Confidence Intervals: Calculating Sample Sizes for Heights

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

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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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Question 9 1/ 1 point A random sample of 150 people was selected and 12% of them were left handed. Find the 90% confidence interval for the proportion of left-handed people. (12, .88) ) (-1.645, 1.645) v (0.0764,0.1636) ~(0.0436,0.1164) ~(0.068,0.172) w Hide question 9 feedback Z-Critical Value = NORM.S.INV(.95) = 1.645 LL=.12 - 1.645" 12 % .88 150 UL=.12 + 1.645" 12 % .88 150 Question 10 1/ 1 point

A teacher wanted to estimate the proportion of students who take notes in her class. She used data from a random sample of size n = 82 and found that 50 of them took notes. The 99% confidence interval for the proportion of student that take notes is: 0.471 V(50 %) <p< 0.749 v (50 %) . Round answers to 3 decimal places. w Hide question 10 feedback Z-Critical Value =NORM.S.INV(.995) =2.575 LL =.609756 - 2.575% 609756 * .390244 82 UL = .609756 + 2.575* 609756 x .390244 82 Question 11 1/ 1 point 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 > We are 94% confident that the true proportion of customers who click ~ on ads on their smartphones is 0.59. > 94% of customers click on ads on their smartphones. > There is a 94% chance that the true proportion of customers who click ~on ads on their smartphones is between 0.56 and 0.62.

¢<> We are 94% confident that the true proportion of customers who click ~on ads on their smartphones is between 0.56 and 0.62. Question 12 0/ 1 point The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the feasibility of providing a dental insurance plan. A random sample of 12 employees reveals the following family dental expenses (in dollars). See Attached Excel for Data. dental expense data.xlsx Construct a 20% confidence interval estimate for the mean of family dental expenses for all employees of this corporation. Place your LOWER limit, in dollars rounded to 1 decimal place, in the first blank. Do not use a dollar sign, a comma, or any other stray mark. For example, 123.4 would be a legitimate entry.___ Place your UPPER limit, in dollars rounded to 1 decimal place, in the second blank. Do not use a dollar sign, a comma, or any other stray mark. For example, 567.8 would be a legitimate entry.___ Answer for blank # 1: 110.6 % (231.5) Answer for blank # 2: 229.4 3 (384.9) w Hide question 12 feedback T-Critical Value = T.INV.2T(.10,11) = 1.795885 LL =308.1667 - 1.795885 *

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147.928 V12 UL = 308.1667 +1.795885 * 147.928 V12 Question 13 1/ 1 point Senior management of a consulting services firm is concerned about a growing decline in the firm's weekly number of billable hours. The firm expects each professional employee to spend at least 40 hours per week on work. In an effort to understand this problem better, management would like to estimate the standard deviation of the number of hours their employees spend on work-related activities in a typical week. Rather than reviewing the records of all the firm's full-time employees, the management randomly selected a sample of size 51 from the available frame. The sample mean and sample standard deviations were 48.5 and 7.5 hours, respectively. Construct a 88% confidence interval for the mean of the number of hours this firm's employees spend on work-related activities in a typical week. Place your LOWER limit, in hours, rounded to 1 decimal place, in the first blank. For example, 6.7 would be a legitimate entry.___ Place your UPPER limit, in hours, rounded to 1 decimal place, in the second blank. For example, 12.3 would be a legitimate entry.___ Answer for blank # 1: 46.8 (50 %) Answer for blank # 2: 50.2 (50 %)

W Hide question 13 feedback T-Ciritical Value = T.INV.2T(.12,50) = 1.581805 LL=48.5-1.581805 * 7.5 Vv 5l UL =48.5 +1.581805 * 7.9 Vv 51l Question 14 1/ 1 point The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the feasibility of providing a dental insurance plan. A random sample of 12 employees reveals the following family dental expenses (in dollars). See Attached Excel for Data. dental expense data.xlsx Construct a 93% confidence interval estimate for the mean of family dental expenses for all employees of this corporation. Place your LOWER limit, in dollars rounded to 1 decimal place, in the first blank. Do not use a dollar sign, a comma, or any other stray mark. For example, 98.4 would be a legitimate entry.___

Place your UPPER limit, in dollars rounded to 1 decimal place, in the second blank. Do not use a dollar sign, a comma, or any other stray mark. For example, 567.8 would be a legitimate entry.___ Answer for blank # 1: 222.5 (50 %) Answer for blank # 2: 393.9 (50 %) w Hide question 14 feedback T-Ciritical Value = T.INV.2T(.07,11) = 2.006663 LL = 308.1667 - 2.006663" 147.928 V12 UL = 308.1667 +2.006663* 147.928 V12 Question 15 1/ 1 point After calculating the sample size needed to estimate a population proportion to within 0.05, you have been told that the maximum allowable error (E) must be reduced to just 0.025. If the original calculation led to a sample size of 1000, the sample size will now have to be___ Place your answer, as a whole number in the blank. For example, 2345 would be a legitimate entry.

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Answer: 4000 w Hide question 15 feedback Use n=pp Z 2 ‘ME) 1000= .5*.5 * Z 2 (5) 4000 = Z 2 ( .05 ) VA v4000 = — .05 v4000 % .00 = Z 3.16227 =Z Now solve for n, using Z n=.5*5*

3.16227 \ 2 025 Question 16 1/ 1 point A sample of 40 country CD recordings of Willie Nelson has been examined. The average playing time of these recordings is 51.3 minutes, and the standard deviation is s = 5.8 minutes. Using an appropriate t-multiplier, construct a 25% confidence interval for the mean playing time of all Willie Nelson recordings. Place your LOWER limit, in minutes, rounded to 2 decimal places, in the first blank. For example, 56.78 would be a legitimate entry.___. Place your UPPER limit, in minutes, rounded to 2 decimal places, in the second blank. For example, 67.89 would be a legitimate entry.___ Answer for blank # 1: 49.45 (50 %) Answer for blank # 2: 53.15 (50 %) w Hide question 16 feedback T-Critical Value =T.INV.2T(.05,39) = 2.022691 LL=51.3 - 2.022691* 5.8 v 40

UL =51.3 + 2.022691" 0.8 v 40 Question 17 0.5/ 1 point A randomly selected sample of college basketball players has the following heights in inches. See Attached Excel for Data. height data.xlsx Compute a 99% confidence interval for the population mean height of college basketball players based on this sample and fill in the blanks appropriately. 63.385 V(50 %) << 66.303 % (66.302) (roundto 3 decimal places) w Hide question 17 feedback T-Critical Value =T.INV.2T(.01,31) = 2.744042 3.006542 V32 3.006542 V32 LL = 64.84375 — 2.744042 * UL = 64.84375 + 2.744042 Question 18 1/ 1 point

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A random sample of college football players had an average height of 64.55 inches. Based on this sample, (63.2, 65.9) found to be a 92% confidence interval for the population mean height of college football players. Select the correct answer to interpret this interval. v We are 92% confident that the population mean height of college ~ football players is between 63.2 and 65.9 inches. > We are 92% confident that the population mean height of college ~ football palyers is 64.55 inches. > A 92% of college football players have height between 63.2 and 65.9 "~ inches. > There is a 92% chance that the population mean height of college ~ football players is between 63.2 and 65.9 inches. Question 19 1/ 1 point A random sample of stock prices per share (in dollars) is shown. Find the 90% confidence interval for the mean stock price. Assume the population of stock prices is normally distributed. See Attached Excel for Data stock price data.xlsx > (17.884, 40.806) ) (27.512, 31.178) ) (13.582, 45.108) v( ) (16.572,42.118) ) (-1.833, 1.833) ¥ Hide question 19 feedback T-Critical Value = T.INV.2T(.10,9) = 1.833113

LL =29.345 - 1.833113" 22.03462 v 10 UL = 29.345 + 1.833113 22.03462 v 10 Question 20 0/ 1 point A tire manufacturer wants to estimate the average number of miles that may be driven in a tire of a certain type before the tire wears out. Assume the population is normally distributed. A random sample of tires is chosen and are driven until they wear out and the number of thousands of miles is recorded, find the 99% confidence interval using the sample data. See Attached Excel for Data miles data.xlsx x () (27.144,33.356) ) (27.256, 33.244) (26746, 33.754) ) (-3.106, 3.106) = () (26.025, 34.475) w Hide question 20 feedback

T-Critical Value = T.INV.2T(.01,11) = 3.105807 4.71217 LL = 30.25 — 3.105807 x* V12 4.71217 UL = 30.25 + 3.105807 x* V12 Done

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