33. 0.25/0.69 points I Previous Answers Mintrostat88.E.512.XP My Notes C Ask You

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33. 0.25/0.69 points I Previous Answers Mintrostat88.E.512.XP My Notes C Ask Your Teacher 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 1401. 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 you P- r value to four decimal places.) P-value Summarize your conclusio O We conclude that the means are not different O We conclude that the proportions are different. O We conclude that the proportions are not different. We conclude that the means are different. O We cannot draw any conclusions using a significance test for this data. Also report a 95% confidence interval for the difference in proportions. (Round your answers to four decimal places.)

Explanation / Answer

We are given,

P1 = 0.27

P2 = 0.20

n1 = 1401

n2 = 1401

z = (p1^ - p2^)/sqrt((PQ*((1/n1)+(1/n2))

P = ((x1+x2)/(n1+n2))

Q = 1-P

x1 = n1*p1

x2 = n2*p2

P = 0.235

Q = 1 – 0.235

   =0.765

z= (0.20 – 0.27)/sqrt((0.235*(1-0.235))*((1/1401)+(1/1401))

   =

Answer:

z =-4.37

P-value Formula:

=2*normsdist(-4.37)

= 2*0.0000

= 0.0000

Answer:

P-value = .0000

Summarize your conclusion:

Answer:

We conclude that the proportions are different.

95% confidence interval:

(p1^-p2^) (-/+) E

E = zc * sqrt (((p1*(1-p1))/n1) + ((p2*(1-p2))/n2)

zc is the critical value at 5% level of significance is 1.96

E = 1.96 * sqrt (((0.20*(1-0.20))/1401)+((0.27*(1-0.27))/1401)

   = 0.0313

(0.20-0.27) (-/+) 0.0313

= -0.07 (-/+) 0.0313

= -0.1013 and -0.0387

Answer:

-0.1013 and -0.0387

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