Question

[Google's free Internet project] According to the Pew Internet & American Life Project, many American adults spend significant amount of time on the Internet every day (Pew Internet Project, 2008). Google has a project that provides free wireless Internet in a medium-sized city. Among several factors that affect Google's decision, one factor is the percentage of adult smartphone users who are actively using social network apps. Google picked a random sample of 400 residents in one of the candidate cities, Rich- Addison, and collected the following data. 6. [2 pt] Construct a 95% confidence interval for the true proportion of Rich-Addison residents who are actively using Facebook. 95% confidence interval: 7. [2 pt] If everything else remains the same, what is the margin of error in a 99% confidence interval? Margin of error: 8. [3 pts] Google wants to know what percentage of users is actively using all three medias - Facebook, Twitter, and an online chat/video tool (Skype, Facetime, etc.). What is 95% confidence interval for the true proportion of smartphone users who are actively using all three? 95% confidence interval:

Answer 1

6)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   211          
Sample Size,   n =    425          
                  
Sample Proportion ,    p̂ = x/n =    0.4965          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.024253          
margin of error , E = Z*SE =    1.960   *   0.02425   =   0.0475
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.49647   -   0.04753   =   0.4489
Interval Upper Limit = p̂ + E =   0.49647   +   0.04753   =   0.5440
                  
95%   confidence interval is (   0.449   < p <    0.544   )

7)

Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.024253          
margin of error , E = Z*SE =    2.576   *   0.02425   =   0.0625

8)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   39          
Sample Size,   n =    425          
                  
Sample Proportion ,    p̂ = x/n =    0.0918          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.014004          
margin of error , E = Z*SE =    1.960   *   0.01400   =   0.0274
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.09176   -   0.02745   =   0.0643
Interval Upper Limit = p̂ + E =   0.09176   +   0.02745   =   0.1192
                  
95%   confidence interval is (   0.064   < p <    0.119   )

Please let me know in case of any doubt.

Thanks in advance!

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