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9. 16 pts) Google wants to know if there is strong evidence that more than 30% of smartphone users actively use Twitter. Use the sample data to perform the hypothesis test at the 1% significance level. For full credits, you need to specify the null and alternative hypotheses, appropriate test- statistic value, p- value and conclusion Parameter to be tested: Ho: LH: Test-statistics: p-value: Conclusion:
Call Style fo card Font Ant Stiri CH Facebook Twitter Online chat/Video M N Column Facebook Twitter Online All Yes Yes 221 No 223 291 172 Total 450 150 0 1 1 0 1 O 1 0 O 1 0 1 0 0 D 1 1 0 1 0 0 ves 0 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 0 0 0 1 0 1 0 1 1 1 0 0 1 O 1 1 0 cebook Twitter Online Chat/Video . Yes Ives NO Yes NO NO NO Yes NO ves 0 [ Yes NO NO NO D Yes Yes NO NO o NO Yes NO Yes No NO Yes NO Yes No Yes Yes Yes NO Yes Yes NO NO Yes No NO INO Yes Yes lyes 0 res NG NO D No Yes Ires NO Yes NO lo Yes Tres NO NO D ves fres Yes NO NO NO NO Yer NO Tres NO NO Yes Ives NO Yes NO NO Yes lyes D NO NO o ves Yes NO Yes NO Yes Yes Yes 0 NO ve NO Yes lyes Yes 23 No Yes NO VE D NO 2 NO Yes 25 NO Yes o NO Yes One Night in Man Google YE 0 0 1 0 1 1 T ves 1 1 1 1 0 0 1 1 1 0 1 0 0 1 1 0 1 1 0 1 1 1 o 1 1 0 0 0 1 0 0 0 1 0 1 0 1 0 Yes 1 1 1 1 1 1 1 1 1 1 O 3 1 1 1 1 1 1 1 0 1 1 0 O
[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 450 residents in one of the candidate cities, Rich-Addison, and collected the data in the "Google" sheet of "Part A - Data Sets for Assessment Assignment.xlsx". Use this data set to answer questions 6 to 9. 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. 12 pt] If everything else remains the same, what is the margin of error in a 98% 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 90% confidence interval for the true proportion of smartphone users who are actively using all three? 90% confidence interval:
9. 16 pts] Google wants to know if there is strong evidence that more than 30% of smartphone users actively use Twitter. Use the sample data to perform the hypothesis test at the 1% significance level. For full credits, you need to specify the null and alternative hypotheses, appropriate test- statistic value, p- value and conclusion. Parameter to be tested: Ho: HA: Test-statistics: p-value: Conclusion: Activate Windows
File Home û X D Paste Cipboard M 05 Insert Page Layout Formulas Calen BIU. Foot C Facebook Twitter No Yes No No Yes No 5 Yes No No No 7 No Yes No No 9 No Ves 10 Yes No 11 No No 12 No No 3 Yes No 14 Yes No 15 No No 16 Yes No 7 Yes Yes yes ING -11-AA== A- f ✓ D Online Chat/Video Yes Yes No Yes Yes No No Yes No Yes ves Yes Yes Ves Yes Yes lyes Data Review D. FOR- Alignment Number G Facebook Twitter Online Chat/Video 0 0 1 0 0 1 0 1 0 0 1 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1 23 19 0 1 View Help General 0 0 0 0 0 0 0 1 O -% 1 0 1 1 1 1 1 B Delete- Format- Cels Conditional Format as Cell Formatting Table Styles Styles M Columnt Facebook Twitter Yes 226 No 223 Total 450" Σ. DO Online 159 291 450 270 Sort & Find & Fiter Sled faring All Yes 273 177 450 Comments KD) Analyse Dala Analysn D Q Share