A survey of Internet users reported that 15% downloaded music onto their compute
ID: 2934692 • Letter: A
Question
A survey of Internet users reported that 15% 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 33% 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 33% is 1000 and the sample size for the survey reporting 15% is 1600.
Explanation / Answer
a.
Hypothesis
Given that,
sample one, x1 =150, n1 =1000, p1= x1/n1=0.15
sample two, x2 =330, n2 =1000, p2= x2/n2=0.33
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.15-0.33)/sqrt((0.24*0.76(1/1000+1/1000))
zo =-9.424
| zo | =9.424
CI
given that,
sample one, x1 =150, n1 =1000, p1= x1/n1=0.15
sample two, x2 =330, n2 =1000, p2= x2/n2=0.33
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 = [ (0.15-0.33) ± 1.96 * 0.0187]
= [ -0.2166 , -0.1434 ]
b.
Hypothesis
sample one, x1 =240, n1 =1600, p1= x1/n1=0.15
sample two, x2 =528, n2 =1600, p2= x2/n2=0.33
zo =(0.15-0.33)/sqrt((0.24*0.76(1/1600+1/1600))
zo =-11.921
| zo | =11.921
CI
When n = 1600
CI = [ (0.15-0.33) ± 1.96 * 0.0148]
= [ -0.2089 , -0.1511 ]
c.
Hypothesis,
sample one, x1 =240, n1 =1600, p1= x1/n1=0.15
sample two, x2 =330, n2 =1000, p2= x2/n2=0.33
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.05
from standard normal table, two tailed z /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 =(0.15-0.33)/sqrt((0.219*0.781(1/1600+1/1000))
zo =-10.793
CI
given that,
sample one, x1 =240, n1 =1600, p1= x1/n1=0.15
sample two, x2 =330, n2 =1000, p2= x2/n2=0.33
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 = [ (0.15-0.33) ± 1.96 * 0.0173]
= [ -0.214 , -0.146 ]
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