A survey of Internet users reported that 15% downloaded music onto their compute
ID: 3059703 • 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 25% 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 25% is 1000 and the sample size for the survey reporting 15% is 1600.
Summarize the effects of the sample sizes on the results.
A) We see in (i) and (ii) that smaller samples result in smaller z (stronger evidence) and wider intervals, while larger samples have the reverse effect. The results of (iii) show that the effect of varying unequal sample sizes is more complicated.
B) We see in (i) and (ii) that smaller samples result in larger z (stronger evidence) and narrower intervals, while larger samples have the reverse effect. The results of (iii) show that the effect of varying unequal sample sizes is more complicated.
C) We see in (i) and (ii) that smaller samples result in larger z (weaker evidence) and smaller intervals, while larger samples have the reverse effect. The results of (iii) show that the effect of varying unequal sample sizes is more complicated.
D) We see in (i) and (ii) that smaller samples result in smaller z (weaker evidence) and narrower intervals, while larger samples have the reverse effect. The results of (iii) show that the effect of varying unequal sample sizes is more complicated.
E) We see in (i) and (ii) that smaller samples result in smaller z (weaker evidence) and wider intervals, while larger samples have the reverse effect. The results of (iii) show that the effect of varying unequal sample sizes is more complicated.
z = 95% C.I. ( ), ( )Explanation / Answer
Pooled sample proportion(P) = (p1 * n1 + p2 * n2)/(n1 + n2)
= (0.25 * 1000 + 0.15 * 1000)/2000
= 0.2
SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))
= sqrt(0.2 * 0.8 * (1/1000 + 1/1000))
= 0.018
The test statistic z = (p1 - p2)/SE
= (0.25 - 0.15)/0.018 = 5.55
The 95% confidence interval is
p1 - p2 +/- z0.025 * SE
= (0.25 - 0.15) +/- 1.96 * 0.018
= 0.1 +/- 0.0353
= 0.0647, 0.1353
b) Pooled sample proportion(P) = (p1 * n1 + p2 * n2)/(n1 + n2)
= (0.25 * 1600 + 0.15 * 1600)/3200
= 0.2
SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))
= sqrt(0.2 * 0.8 * (1/1600 + 1/1600))
= 0.014
The test statistic z = (p1 - p2)/SE
= (0.25 - 0.15)/0.014 = 7.14
The 95% confidence interval is
p1 - p2 +/- z0.025 * SE
= (0.25 - 0.15) +/- 1.96 * 0.014
= 0.1 +/- 0.0274
= 0.0726, 0.1274
c) Pooled sample proportion(P) = (p1 * n1 + p2 * n2)/(n1 + n2)
= (0.25 * 1000 + 0.15 * 1600)/3200
= 0.15
SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))
= sqrt(0.15 * 0.85 * (1/1000 + 1/1600))
= 0.014
The test statistic z = (p1 - p2)/SE
= (0.25 - 0.15)/0.014 = 7.14
The 95% confidence interval is
p1 - p2 +/- z0.025 * SE
= (0.25 - 0.15) +/- 1.96 * 0.014
= 0.1 +/- 0.0274
= 0.0726, 0.1274
OPtion - E is the correct answer.
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