WK 5(1)

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Apr 26, 2024

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Week 5 Knowledge Check Homework Practice Questions - Re… Attempt 1 of 4 Written Apr 3, 2024 2:31 PM - Apr 3, 2024 4:33 PM Attempt Score 14 / 20 - 70 % Overall Grade (Highest Attempt) 15.5 / 20 - 77.5 % Question 1 1 / 1 point ___ 231___ Hide ques±on 1 feedback The population standard deviation for the height of college basketball players is 3.5 inches. If we want to estimate 97% confidence interval for the population mean height of these players with a 0.5 margin of error, how many randomly selected players must be surveyed? (Round up your answer to nearest whole number) Answer: Z-Critical Value = NORM.S.INV(.985) = 2.17009 n = n =
Question 2 1 / 1 point A random sample of college basketball players had an average height of 66.35 inches. Based on this sample, (65.6, 67.1) found to be a 94% confidence interval for the population mean height of college basketball players. Select the correct answer to interpret this interval. Question 3 1 / 1 point ___ 20___ Hide ques±on 3 feedback 94% of college basketball players have height between 65.6 and 67.1 inches. There is a 94% chance that the population mean height of college basketball players is between 65.6 and 67.1 inches. We are 94% confident that the population mean height of college basketball players is between 65.6 and 67.1 inches. We are 94% confident that the population mean height of college basketball players is 66.35 inches. 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) Z-Critical Value =NORM.S.INV(.96) = 1.750686 n = n =
Question 4 0 / 1 point ___ 74___ (8) Hide ques±on 4 feedback Question 5 1 / 1 point There is no prior information about the proportion of Americans who support gun control in 2018. If we want to estimate 95% confidence interval for the true proportion of Americans who support gun control in 2018 with a 0.36 margin of error, how many randomly selected Americans must be surveyed? Answer: (Round up your answer to nearest whole number) Z-Critical Value =NORM.S.INV(.975) = 1.96 n = n = The FDA regulates that fresh Albacore tuna fish that is consumed is allowed to contain 0.82 ppm of mercury or less. A laboratory is estimating the amount of mercury in tuna fish for a new company and needs to have a margin of error within 0.03 ppm of mercury with 95% confidence. Assume the population standard deviation is 0.138 ppm
___ 82___ Hide ques±on 5 feedback Question 6 0 / 1 point ___ 49___ (50) Hide ques±on 6 feedback of mercury. What sample size is needed? Round up to the nearest integer. Answer: Z-Critical Value = NORM.S.INV(.975) = 1.96 n = n = There is no prior information about the proportion of Americans who support free trade in 2018. If we want to estimate a 97.5% confidence interval for the true proportion of Americans who support free trade in 2018 with a 0.16 margin of error, how many randomly selected Americans must be surveyed? Answer: (Round up your answer to nearest whole number) Z-Critical Value = NORM.S.INV(.9875) = 2.241403 n =
Question 7 1 / 1 point ___ 87___ Hide ques±on 7 feedback Question 8 1 / 1 point From a sample of 500 items, 30 were found to be defective. The point estimate of the population proportion defective will be: n = 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: Z-Critical Value = NORM.SINV(.95) = 1.645 n = n =
Hide ques±on 8 feedback Question 9 1 / 1 point ___ .274___ (50 %) ___ .526___ (50 %) Hide ques±on 9 feedback 0.94 30 0.60 0.06 30/500 A random sample found that forty percent of 100 Americans were satisfied with the gun control laws in 2017. Compute a 99% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2017. Fill in the blanks appropriately. A 99% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2017 is ( ) (round to 3 decimal places) Z-Critical Value = NORM.S.INV(.995) = 2.575 LL = 0.4 - 2.575* UL = 0.4 -+2.575*
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