The results of a recent poll on the preference of shoppers regarding two products are shown below. Product shoppers surveyed shoppers favoring a 800 560 b 900 612 the point estimate for the difference between the two population proportions is
Answers
Product Shoppers Surveyed This Product
A 800 560
B 900 612
Let p1 be in proportion preferring product A and p2 be in proportion preferring product B.
Improve a 90% confidence interval estimate for the alteration
p1-p2
between the proportions favoring each product.
(b) Test H0 : p1 = p2 at ∝ = 0.05 based on classical approach
(c) Test H0 : p1 = p2 at ∝ = 0.05 based on p- value method
[solution:]
J n1 =800 ,n2 = 900, p1 = 560/800 =0.7, p2 = 612/900 = 0.68
Sp1-p2 = √((p1(1-p1)/n1)+(p2(1-p2)/n2) ) = √((0.7(1-0.7)/800)+(.68(1-0.68)/900) )
= 02246
.
Thus, a 90% confidence interval is
(p1-p2) ± Zα/2 Sp1-p2 = (0.7-0.68) ±z0.05.0.02246 = 0.02± 0.03695
= [-0.01695,0.5695]
P =(n1p1+n2p2)/(n1+n2) = (560+612)/(800+900 ) = 0689
S*p1-p2 = √(p(1-p)(1/n1+1/n2) ) = √(0.689(1-0.689)(1/800+1/900) )
=
=0.02249
Therefore,
Z = (p1-p2)/(Sp1-p2) = (0.7-0.68)/0.02249 = 0.89
Z = 0.89<1.96 = Z0.025 =Z∝/2
We don’t discard H0
p – value = P(Z >0.89) = 0.3734>0.05 =>not rejecting H0