Question

10.73 The lottery commissioner office in a state wanted to find if the percentages of men and women who play the lottery often are differentk A sample of 500 men taken by the commissioner's office showed that 160 of them play the lottery often. Another sample of 300 women showed that 66 of them play the lottery often. a. What is the point estimate of the difference between the two population proportions? b. Construct a 99% confidence interval for the difference between the proportions of all men and all women who play the lottery often. C. Testing at the 1% significance level, can you conclude that the proportions of all men and all women who play the lottery often are different?

Answer 1

SOLUTION

a)

p1=160/500=0.32

p2=66/300=0.22

p1-p2=0.32-0.22=0.1

b)

(p1-p2)+/-Za/2$\sqrt$p1(1-p1)/n1+p2(1-p2)/n2

0.1+/-2.58*$\sqrt$0.32(0.68)/500+0.22(0.78)/300

0.1+/-0.08188

0.01812to 0.1819

c)

H0: π1 – π2 = 0   (the two proportions are equal)

H1: π1 – π2 ≠ 0   (there is a significant difference between proportions)

p=160+66/500+300=0.2825

Z_STAT=(p1-p2)-(π1 – π2)/$\sqrt$p(1-p)(1/n1+1/n2)

=0.1-0/$\sqrt$0.2825(0.7175)(1/500+1/300)

=0.1/0.00365

=2.74

critical values of z would be +/-1.645 and hence reject the null hypothesis because 2.74>1.645 and it lies in the rejection region