MATH302 Wk5 Knowledge Test

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School

American Military University *

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302

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Management

Date

Jan 9, 2024

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pdf

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Uploaded by DrQuetzalMaster563

Week 5 Knowledge Check Homework Practice Questions - Results Attempt 1 of 4 Written Jan 6, 2024 13:09 - Jan 6, 2024 15:30 Attempt Score 19 / 20 - 95 % Overall Grade (Highest Attempt) 19 / 20 - 95 % Question 1 1 / 1 point ___ 118___ Hide ques±on 1 feedback Question 2 1 / 1 point ___ 239___ Hide ques±on 2 feedback The population standard deviation for the height of college basketball players is 3 inches. If we want to estimate 97% 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.S.INV(.985) = 2.17009 n = n = The population standard deviation for the height of college basketball players is 3 inches. If we want to estimate a 99% 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:

Question 3 1 / 1 point Select the correct answer for the blank: If everything else stays the same, the required sample size ____ as the confidence level decreases to reach the same margin of error. Answer: Question 4 1 / 1 point ___ 11___ Hide ques±on 4 feedback Z-Critical Value = NORM.SINV(.995) = 2.575 n = n = Increases Decreases Remains the same 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) Z-Critical Value = NORM.S.INV(.975) = 1.96 n = n =

Question 5 1 / 1 point ___ 178___ Hide ques±on 5 feedback Question 6 1 / 1 point ___ 20___ Hide ques±on 6 feedback 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: Z-Critical Value = NORM.S.INV(.975) = 1.96 n = n = 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 =

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Question 7 1 / 1 point ___ 61___ 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: Hide ques±on 8 feedback Question 9 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: Z-Critical Value =NORM.S.INV(.95) = 1.645 n = n = 0.60 0.06 30 0.94 30/500

A recent study of 750 Internet users in Europe found that 35% of Internet users were women. What is the 95% confidence interval estimate for the true proportion of women in Europe who use the Internet? Hide ques±on 9 feedback Question 10 1 / 1 point ___ 0.471___ (50 %) ___ 0.749___ (50 %) Hide ques±on 10 feedback 0.321 to 0.379 0.316 to 0.384 0.309 to 0.391 0.305 to 0.395 Z-Critical Value = NORM.S.INV(.975) = 1.96 LL = 0.35 - 1.96* UL = 0.35 +1.96* 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: < p < . Round answers to 3 decimal places. Z-Critical Value =NORM.S.INV(.995) =2.575 LL = .609756 - 2.575* UL = .609756 + 2.575*

Question 11 1 / 1 point In a random sample of 200 people, 135 said that they watched educational TV. Find the 95% confidence interval of the true proportion of people who watched educational TV. Hide ques±on 11 feedback Question 12 0 / 1 point Z - Critical Value =NORM.S.INV(.975) = 1.96 LL = .675 - 1.96* UL = .675 + 1.96*

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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: 0.322 (46.8) Answer for blank # 2: 0.478 (50.2) Hide ques±on 12 feedback Question 13 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. ___ T-Critical Value = T.INV.2T(.12,50) = 1.581805 LL = 48.5 - 1.581805 * UL = 48.5 +1.581805 *

Answer: 4000 Hide ques±on 13 feedback 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). Use n = p*p* 1000= .5*.5 * 4000 = 3.16227 = Z Now solve for n, using Z n = .5*.5 *

See Attached Excel for Data. dental expense data.xlsx Construct a 90% 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: 231.5 (50 %) Answer for blank # 2: 384.9 (50 %) Hide ques±on 14 feedback Question 15 1 / 1 point Suppose you compute a confidence interval with a sample size of 25. What will happen to the confidence interval if the sample size increases to 50? Hide ques±on 15 feedback T-Critical Value = T.INV.2T(.10,11) = 1.795885 LL = 308.1667 - 1.795885 * UL = 308.1667 +1.795885 * Get larger Nothing Get smaller

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Question 16 1 / 1 point If a sample has 25 observations and a 99% confidence estimate for is needed, the appropriate value of the t-multiple required is___. Place your answer, rounded to 3 decimal places, in the blank. For example, 3.456 would be an appropriate entry. ___ Answer: 2.797 Hide ques±on 16 feedback Question 17 1 / 1 point In a certain town, a random sample of executives have the following personal incomes (in thousands); Assume the population of incomes is normally distributed. Find the 95% confidence interval for the mean income. See Attached Excel for Data. income data.xlsx Hide ques±on 17 feedback As the Sample Size increases, the Margin of Error decreases. Making the interval smaller. In Excel, =T.INV.2T(0.01,24) 32.180 < μ < 55.543 35.132 < μ < 50.868 40.840 < μ < 45.160 39.419 < μ < 46.581 35.862 < μ < 50.138 T- Critical Value = T.INV.2T(.05,13) = 2.160369 LL = 43 - 2.160369*

Question 18 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 Hide ques±on 18 feedback Question 19 1 / 1 point UL = 43 + 2.160369* (17.884, 40.806) (27.512, 31.178) (13.582, 45.108) (16.572, 42.118) (–1.833, 1.833) T-Critical Value = T.INV.2T(.10,9) = 1.833113 LL = 29.345 - 1.833113* UL = 29.345 + 1.833113*

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. Question 20 1 / 1 point ___ 63.846___ (50 %) ___ 65.841___ (50 %) Hide ques±on 20 feedback Done 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. A randomly selected sample of college basketball players has the following heights in inches. See Attached Excel for Data. height data.xlsx Compute a 93% confidence interval for the population mean height of college basketball players based on this sample and fill in the blanks appropriately. < μ < (round to 3 decimal places) T-Critical Value =T.INV.2T(.07,31) = 1.876701

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MATH302 Wk5 Knowledge Test

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