find a 95% CI for p_1 - p_2

ve Is someone who switches brands because of a financial inducement less likely to remain loyal than someone who switches without induce- ment? Let P1 and P2 denote the true proportions of switchers to a certain brand with and without inducement, respectively, who subsequently make a repeat purchase. Test Ho: Pi - P2 = 0 versus Ha: P1 - P2 < 0 using a = .01 and the following data: m=200 n=600 number of successes = 30 number of successes = 180 (Similar data is given in "Impact of Deals and Deal Retraction on Brand Switching," J. Market- ing, 1980: 62-70.)

Answer 1

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 = ( 30 + 180 ) / ( 200 + 600 )
p̂ = 0.2625
q̂ = 1 - p̂ = 0.7375
Z = ( 0.15 - 0.3) / √( 0.2625 * 0.7375 * (1/200 + 1/600) )
Z = -4.18

Test Criteria :-
Reject null hypothesis if Z < -Z(α)
Critical value * Z(α) = Z(0.01) = 2.326 ( From Z table )
Z < -Z(α) = -4.1753 < - 2.326, hence we reject the null hypothesis
Conclusion :- We Reject H0

There is sufficient evidence to support the claim that P1 < P2

Confidence interval :-

(p̂1 - p̂2) ± Z(α/2) * √( ((p̂1 * q̂1)/ n1) + ((p̂2 * q̂2)/ n2) )
Critical value Z(α/2) = Z(0.05 /2) = 1.96 ( From Z table )
Lower Limit = ( 0.15 - 0.3 )- Z(0.05/2) * √(((0.15 * 0.85 )/ 200 ) + ((0.3 * 0.7 )/ 600 ) = -0.2116
upper Limit = ( 0.15 - 0.3 )+ Z(0.05/2) * √(((0.15 * 0.85 )/ 200 ) + ((0.3 * 0.7 )/ 600 )) = -0.0884
95% Confidence interval is ( -0.2116 , -0.0884 )
( -0.2116 < ( P1 - P2 ) < -0.0884 )