A survey of Internet users reported that 18% downloaded music onto their compute

Question

A survey of Internet users reported that 18% 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. Assume that the sample sizes are both 1361. 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.)

Summarize your conclusion.

We conclude that the proportions are not different.

We conclude that the means are not different.    

We cannot draw any conclusions using a significance test for this data.We conclude that the proportions are different.

We conclude that the means are different.

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

( , )

Explain what information is provided in the interval that is not in the significance test results.

The interval tells us there was a significant change in music downloads, but the test statistic is inconclusive.

The interval does not provide any more information than the significance test would tell us.

The interval gives us an idea of how large the difference is between the first survey and the second survey.

The significance test does not indicate the direction of change, but the interval shows that the music downloads decreased.

The interval shows no significant change in music downloads.

Explanation / Answer

A)

Formulating the hypotheses          
Ho: p1 - p2   =   0  
Ha: p1 - p2   =/=   0  
Here, we see that pdo =    0   , the hypothesized population proportion difference.  
          
Getting p1^ and p2^,          
          
p1^ = x1/n1 =    0.18      
p2 = x2/n2 =    0.25      
          
Also, the standard error of the difference is          
          
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] =    0.015691272      
          
Thus,          
          
z = [p1 - p2 - pdo]/sd =    -4.461078739   [ANSWER, Z]

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Also, the p value is, as this is two tailed,          
          
P =    8.15481*10^-6   [ANSWER, P VALUE]

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As P is very small,

We conclude that the proportions are different. [ANSWER, D]

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Also report a 95% confidence interval for the difference in proportions. (Round your answers to four decimal places.)  
          
For the   95%   confidence level, then  
          
alpha/2 = (1 - confidence level)/2 =    0.025      
z(alpha/2) =    1.959963985      
Margin of error = z(alpha/2)*sd =    0.030754328      
lower bound = p1^ - p2^ - z(alpha/2) * sd =    -0.100754328      
upper bound = p1^ - p2^ + z(alpha/2) * sd =    -0.039245672      
          
Thus, the confidence interval is          
          
(   -0.100754328   ,   -0.039245672 ) [ANSWER]

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The interval gives us an idea of how large the difference is between the first survey and the second survey. [ANSWER, C]

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