A survey of Internet users reported that 20% downloaded music onto their compute

Question

A survey of Internet users reported that 20% 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 29% from a survey taken two years before. Assume that the sample sizes are both 1431. 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 zto two decimal places. Round your P-value to four decimal places.)

z =

P Value=

Summarize your conclusion.

We conclude that the means are different.

We cannot draw any conclusions using a significance test for this data.     

We conclude that the proportions are different.

We conclude that the means are not different.

We conclude that the proportions are not different.

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

Explain what information is provided in the interval that is not in the significance test results.

The interval tells us there was a significant change in music downloads, but the test statistic is inconclusive.

The interval shows no significant change in music downloads.     

The interval gives us an idea of how large the difference is between the first survey and the second survey.

The interval does not provide any more information than the significance test would tell us.

The significance test does not indicate the direction of change, but the interval shows that the music downloads decreased.

Explanation / Answer

PART A.
Given that,
sample one, x1 =286.2, n1 =1431, p1= x1/n1=0.2
sample two, x2 =414.99, n2 =1431, p2= x2/n2=0.29
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.05
from standard normal table, two tailed z /2 =
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.2-0.29)/sqrt((0.245*0.755(1/1431+1/1431))
zo =-5.597
| zo | =5.597
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =5.597 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -5.5975 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: -5.597
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0
we conclude that the proportions are different.

PART 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 chances( X1 )=286.2
No.Of Observed (n1)=1431
P1= X1/n1=0.2
Proportion 2
No. of chances(X2)=414.99
No.Of Observed (n2)=1431
P2= X2/n2=0.29
C.I = (0.2-0.29) ±Z a/2 * Sqrt( (0.2*0.8/1431) + (0.29*0.71/1431) )
=(0.2-0.29) ± 2.33* Sqrt(0)
=-0.09-0.037,-0.09+0.037
=[-0.127,-0.053]

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