A survey of Internet users reported that 19% downloaded music onto their compute
ID: 3131516 • Letter: A
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 27% 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. (Round your test statistics to two decimal places and your confidence intervals to four decimal places.) (i) Both sample sizes are 1000. z = 95% C.I. , (ii) Both sample sizes are 1600. z = 95% C.I. , (iii) The sample size for the survey reporting 27% is 1000 and the sample size for the survey reporting 19% is 1600. z = 95% C.I. ,
Explanation / Answer
i.
Formulating the hypotheses
Ho: p1 - p2 >= 0
Ha: p1 - p2 < 0
Here, we see that pdo = 0 , the hypothesized population proportion difference.
Getting p1^ and p2^,
p1^ = x1/n1 = 0.19
p2 = x2/n2 = 0.27
Also, the standard error of the difference is
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] = 0.018734994
Thus,
z = [p1 - p2 - pdo]/sd = -4.270084101 [ANSWER]
For the 95% confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.025
z(alpha/2) = 1.959963985
Margin of error = z(alpha/2)*sd = 0.036719913
lower bound = p1^ - p2^ - z(alpha/2) * sd = -0.116719913
upper bound = p1^ - p2^ + z(alpha/2) * sd = -0.043280087
Thus, the confidence interval is
( -0.116719913 , -0.043280087 ) [ANSWER]
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ii.
Getting p1^ and p2^,
p1^ = x1/n1 = 0.19
p2 = x2/n2 = 0.27
Also, the standard error of the difference is
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] = 0.014811313
Thus,
z = [p1 - p2 - pdo]/sd = -5.401276624 [ANSWER, Z]
For the 95% confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.025
z(alpha/2) = 1.959963985
Margin of error = z(alpha/2)*sd = 0.029029641
lower bound = p1^ - p2^ - z(alpha/2) * sd = -0.109029641
upper bound = p1^ - p2^ + z(alpha/2) * sd = -0.050970359
Thus, the confidence interval is
( -0.109029641 , -0.050970359 ) [ANSWER]
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iii.
Getting p1^ and p2^,
p1^ = x1/n1 = 0.19
p2 = x2/n2 = 0.27
Also, the standard error of the difference is
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] = 0.017125639
Thus,
z = [p1 - p2 - pdo]/sd = -4.671358629 [ANSWER, Z]
For the 95% confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.025
z(alpha/2) = 1.959963985
Margin of error = z(alpha/2)*sd = 0.033565635
lower bound = p1^ - p2^ - z(alpha/2) * sd = -0.113565635
upper bound = p1^ - p2^ + z(alpha/2) * sd = -0.046434365
Thus, the confidence interval is
( -0.113565635 , -0.046434365 ) [ANSWER]
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