Question

A 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 decreased from the estimate of 30% 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 z to two decimal places. Round your P-value to four decimal places.)

z =

P-value = 0 Correct:

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

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

The interval shows no significant change in music downloads. The interval does not provide any more information than the significance test would tell us.

Answer 1

a.
Given that,
sample one, x1 =3.08, n1 =14, p1= x1/n1=0.22
sample two, x2 =9.3, n2 =31, p2= x2/n2=0.3
finding a p^ value for proportion p^=(x1 + x2 ) / (n1+n2)
p^=0.2751
q^ Value For Proportion= 1-p^=0.7249
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
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 =(0.22-0.3)/sqrt((0.275*0.7249(1/14+1/31))
zo =-0.5563
| zo | =0.5563
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =0.556 & | z α | =1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.5563 ) = 0.578
hence value of p0.05 < 0.578,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: -0.5563
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.578

b.
TRADITIONAL METHOD
given that,
sample one, x1 =3.08, n1 =14, p1= x1/n1=0.22
sample two, x2 =9.3, n2 =31, p2= x2/n2=0.3
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.22*0.78/14) +(0.3 * 0.7/31))
=0.138
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.138
=0.2704
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.22-0.3) ±0.2704]
= [ -0.3504 , 0.1904]
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DIRECT METHOD
given that,
sample one, x1 =3.08, n1 =14, p1= x1/n1=0.22
sample two, x2 =9.3, n2 =31, p2= x2/n2=0.3
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.22-0.3) ± 1.96 * 0.138]
= [ -0.3504 , 0.1904 ]
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interpretations:
1) we are 95% sure that the interval [ -0.3504 , 0.1904] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean P1-P2

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