MATH 302 Week 5 Quiz - Question and Answers

MATH 302 Week 5 Quiz - Question and Answers Question 1 1 / 1 point A researcher would like to estimate the proportion of all children that have been diagnosed with Autism Spectrum Disorder (ASD) in their county. They are using 97% confidence level and the CDC national estimate that 1 in 68 ≈ 0.0147 children are diagnosed with ASD. What sample size should the researcher use to get a margin of error to be within 1.5%? Round up to the nearest integer, do not include decimals. Answer ___304___ Hide Feedback Z-Critical Value = NORM.S.INV(.985) = 2.17009 n = .0147∗.9853∗2..0152{"version":"1.1","math":"displaystyle frac{.0147*.9853*2.17009^2}{.015^2} "} Question 2 1 / 1 point The population standard deviation for the height of college basketball players is 2.9 inches. If we want to estimate a 99% confidence interval for the population mean height of these players with a 0.45 margin of error, how many randomly selected players must be surveyed? (Round up your answer to nearest whole number, do not include any decimals) Answer: ___276___ Hide Feedback MATH 302 Week 5 Quiz - Question and Answers Z-Critical Value = NORM.SINV(.995) = 2.575 n = 2.92∗2.5752.452{"version":"1.1","math":"displaystyle frac{2.9^2*2.575^2}{.45^2}"} Question 3 1 / 1 point There is no prior information about the proportion of Americans who support gun control in 2019. If we want to estimate 93% confidence interval for the true proportion of Americans who support gun control in 2019 with a 0.18 margin of error, how many randomly selected Americans must be surveyed? Answer: (Round up your answer to nearest whole number, do not include any decimals) ___26___ Hide Feedback Z-Critical Value = NORM.S.INV(.965) = 1. n = .5∗.5∗1..182{"version":"1.1","math":"displaystyle frac{.5*.5*1.^2}{.18^2}"} Question 4 1 / 1 point The population standard deviation for the height of college basketball players is 2.4 inches. If we want to estimate 97% confidence interval for the population mean height of these players with a 0.35 margin of error, how many randomly MATH 302 Week 5 Quiz - Question and Answers selected players must be surveyed? (Round up your answer to nearest whole number, do not include any decimals.) Answer: ___222___ Hide Feedback Z-Critical Value = NORM.S.INV(.985) = 2.17009 n = 2.42∗2..352{"version":"1.1","math":"displaystyle frac{2.4^2*2.17009^2}{.35^2}"} Question 5 1 / 1 point There is no prior information about the proportion of Americans who support free trade in 2019. If we want to estimate a 98% confidence interval for the true proportion of Americans who support free trade in 2019 with a 0.21 margin of error, how many randomly selected Americans must be surveyed? Answer: (Round up your answer to nearest whole number, do not include any decimals) ___31___ Hide Feedback Z-Critical Value = NORM.S.INV(.99) = 2. n = .5∗.5∗2..212{"version":"1.1","math":"displaystyle frac{.5*.5*2.^2}{.21^2}"} Question 6 1 / 1 point MATH 302 Week 5 Quiz - Question and Answers The population standard deviation for the height of college baseball players is 3.0 inches. If we want to estimate 95% confidence interval for the population mean height of these players with a 0.68 margin of error, how many randomly selected players must be surveyed? (Round up your answer to nearest whole number, do not include any decimals) Answer: ___75___ Hide Feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 n = 32∗1.962.682{"version":"1.1","math":"displaystyle frac{3^2*1.96^2}{.68^2}"} Question 7 1 / 1 point There is no prior information about the proportion of Americans who support Medicare-for-all in 2019. If we want to estimate 95% confidence interval for the true proportion of Americans who support Medicare-for-all in 2019 with a 0.175 margin of error, how many randomly selected Americans must be surveyed? Answer: (Round up your answer to nearest whole number, do not include any decimals) ___32___ Hide Feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 n = MATH 302 Week 5 Quiz - Question and Answers .5∗.5∗1.962.1752{"version":"1.1","math":"displaystyle frac{.5*.5*1.96^2}{.175^2} "}