Estimating Sample Sizes for Confidence Intervals in Research

Question 1 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 ___ Hide question 1 feedback Z-Critical Value = NORM.S.INV(.985) = 2.17009 n = .0147 ∗ .9853 ∗ 2.170092.0152 Question 2 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 ___ Hide question 2 feedback

Z-Critical Value = NORM.SINV(.995) = 2.575 n =3.42 ∗ 2.5752.432 Question 3 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 ___ Hide question 3 feedback Z-Critical Value = NORM.SINV(.995) = 2.575 n = 2.92 ∗ 2.5752.452 Question 4 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 __ _ Hide question 4 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 n =2.72 ∗ 1.962.652 Question 5 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 ___ Hide question 5 feedback Z-Critical Value =NORMS.INV(.995) = 2.575 n = 3.12 ∗ 2.5752.582 Question 6 1 / 1 point There is no prior information about the proportion of Americans who support gun control in 2019. If we want to estimate 93% confidence interval for the true proportion of Americans who support gun control in 2019 with a 0.18 margin of error, how many

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randomly selected Americans must be surveyed? Answer: (Round up your answer to nearest whole number, do not include any decimals) ___ 26 ___ Hide question 6 feedback Z-Critical Value = NORM.S.INV(.965) = 1.811911 n =.5 ∗ .5 ∗ 1.8119112.182 Question 7 1 / 1 point There is no prior information about the proportion of Americans who support gun control in 2019. If we want to estimate 97% confidence interval for the true proportion of Americans who support gun control in 2019 with a 0.27 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) ___ 17 ___ Hide question 7 feedback Z-Critical Value = NORM.S.INV(.985) = 2.17009 n =.5 ∗ .5 ∗ 2.170092.272 Question 8 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. (0.069, 0.191) (0.1125, .1576) (.13, .87) (0.0764, 0.1636) (0.0836, 0.1764) Hide question 8 feedback Z-Critical Value = NORM.S.INV(.985) = 2.17009 LL = .13 - 2.17009*.13 ∗ .87145 UL = .13 +2.17009*.13 ∗ .87145 Question 9 0.5 / 1 point A random sample found that 30% of 150 Americans were satisfied with the gun control laws in 2018. Compute a 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 Fill in the blanks appropriately. A 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 is ( ___ 0.213 ___ ( 50 %) , ___0.398 ___ ( 0.387, .387 )

) (round to 3 decimal places) Hide question 9 feedback Z-Critical Value = NORM.S.INV(.99) = 2.326348 LL = 0.3 - 2.326348*.30 ∗ .70150 UL = 0.3 +2.326348*.30 ∗ .70150 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 = 76 and found that 45 of them took notes. The 97% confidence interval for the proportion of student that take notes is: ___ 0.470 ___ ( 50 %) < p < ___ 0.714 ___ (50 %) . Round answers to 3 decimal places. Hide question 10 feedback Z-Critical Value =NORM.S.INV(.985) =2.17009

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LL = .5921 - 2.17009*.5921 ∗ .407976 UL = .5921 + 2.17009*.5921 ∗ .407976 Question 11 1 / 1 point From a sample of 550 items, 52 were found to be defective. The point estimate of the population proportion defective will be: 0.10 0.0855 52 0.0945 Hide question 11 feedback 52/550 Question 12 1 / 1 point The percent defective for parts produced by a manufacturing process is targeted at 4%. The process is monitored daily by taking samples of sizes n = 160 units. Suppose that today's sample contains 14 defectives. Determine a 88% confidence interval for the proportion defective for the process today. Place your LOWER limit, rounded to 3 decimal places, in the first blank. For example, 0.123 would be a legitimate answer.___ Place your UPPER limit, rounded to 3 decimal places, in the second blank. For example, 0.345 would be a legitimate entry.___

Make sure you include the 0 before the decimal. ___ Answer for blank # 1: 0.053 (50 %) Answer for blank # 2: 0.122 (50 %) Hide question 12 feedback Z-Critical Value = NORM.S.INV(.94) = 1.554774 LL = .0875 - 1.554774 *.0875 ∗ .9125160 UL = .0875 +1.554774 *.0875 ∗ .9125160 Question 13 0 / 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: 43.6 ( 44.1) Answer for blank # 2: 47.4 ( 48.7) Hide question 13 feedback T-Critical Value = T.INV.2T(.03,49) = 2.235124 LL = 46.4 - 2.235124*7.250 UL = 46.4 +2.235124*7.250 Question 14 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 σ = $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 Hide question 14 feedback Use Z-Critical Value = NORM.S.INV(.995) = 2.575

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� =( � ∗ �� )2 � =(2.575 ∗ 120060)2 You will need to round up. Question 15 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 97% confidence interval estimate for the average family dental expenses for all employees of this corporation. Place your LOWER limit, in dollars rounded to 2 decimal places, in the first blank. Do not use a dollar sign, a comma, or any other stray mark. For example, 123.45 would be a legitimate entry.___ Place your UPPER limit, in dollars rounded to 2 decimal places, in the second blank. Do not use a dollar sign, a comma, or any other stray mark. For example, 567.89 would be a legitimate entry.___ ___

Answer for blank # 1: 201.81 (50 %) Answer for blank # 2: 414.53 (50 %) Hide question 15 feedback T-Critical Value = T.INV.2T(.03, 11) = 2.490664 LL = 308.1667 - 2.490664*147.92812 UL = 308.1667 +2.490664*147.92812 Question 16 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 45.5 and 7.5 hours, respectively. Construct a 92% confidence interval for the average 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: 43.6 (50 %) Answer for blank # 2: 47.4 (50 %) Hide question 16 feedback T-Critical Value = T.INV.2T(.08,49) = 1.787758 LL = 45.5 - 1.787758*7.550 UL = 45.5 + 1.787758*7.550 Question 17 1 / 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 97% confidence interval using the sample data. See Attached Excel for Data miles data.xlsx (27.144, 33.356) (26.862,33.638) (28.746, 35.754) (26, 34) (26.025, 34.475) Hide question 17 feedback

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T-Critical Value = T.INV.2T(.03,11) = 2.490664 �� =30.25−2.490664 ∗ 4.7121712 �� =30.25+2.490664 ∗ 4.7121712 Question 18 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 ___ (50 %) < μ < ___ 65.928 ___ ( 50 %) (round to 3 decimal places) Hide question 18 feedback T-Critical Value =T.INV.2T(.05,31) = 2.039513

�� =64.84375−2.039513 ∗ 3.00654232 �� =64.84375+2.039513 ∗ 3.00654232 Question 19 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. (13.065, 18.535) (13.063, 18.537) (13.550, 21.050) (12.347, 19.253) (14.396, 19.204) Hide question 19 feedback T-Critical Value = T.INV.2T(.03,9) = 2.573804 LL = 15.8 - 2.573804*3.35989410 UL = 15.8 + 2.573804*3.35989410 Question 20 1 / 1 point

A random sample of college football players had an average height of 66.5 inches. Based on this sample, (64.2, 67.1) found to be a 95% confidence interval for the population mean height of college football players. Select the correct answer to interpret this interval. We are 95% confident that the population mean height of college football players is between 64.2 and 67.1 inches. We are 95% confident that the population mean height of college football palyers is 66.5 inches. A 95% of college football players have height between 64.2 and 67.1 inches. There is a 95% chance that the population mean height of college football players is between 64.2 and 67.1 inches.

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Quiz 5 Attempt 1

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