A survey of Internet users reported that 15% downloaded music onto their compute
ID: 3174363 • 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 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.
z =
95% C.I.= ( , )
(ii) Both sample sizes are 1600.
z =
95% C.I.= ( , )
(iii) The sample size for the survey reporting 31% is 1000 and the sample size for the survey reporting 15% is 1600.
z =
95% C.I. = ( , )
Explanation / Answer
(i)
Data:
n1 = 1000
n2 = 1000
p1 = 0.031
p2 = 0.015
Hypotheses:
Ho: p1 p2
Ha: p1 < p2
Decision Rule:
= 0.05
Lower Critical z- score = -1.644853627
Reject Ho if z < -1.644853627
Test Statistic:
Average proportion, p = (n1p1 + n2p2)/(n1 + n2) = (1000 * 0.031 + 1000 * 0.015)/(1000 + 1000) = 0.023
q = 1 - p = 1 - 0.023 = 0.977
SE = [pq * {(1/n1) + (1/n2)}] = (0.023 * 0.977 * ((1/1000) + (1/1000))) = 0.006703879
z = (p1 - p2)/SE = (0.031 - 0.015)/0.00670387947385691 = 2.39
n1 = 1000
n2 = 1000
p1 = 0.031
p2 = 0.015
% = 95
Pooled Proportion, p = (n1 p1 + n2 p2)/(n1 + n2) = (1000 * 0.031 + 1000 * 0.015)/(1000 + 1000) = 0.023
q = 1 - p = 1 - 0.023 = 0.977
SE = (pq * ((1/n1) + (1/n2))) = (0.023 * 0.977 * ((1/1000) + (1/1000))) = 0.006703879
z- score = 1.959963985
Width of the confidence interval = z * SE = 1.95996398454005 * 0.00670387947385691 = 0.013139362
Lower Limit of the confidence interval = (p1 - p2) - width = 0.016 - 0.0131393623254569 = 0.002860638
Upper Limit of the confidence interval = (p1 - p2) + width = 0.016 + 0.0131393623254569 = 0.029139362
The 95% confidence interval is [0.0029, 0.0291]
(ii)
Data:
n1 = 1600
n2 = 1600
p1 = 0.031
p2 = 0.015
Hypotheses:
Ho: p1 ³ p2
Ha: p1 < p2
Decision Rule:
= 0.05
Lower Critical z- score = -1.644853627
Reject Ho if z < -1.644853627
Test Statistic:
Average proportion, p = (n1p1 + n2p2)/(n1 + n2) = (1600 * 0.031 + 1600 * 0.015)/(1600 + 1600) = 0.023
q = 1 - p = 1 - 0.023 = 0.977
SE = [pq * {(1/n1) + (1/n2)}] = (0.023 * 0.977 * ((1/1600) + (1/1600))) = 0.005299882
z = (p1 - p2)/SE = (0.031 - 0.015)/0.00529988207415976 = 3.02
n1 = 1600
n2 = 1600
p1 = 0.031
p2 = 0.015
% = 95
Pooled Proportion, p = (n1 p1 + n2 p2)/(n1 + n2) = (1600 * 0.031 + 1600 * 0.015)/(1600 + 1600) = 0.023
q = 1 - p = 1 - 0.023 = 0.977
SE = (pq * ((1/n1) + (1/n2))) = (0.023 * 0.977 * ((1/1600) + (1/1600))) = 0.005299882
z- score = 1.959963985
Width of the confidence interval = z * SE = 1.95996398454005 * 0.00529988207415976 = 0.010387578
Lower Limit of the confidence interval = (p1 - p2) - width = 0.016 - 0.0103875779876626 = 0.005612422
Upper Limit of the confidence interval = (p1 - p2) + width = 0.016 + 0.0103875779876626 = 0.026387578
The 95% confidence interval is [0.0056, 0.0264]
(iii)
Data:
n1 = 1000
n2 = 1600
p1 = 0.031
p2 = 0.015
Hypotheses:
Ho: p1 ³ p2
Ha: p1 < p2
Decision Rule:
= 0.05
Lower Critical z- score = -1.644853627
Reject Ho if z < -1.644853627
Test Statistic:
Average proportion, p = (n1p1 + n2p2)/(n1 + n2) = (1000 * 0.031 + 1600 * 0.015)/(1000 + 1600) = 0.021153846
q = 1 - p = 1 - 0.0211538461538462 = 0.978846154
SE = [pq * {(1/n1) + (1/n2)}] = (0.0211538461538462 * 0.978846153846154 * ((1/1000) + (1/1600))) = 0.005800676
z = (p1 - p2)/SE = (0.031 - 0.015)/0.00580067552432142 = 2.76
n1 = 1000
n2 = 1600
p1 = 0.031
p2 = 0.015
% = 95
Pooled Proportion, p = (n1 p1 + n2 p2)/(n1 + n2) = (1000 * 0.031 + 1600 * 0.015)/(1000 + 1600) = 0.021153846
q = 1 - p = 1 - 0.0211538461538462 = 0.978846154
SE = (pq * ((1/n1) + (1/n2))) = (0.0211538461538462 * 0.978846153846154 * ((1/1000) + (1/1600))) = 0.005800676
z- score = 1.959963985
Width of the confidence interval = z * SE = 1.95996398454005 * 0.00580067552432142 = 0.011369115
Lower Limit of the confidence interval = (p1 - p2) - width = 0.016 - 0.011369115113673 = 0.004630885
Upper Limit of the confidence interval = (p1 - p2) + width = 0.016 + 0.011369115113673 = 0.027369115
The 95% confidence interval is 0.0046, 0.0274].
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