Question

1) In the Health ABC Study, 522 subjects owned a pet and 1982 subjects did not. Among the pet owners, there were 295 women; 968 of the non-pet owners were women. Find the proportion of pet owners who were women. Do the same for the non-pet owners. (Be sure to let Population 1 correspond to the group with the higher proportion so that the difference will be positive. Round your answers to three decimal places.)

p̂1	=
p̂2	=

Give a 95% confidence interval for the difference in the two proportions. (Do not use rounded values. Round your final answers to three decimal places.)

( , )

2) 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 28% from a survey taken two years before. Assume that the sample sizes are both 1441. 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.)

z	=
P-value	=

Summarize your conclusion.

1 We conclude that the means are different.

2 We conclude that the means are not different.

3 We conclude that the proportions are not different.

4 We cannot draw any conclusions using a significance test for this data.

5 We conclude that the proportions 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.

1 The interval shows no significant change in music downloads.

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

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

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

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

Thanks so very much ive been struggling with these 2 all day

Answer 1

1)

sample #1   ----->   pet           
first sample size,     n1=   522          
number of successes, sample 1 =     x1=   195          
proportion success of sample 1 , p̂1=   x1/n1=   0.37
                  
sample #2   ----->   non pet          
second sample size,     n2 =    1982          
number of successes, sample 2 =     x2 =    968          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.49
                  
difference in sample proportions, p̂1 - p̂2 =     0.3736   -   0.4884   =   -0.12

level of significance, α =   0.05              
Z critical value =   Z α/2 =    1.960   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0240          
margin of error , E = Z*SE =    1.960   *   0.0240   =   0.0470
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    -0.115   -   0.0470   =   -0.1618
upper limit = (p̂1 - p̂2) + E =    -0.115   +   0.0470   =   -0.0679
                  
so, confidence interval is (   -0.162 < p1 - p2 <   -0.068 )