A survey of Internet users reported that 19% downloaded music onto their compute

Question

A survey of Internet users reported that 19% 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 31% from a survey taken two years before. Suppose we are not exactly sure about the sizes of the samples. Perform the calculations for the significance tests and 95% confidence intervals under each of the following assumptions. (Use previous recent. Round your test statistics to two decimal places and your confidence intervals to four decimal places.)

(i) Both sample sizes are 1000.

(ii) Both sample sizes are 1600.

(iii) The sample size for the survey reporting 31% is 1000 and the sample size for the survey reporting 19% is 1600.

Explanation / Answer

a.
Given that,
sample one, n1 =1000, p1= x1/n1=0.19
sample two, n2 =1000, p2= x2/n2=0.31
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 =(0.19-0.31)/sqrt((0.25*0.75(1/1000+1/1000))
zo =-6.197

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)=1000
P1= X1/n1=0.19
Proportion 2
No.Of Observed (n2)=1000
P2= X2/n2=0.31
C.I = (0.19-0.31) ±Z a/2 * Sqrt( (0.19*0.81/1000) + (0.31*0.69/1000) )
=(0.19-0.31) ± 1.96* Sqrt(0)
=-0.12-0.038,-0.12+0.038
=[-0.158,-0.082]

b.

Given that,
sample one, n1 =1600, p1= x1/n1=0.19
sample two, n2 =1600, p2= x2/n2=0.31
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 =(0.19-0.31)/sqrt((0.25*0.75(1/1600+1/1600))
zo =-7.838

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)=1600
P1= X1/n1=0.19
Proportion 2
No.Of Observed (n2)=1600
P2= X2/n2=0.31
C.I = (0.19-0.31) ±Z a/2 * Sqrt( (0.19*0.81/1600) + (0.31*0.69/1600) )
=(0.19-0.31) ± 1.96* Sqrt(0)
=-0.12-0.03,-0.12+0.03
=[-0.15,-0.09]

c.
Given that,
sample one, n1 =1600, p1= x1/n1=0.19
sample two, n2 =1000, p2= x2/n2=0.31
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 =(0.19-0.31)/sqrt((0.236*0.764(1/1600+1/1000))
zo =-7.009

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)=1600
P1= X1/n1=0.19
Proportion 2
No.Of Observed (n2)=1000
P2= X2/n2=0.31
C.I = (0.19-0.31) ±Z a/2 * Sqrt( (0.19*0.81/1600) + (0.31*0.69/1000) )
=(0.19-0.31) ± 1.96* Sqrt(0)
=-0.12-0.035,-0.12+0.035
=[-0.155,-0.085]

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