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(A) In a chili cookoff, 66 people sample all of the chili and 30% vote to give the "Fire Hot Mouth Burner" the best chili in the contest. Since we haven't sampled everyone at the entire cook off we have to use a confidence interval to estimate the proportion of chili eating contest-goers that might choose this as the best chili of the contest. (1) p .3 (ii) n = 66 (ii) nô (iv) n (1 – p) = 19.8 (this is the number that voted for the "Fire Hot Mouth Burner") || 46.2 (this is the number that didn't vote for the "Fire Hot Mouth Burner") (v) Since both of the previous answers are greater than or equal to 10 we can proceed with a normal model for the confidence interval. We now need to get the critical z-score. We will use a 92% confidence level. In MS Excel we need to use the "=norm.s.inv()" command to determine the appropriate critical z-score. =norm.s.inv( .96 (vi) This gives a critical z-score of 1.75 (vii) The standard error is: SE = (viii) The margin of error is: МОЕ (ix) Hence, we are 92% confident that the true proportion of people that will vote for "Fire Hot Mouth Burner" is between and

(A) In a chili cookoff, 66 people sample all of the chili and 30% vote to give the "Fire Hot Mouth Burner" the best chili in the contest. Since we haven't sampled everyone at the entire cook off we have to use a confidence interval to estimate the proportion of chili eating contest-goers that might choose this as the best chili of the contest. (1) p .3 (ii) n = 66 (ii) nô (iv) n (1 – p) = 19.8 (this is the number that voted for the "Fire Hot Mouth Burner") || 46.2 (this is the number that didn't vote for the "Fire Hot Mouth Burner") (v) Since both of the previous answers are greater than or equal to 10 we can proceed with a normal model for the confidence interval. We now need to get the critical z-score. We will use a 92% confidence level. In MS Excel we need to use the "=norm.s.inv()" command to determine the appropriate critical z-score. =norm.s.inv( .96 (vi) This gives a critical z-score of 1.75 (vii) The standard error is: SE = (viii) The margin of error is: МОЕ (ix) Hence, we are 92% confident that the true proportion of people that will vote for "Fire Hot Mouth Burner" is between and

College Algebra
College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter9: Sequences, Probability And Counting Theory
Section9.7: Probability
Problem 55SE: Use the following scenario for the exercises that follow: In the game of Keno, a player starts by...
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(A) In a chili cookoff, 66 people sample all of the chili and 30% vote to give the "Fire Hot Mouth Burner" the best chili in the contest. Since we haven't sampled everyone at the entire cook off we have to use a confidence interval to estimate the proportion of chili eating contest-goers that might choose this as the best chili of the contest. (1) P .3 || (ii) n = 66 (ii) nộ = (м n(1 — р) 19.8 (this is the number that voted for the "Fire Hot Mouth Burner") 46.2 (this is the number that didn't vote for the "Fire Hot Mouth Burner") (v) Since both of the previous answers are greater than or equal to 10 we can proceed with a normal model for the confidence interval. We now need to get the critical z-score. We will use a 92% confidence level. In MS Excel we need to use the "=norm.s.inv()" command to determine the appropriate critical z-score. =norm.s.inv( .96 (vi) This gives a critical z-score of z* = 1.75 (vii) The standard error is: SE (viii) The margin of error is: МОЕ (ix) Hence, we are 92% confident that the true proportion of people that will vote for "Fire Hot Mouth Burner" is between and
Transcribed Image Text:(A) In a chili cookoff, 66 people sample all of the chili and 30% vote to give the "Fire Hot Mouth Burner" the best chili in the contest. Since we haven't sampled everyone at the entire cook off we have to use a confidence interval to estimate the proportion of chili eating contest-goers that might choose this as the best chili of the contest. (1) P .3 || (ii) n = 66 (ii) nộ = (м n(1 — р) 19.8 (this is the number that voted for the "Fire Hot Mouth Burner") 46.2 (this is the number that didn't vote for the "Fire Hot Mouth Burner") (v) Since both of the previous answers are greater than or equal to 10 we can proceed with a normal model for the confidence interval. We now need to get the critical z-score. We will use a 92% confidence level. In MS Excel we need to use the "=norm.s.inv()" command to determine the appropriate critical z-score. =norm.s.inv( .96 (vi) This gives a critical z-score of z* = 1.75 (vii) The standard error is: SE (viii) The margin of error is: МОЕ (ix) Hence, we are 92% confident that the true proportion of people that will vote for "Fire Hot Mouth Burner" is between and
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