Week 5 Test: Confidence Intervals and Sample Sizes Explained

Week 5 Test - Results X Attempt 1 of 2 Written Oct 3, 2023 11:13 PM - Oct 4, 2023 12:55 AM Attempt Score 20/ 20 - 100 % Overall Grade (Highest Attempt) 20 /20 - 100 % Question 1 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 v w Hide question 1 feedback Z-Critical Value = NORM.SINV(.995) = 2.575 n= 2.92 % 2.5752 452 Question 2 1/ 1 point

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 0.023 ppm of mercury with 97% confidence. Assume the population standard deviation is 0.143 ppm of mercury. What sample size is needed? Round up to the nearest integer, do not include any decimals. Answer: 183 v w Hide question 2 feedback Z-Critical Value = NORM.S.INV(.985) = 2.17009 n: 1432 % 2.170092 .0232 Question 3 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 v w Hide question 3 feedback Z-Critical Value = NORM.S.INV(.985) = 2.17009 n:

0147 * .9853 % 2.170092 0152 Question 4 1/ 1 point The population standard deviation for the height of college basketball players is 3.1 inches. If we want to estimate 99% confidence interval for the population mean height of these players with a 0.58 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: __ 190 w Hide question 4 feedback Z-Critical Value =NORMS.INV(.995) = 2.575 n= 3.12 % 2.5752 582 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 v w Hide question 5 feedback Z-Critical Value = NORM.S.INV(.99) = 2.326348

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5% .5 % 2.3263482 217 Question 6 1/ 1 point The population standard deviation for the height of college football players is 2.7 inches. If we want to estimate a 95% confidence interval for the population mean height of these players with a 0.65 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: ___67 v w Hide question 6 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 n: 2.72 % 1.962 652 Question 7 1/ 1 point The population standard deviation for the height of college basketball players is 3.4 inches. If we want to estimate 99% confidence interval for the population mean height of these players with a 0.43 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: ___ 415 v

W Hide question 7 feedback Z-Critical Value = NORM.SINV(.995) = 2.575 n: 3.42 % 2.5752 432 Question 8 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 = 76 and found that 45 of them took notes. The 97% confidence interval for the proportion of student that take notesis: __ 0.470 v(50%) <p< 0.714 v (50 %) . Round answers to 3 decimal places. w Hide question 8 feedback Z-Critical Value =NORM.S.INV(.985) =2.17009 LL=.5921 - 2.17009* 5921 % .4079 76 UL =.5921 + 2.17009* 5921 *x .4079 76

Question 9 1/ 1 point In a random sample of 145 people, 112 said that they watched educational TV. Find the 93% confidence interval of the true proportion of people who watched educational TV. v ).7093 < P< 8355 ©0.4578 < 7 < 0.8924 07542 < P < 0.8356 ) 0.6101 < 7< 0.7399 w Hide question 9 feedback Z - Critical Value =ENORM.S.INV(.965) = 1.811911 LL=.7724 - *1.811911 7724 % 2276 145 UL=.7724 + *1.811911 7724 x . 2276 145 Question 10 1/ 1 point In a certain state, a survey of 434 workers showed that 26% belonged to a union. Find the 291% confidence interval of true proportion of workers who

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belong to a union. v (0.2243,0.2957) (03118, 0.3882) ~(0.1983,0.2798) (1578, .3511) 1 (0.24, 0.45) w Hide question 10 feedback Z - Critical Value =NORM.S.INV(.955) = 1.695398 LL = .26 -1.695398" 20 x .74 434 UL=.26 +1.695398" .20 x .74 434 Question 11 1/ 1 point A random sample of 145 people was selected and 13% of them were left handed. Find the 97% confidence interval for the proportion of left-handed people. (0.0764,0.1636) v (0.069,0.191)

(13, .87) (0.1125, .1576) " (0.0836, 0.1764) w Hide question 11 feedback Z-Critical Value = NORM.S.INV(.985) = 2.17009 LL=.13 - 2.17009* 13 % .87 145 UL=.13 +2.17009" 13 % .87 145 Question 12 1/ 1 point You are trying to estimate the average amount a family spends on food during a year. In the past, the standard deviation of the amount a family has spent on food during a year has been o = $1200. If you want to be 99% sure that you have estimated average family food expenditures within $60, how many families do you need to survey? Place your answer, a whole number, do not use any decimals, in the blank__ . For example, 1234 would be a legitimate entry.

Answer: 2655 v w Hide question 12 feedback Use Z-Critical Value = NORM.S.INV(.995) = 2.575 Z%8D)\° ME n — 2.575 % 1200 \ 2 60 N = You will need to round up. Question 13 1/ 1 point Assume 50 random samples of the same sample size are taken from a population, and a 90% confidence interval is constructed from each sample. How many of the intervals would you expect to contain the true population mean? Answer:___Round your answer to a whole number value as necessary, do not include any decimals. For example, 37 would be a legitimate entry. Answer: 45 Ww Hide question 13 feedback 90% of 50 is 45, so we would expect 45 of the intervals to contain the true population mean.

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Question 14 1/ 1 point Which confidence level would give the narrowest margin of error? () 90% () 99% ) 95% v( ) 80% Question 15 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 50 from the available frame. The sample mean and sample standard deviations were 46.4 and 7.2 hours, respectively. Construct a 97% confidence interval for the average 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: 44.1 (50 %) Answer for blank # 2: 48.7 (50 %) w Hide question 15 feedback

T-Critical Value = T.INV.2T(.03,49) = 2.235124 LL=46.4 - 2.235124* 7.2 v 80 UL = 46.4 +2.235124* 7.2 v 50 Question 16 1/ 1 point A lawyer researched the average number of years served by 45 different justices on the Supreme Court. The average number of years served was 13.8 years with a standard deviation of 7.3 years. What is the 95% confidence interval estimate for the average number of years served by all Supreme Court justices? Place your limits, rounded to 1 decimal place, in the blanks. Place you lower limit in the first blank.___ Place your upper limit in the second blank. ___ When entering your answer do not use any labels or symbols. Simply provide the numerical value. For example, 12.3 would be a legitimate entry. Answer for blank # 1: 11.6 (50 %) Answer for blank # 2: 16.0 (50 %) w Hide question 16 feedback T-Critical Value = T.INV.2T(.05,44) = 2.015368

LL=13.8 - 2.015368" UL =13.8 +2.015368" Question 17 1/ 1 point Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean. See Attached Excel for Data. germination time data.xlsx Assume that the population germination time is normally distributed. Find the 97% confidence interval for the mean germination time. v (13.065, 18.535) ) (13.063, 18.537) " (13.550, 21.050) ) (12.347,19.253) ) (14.396, 19.204) w Hide question 17 feedback

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T-Critical Value = T.INV.2T(.03,9) = 2.573804 LL=15.8 - 2.573804* 3.399894 v 10 UL = 15.8 + 2.573804* 3.399894 v 10 Question 18 1/ 1 point The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below. See Attached Excel for Data. ounces data.xlsx Assume that the amount of beverage in a randomly selected 16-ounce beverage can has a normal distribution. Compute a 97% confidence interval for the population mean amount of beverage in 16-ounce beverage cans and fill in the blanks appropriately. A 97% confidence interval for the population mean amount of beverage in 16- ounce beverage cans is ( 15.377 v(50%) , 16.223 v(50 %) ) ounces. (round to 3 decimal places) w Hide question 18 feedback T-Ciritical Value = T.INV.2T(.03,7) = 2.714573

440779 V'8 440779 V8 LL =15.8 —2.714573 UL =15.8 + 2.714573 Question 19 1/ 1 point A random sample of college football players had an average height of 67.5 inches. Based on this sample, (65.9, 69.0) 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 65.9 and 69.0 inches. > There is a 92% chance that the population mean height of college ~ football players is between 65.9 and 69.0 inches. > We are 92% confident that the population mean height of college ~ football palyers is 67.5 inches. > A 92% of college football players have height between 65.9 and 69.0 "~ inches. Question 20 1/ 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 95% confidence interval for the population mean height of college

basketball players based on this sample and fill in the blanks appropriately. ___63.760 v(50%) <mnu< __ 65928 V(50 %) (round to 3 decimal places) w Hide question 20 feedback T-Critical Value =T.INV.2T(.05,31) = 2.039513 3.006542 V32 3.006542 V32 LL = 64.84375 — 2.039513 * UL = 64.84375 + 2.039513 Done

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