A survey of Internet users reported that 20% downloaded music onto their compute

Question

A survey of Internet users reported that 20% 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. Assume that the sample sizes are both 1371. Using a significance test, evaluate whether or not there has been a change in the percent of Internet users who download music. Provide all details for the test. (Round your value for z to two decimal places. Round your P-value to four decimal places.)

Also report a 95% confidence interval for the difference in proportions. (Round your answers to four decimal places.)

Also report a 95% confidence interval for the difference in proportions. (Round your answers to four decimal places.)

Explanation / Answer

here n1=1371, p1=0.2, var(p1)=(p1(1-p1)/n1)=0.2*(1-0.2)/1371=0.000117

n2=1371,p2=0.27,var(p2)=0.27*(1-0.27)/1371=0.000144

SE(p1-p2)=sqrt(var(p1)+var(p2))=sqrt(0.000117+0.000144)=0.0162

we use z-test for significance test of difference in proportions and with

null hypothesis H0:p1=p2 and

alternate hypothesis H1: p1 is not equal to p2 (two tailed test)

z=(p1-p2)/SE(p1-p2)=(0.2-0.27)/0.0162=-4.32

the two tailed critical z(0.05)=1.96 is less than absolute value of calcuated z=4.32, so we failed to accept H0 and conclude that there has been a change in the percent of Internet users who download music.

(1-alpha)*100% confidence interval for difference of propotion =|(p1-p2)|±z(alpha/2)*SE(p1-p2)

95% confidence interval =|(0.2-0.27)|±z(0.05/2)*0.0162=0.07±1.96*0.0162=0.07±0.0318=(0.0382,0.1018)

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