Week 5 Knowledge

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American Public University *

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302

Subject

Statistics

Date

Feb 20, 2024

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pdf

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18

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Week 5 Knowledge Check Homework Practice Questio... X Attempt 1 of 4 Written Oct 2, 2023 10:34 PM - Oct 3, 2023 12:27 AM Attempt Score 17.5/20-87.5% Overall Grade (Highest Attempt) 17.5/20-87.5% Question 1 1/ 1 point The population standard deviation for the height of college baseball players is 3.2 inches. If we want to estimate 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: 57 v w Hide question 1 feedback Z-Critical Value = NORM.S.INV(.95) = 1.645 n: SD? x 72 ME?
3.92 % 1.6452 7 Question 2 1/ 1 point 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 v w Hide question 2 feedback Z-Critical Value = NORM.SINV(.95) = 1.645 SD? x Z?2 ME? 3.42 x 1.6452 .67 Question 3 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.
> 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. v() 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. Question 4 1/ 1 point 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: 231 v w Hide question 4 feedback Z-Critical Value = NORM.S.INV(.985) = 2.17009 n: SD? x 7?2 M E? 3.52 % 2.170092 52
Question 5 1/ 1 point The population standard deviation for the height of college basketball players is 3.4 inches. If we want to estimate 95% 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: 178 w Hide question 5 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 SD? x 7?2 ME-? 3.42 % 1.962 52 Question 6 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) w Hide question 6 feedback
Z-Critical Value = NORM.S.INV(.975) = 1.96 pxq* Z? ME? 5% .5 % 1.967 32 Question 7 1/ 1 point 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 v W Hide question 7 feedback Z-Critical Value =NORM.S.INV(.95) = 1.645 n= SD? x 72 ME?
3.32 % 1.6452 VE Question 8 1/ 1 point In a certain state, a survey of 600 workers showed that 35% belonged to a union. Find the 95% confidence interval of true proportion of workers who belong to a union. (O 210 (0.0396,0.0771) v (0.3118,0.3882) " (34.9618, 35.0382) 1(0.29,0.41) ¥ Hide question 8 feedback Z - Critical Value =NORM.S.INV(.975) = 1.96 LL=.35-1.96* .35 x .65 600 UL =.35+ 1.96* .35 x .65 600
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