Question

Example 12.6 A 2005 survey of Internet users reported that 22% downloaded music onto their computers. The filing of lawsuits by the recording industry may be a reason why this percent has decresased from the estimate of 29% from a survey taken two years before. Assume that the sample sizes are both 1421. o Using a significance test (a = 0.05), evaluate whether or not there has been a change in the percent of Internet users who download music. Report a 95% confidence interval for the difference in proportions.

Answer 1

To Test :-

H0 :- P1 = P2
H1 :- P1 ≠ P2

p̂1 = 312.62 / 1421 = 0.22
p̂2 = 412.09 / 1421 = 0.29

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 = ( 312.62 + 412.09 ) / ( 1421 + 1421 )
p̂ = 0.255
q̂ = 1 - p̂ = 0.745
Z = ( 0.22 - 0.29) / √( 0.255 * 0.745 * (1/1421 + 1/1421) )
Z = -4.2809

Test Criteria :-
Reject null hypothesis if Z < -Z(α/2)
Critical value Z(α/2) = Z(0.05/2) = 1.96
Z < -Z(α/2) = -4.2809 < -1.96, hence we reject the null hypothesis
Conclusion :- We Reject H0

There is sufficient evidence to support the claim that percentage has changed.

Part b)

(p̂1 - p̂2) ± Z(α/2) * √( ((p̂1 * q̂1)/ n1) + ((p̂2 * q̂2)/ n2) )
Critical value Z(α/2) = Z(0.05 /2) = 1.96
Lower Limit = ( 0.22 - 0.29 )- Z(0.05/2) * √(((0.22 * 0.78 )/ 1421 ) + ((0.29 * 0.71 )/ 1421 ) = -0.1019
upper Limit = ( 0.22 - 0.29 )+ Z(0.05/2) * √(((0.22 * 0.78 )/ 1421 ) + ((0.29 * 0.71 )/ 1421 )) = -0.0381
95% Confidence interval is ( -0.1019 , -0.0381 )
( -0.1019 < ( P1 - P2 ) < -0.0381 )