Question

This Question: 1 pt 5 of 9 (0 complete) This Test: 9 pts possible A website reported that 31% oft drivers 18 and okder admitted to texting whle driving in 2009 In a random sample of 200 drivers 18 years and older drawn in 2010, 49 of the drivers said they texted whle driving Complete parts a through c below a. Construct a 99% contidence interval to estmate the actual proportion of people who texted whle driving in 2010 A 99% contidence interval to estimate the actual proportion has a lower imit ofand an upper imit of (Round to three decimal places as needed) b. What is the margin of eror for this sampie? The margin of emor is (Round to three decimal places as needed c. is there any evidence that this peoportion has changed since 2009 based on this sample? Because the contidence interval found in part a the reported proportion from 2009, this sample evidence that this proportion has changed since then

Answer 1

n=200, x=49, C=99.1 Sample Proportion 18 ) = = 49 - 0-245 Critical value: Q = 1-0.99=0.01 9/2=0.005 Ex12= 70.005 = + 324576] (using table] margin of error ME = Expo x P x C1-P) = 2.576 x 0.945 x (1-0:245) 200 = 0.078346 ? 0.078 Arx 99 % Confidence interval . § + ME = 0.245 + 0.078346 = 0,1667, 0.3233 994. CI = lower limit = 10-167) s ubber Limit = 0.393 There is not sufficient evidence to conclude that the Proportion has changed since 2009 based on this Sample because the CI Contein 0.31 (6)