week 5 homework questions

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Statistics

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Jan 9, 2024

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Attempt Score 19.5 / 20 - 97.5 % Overall Grade (Highest Attempt) 19.5 / 20 - 97.5 % stion 1 1 / 1 p 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 95% 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 2%? Round up to the nearest integer. Answer ___140 ___ uestion 1 feedback cal Value = NORM.S.INV(.975) = 1.96 47 ∗ .9853 ∗ 1.962.022 on 2 1 / There is no prior information about the proportion of Americans who support gun control in 2018. If we want to estimate 92% confidence interval for the true proportion of Americans who support gun control in 2018 with a 0.2 margin of error, how many randomly selected Americans must be surveyed? Answer: (Round up your answer to nearest whole number) ___20 ___ uestion 2 feedback cal Value =NORM.S.INV(.96) = 1.750686 ∗ � ∗ � 2 �� 2

.5 ∗ 1.7506862.22 on 3 1 / Select the correct answer for the blank: If everything else stays the same, the required sample size ____ as the confidence level decreases to reach the same margin of error. Answer: Increases Decreases Remains the same Question 4 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 0.3 margin of error, how many randomly selected Americans must be surveyed? Answer: (Round up your answer to nearest whole number) ___11 ___ uestion 4 feedback cal Value = NORM.S.INV(.975) = 1.96 ∗ � ∗ � 2 �� 2 .5 ∗ 1.962.32 on 5 1 / The population standard deviation for the height of college basketball players is 3 inches. If we want to estimate 97% confidence interval for the population mean height of these players with a 0.6 margin of error, how many randomly

selected players must be surveyed? (Round up your answer to nearest whole number) Answer: ___118 ___ uestion 5 feedback cal Value = NORM.S.INV(.985) = 2.17009 � 2 ∗ � 2 �� 2 ∗ 2.170092.62 on 6 1 / The population standard deviation for the height of college football players is 3.3 inches. If we want to estimate a 90% confidence interval for the population mean height of these players with a 0.7 margin of error, how many randomly selected players must be surveyed? (Round up your answer to nearest whole number) Answer: ___61 ___ uestion 6 feedback cal Value =NORM.S.INV(.95) = 1.645 � 2 ∗ � 2 �� 2 2 ∗ 1.6452.72 on 7 1 / The population standard deviation for the height of college hockey players is 3.4 inches. If we want to estimate 90% confidence interval for the population mean height of these players with a 0.6 margin of error, how many randomly selected players must be surveyed? (Round up your answer to nearest whole number) Answer:

___87 ___ uestion 7 feedback cal Value = NORM.SINV(.95) = 1.645 � 2 ∗ � 2 �� 2 2 ∗ 1.6452.62 on 8 1 / In a random sample of 200 people, 135 said that they watched educational TV. Find the 95% confidence interval of the true proportion of people who watched educational TV. 0.6101< � ^<0.7399 0.6704< � ^<0.6796 0.325< � ^<0.675 –1.96< � ^<1.96 Hide question 8 feedback Z - Critical Value =NORM.S.INV(.975) = 1.96 LL = .675 - 1.96*.675 ∗ .325200 UL = .675 + 1.96*.675 ∗ .325200 n 9 1 Suppose a marketing company computed a 94% confidence interval for the true proportion of customers who click on ads on their smartphones to be (0.56 , 0.62). Select the correct answer to interpret this interval 94% of customers click on ads on their smartphones.

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