A survey of Internet users reported that 17% downloaded music onto their compute

Question

A survey of Internet users reported that 17% 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.

Explanation / Answer

i) Sample size are same n1=12=1000

Using Minitab :

Test and CI for Two Proportions

Sample X N Sample p
1 170 1000 0.170000
2 330 1000 0.330000

Difference = p (1) - p (2)
Estimate for difference: -0.16
95% CI for difference: (-0.197301, -0.122699)
Test for difference = 0 (vs 0): Z = -8.41 P-Value = 0.000

z=-8.41 and 95% CI for difference: (-0.197301, -0.122699)

(ii) Both sample sizes are 1600.

Test and CI for Two Proportions

Sample X N Sample p
1 170 1600 0.106250
2 330 1600 0.206250

Difference = p (1) - p (2)
Estimate for difference: -0.1
95% CI for difference: (-0.124921, -0.0750792)
Test for difference = 0 (vs 0): Z = -7.86 P-Value = 0.000

Z=-7.86 and 95% CI for difference: (-0.124921, -0.0750792)

iii)The sample size for the survey reporting 25% is 1000 and the sample size for the survey reporting 17% is 1600.

Test and CI for Two Proportions

Sample X N Sample p
1 170 1000 0.170000
2 250 1600 0.156250

Difference = p (1) - p (2)
Estimate for difference: 0.01375
95% CI for difference: (-0.0155511, 0.0430511)
Test for difference = 0 (vs 0): Z = 0.92 P-Value = 0.358

z=0.92 and 95% CI for difference: (-0.0155511, 0.0430511)

Hope this will be helpful to you. Thanks and God Bless You :-)

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