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Previous Problem Problem List Next Problem (1 point) In 2012, the percentage of cell phone owners who used their cell phone to send or receive text messages was 80 %. A polling firm contacted a simple random sample of 1300 people chosen from the population of cell phone owners. If p is the proportion of cell phone owners the mean for p who used their phone to text, then what was 0.011 Submit Answers Preview My Answers You have attempted this problem 2 times Your overall recorded score is 0 % You have unlimited attempts remaining. Email instructor Page gener ated al 06/18/2019 at 09 53pm EDT 37 h Previous Problem Problem List Next Problem (1 point) In 2012, the percentage of cell phone owners who used their cell phone to send or receive text messages was 80 % A polling firm contacted a simple random sample of 1800 people chosen from the population of cell phone owners. Ifp is the proportion of cell phone owners who used their phone to text, then what was the standard deviation for p? Submit Answers Preview My Answers You have attempted this problem 0 times You have unlimited attempts remaining Email instructor EDT Page genarated at 06/18/2019 at 09 53p 47 to search Previous Problem Problem List Next Problem (1 point) Greeks bucked a global trend in which people in most countries expect their lives in five years to tbe better than their current lives. Even among those countries with much lower current life ratings, greater optimism was found because people cannot fathom their lives getting worse But in Greece, only 25% expect their lives to be better in five years. In a simple random sample of 1800 Greek citizens, what is the probability that less than 27 percent of the sample expect their lives to be better in five years? 0.95 Submit Answers Preview My Answers You have attempted this problem 2 times. Your overall recorded score is 0 % . You have unlimited attempts remaining. Email instructor ch Chapter 16 Quiz Soltys: Problem 1 Previous Problem Problem List Next Problem (1 point) An article in the Washington Post on March 16, 1993 stated that nearly 45 percent of all Americans have brown eyes. A random sample of n= 80 Americans found 24 with brown eyes. Give the numerical value of the "plus 4" statistic p that is an estimate for p. C Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining. Email instructor Page generated at 06/18/2019 at 09 54pm EDT WeBWork 1996-2017) theme math4 /ww.version WeBWork-2 13 pg version PG-2 131 The WeBWork Project 포e Previous Problem Problem List Next Problem esults for this submission Entered Answer Preview Result Message 0.176 0.176 correct 0.2348 0.2348 incorrect Closel Use more accuracy in our calculation t least one of the answers above is NOT correct (4 points) A recent survey showed that among 675 randomly selected subjects who completed 4 years of college, 138 smoke and 537 do not smoke. Determine a 95 % confidence interval for the true proportion of the given population that smokes 95% CI: 0.1760 to 0.2348 Note: You can earn partial credit on this problem. Preview My Answers Submit Answers Your score was recorded. You have attempted this problem 3 times You received a score of 50% for this attempt 37 Previous Problem Problem List Next Problem (1 point) Albert wants to determine a 95 percent confidence interval for the true proportion of times he rolls a 3 (using a fair., 6-sided die). How many rolls must Albert make to get a margin of error less than or equal to 0.05? Use the guessed value p 1/6 for the sample proportion. Preview My Answers Submit Answers You have attempted this problem 0 times You have unlimited attempts remaining Email instructor Page genefated at 06/18/2019 at 10 00pm EDT on Po-2 13 The weBWork Project WeBWork 1996-2017 theme math4/ rersion arch

Answer 1

Gen that P=80% o.8 1 1300 4=1-P -O Mean =0.84 1300 0yo a) Standan d deviation 29 Standard deviahm P (1-P) t08)+ (01) o.009428 (3) 1-P 1-025 o.5 R[PAAT) P(2 0 R3-025 O010

P 6) P(2<196) O.9250 (4) 3 4 υμαλοιό 01 at ag P o.3 (5) 675 $ 39 do notmoke 957 Contidence interval or true aiven Poputation that Citical Value at 138 o. 204 635 Contidena inter Val P P(1-P) + 2. Codou)t(196) 1 G75 (owout to-0155) l01936 0 2344 95% Con-tidena oerval (0174, 0.235

P6 2 1.96 ME PMPL-P) (0.097 2xR.S8 P: MEtO.0 / (2s6 /2(1-1)

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