A survey of Internet users reported that 19% downloaded music onto their compute

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 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.

95% of C.I=

(ii) Both sample sizes are 1600.

95% of C.I=

(iii) The sample size for the survey reporting 31% is 1000 and the sample size for the survey reporting 19% is 1600.

95% of C.I=

Explanation / Answer

here p1 =0.31 ; p2=0.19

std error of difference =(p1(1-p1)/n1 +p2(1-p2)/n2)1/2

for 95% CI, z=1.96

hence confidence interval =p1-p2 +/- z*std errorr

from above procedure

(i)std error =0.0191

CI =0.0824 ; 0.1576

(ii)std error =0.0152

CI =0.0903 ; 0.1497

(iii) std error 0.0176

CI =0.0855 ; 0.1545

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