A survey of Internet users reported that 18% downloaded music onto their compute

Question

A survey of Internet users reported that 18% downloaded music onto their computers. The filing of lawsuits by the recording industry may be a reason why this percent has decreased from the estimate of 25% from a survey taken two years before. Assume that the sample sizes are both 1441. Using a significance test, evaluate whether or not there has been a change in the percent of Internet users who download music. Provide all details for the test. (Round your value for z to two decimal places. Round your P-value to four decimal places.)

Also report a 95% confidence interval for the difference in proportions. (Round your answers to four decimal places.)

  ,

Explanation / Answer

Given that,
sample one, n1 =1441, p1= x1/n1=0.18
sample two, n2 =1441, p2= x2/n2=0.25
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.01
from standard normal table, two tailed z /2 =
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.18-0.25)/sqrt((0.215*0.785(1/1441+1/1441))
zo =-4.574

ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: -4.574
p-value: 0

b.
Confidence Interval for Diffrence of Proportion
CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
No.Of Observed (n1)=1441
P1= X1/n1=0.18
Proportion 2
No.Of Observed (n2)=1441
P2= X2/n2=0.25
C.I = (0.18-0.25) ±Z a/2 * Sqrt( (0.18*0.82/1441) + (0.25*0.75/1441) )
=(0.18-0.25) ± 1.96* Sqrt(0)
=-0.07-0.03,-0.07+0.03
=[-0.1,-0.04]

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.