MATH302 week 5 knowledge check

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American Military University *

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Course

302

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Mechanical Engineering

Date

Feb 20, 2024

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pdf

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18

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Week 5 Knowledge Check Homework Practice Quest… 1 2 Your work has been saved and submitted Written Jan 6, 2024 10:15 PM - Jan 6, 2024 11:29 PM Attempt 1 of 4 Attempt Score 20 / 20 - 100 % Overall Grade (Highest Attempt) 20 / 20 - 100 % Question 1 1 / 1 point 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
___ 20___ Hide ques ! on 1 feedback Question 2 1 / 1 point ___ 231___ Hide ques ! on 2 feedback must be surveyed? Answer: (Round up your answer to nearest whole number) Z-Critical Value =NORM.S.INV(.96) = 1.750686 n = n = p q Z 2 ME 2 .5 .5 1.750686 2 .2 2 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: Z-Critical Value = NORM.S.INV(.985) = 2.17009 n =
Question 3 1 / 1 point ___ 239___ Hide ques ! on 3 feedback n = SD 2 Z 2 ME 2 3.5 2 2.17009 2 .5 2 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: Z-Critical Value = NORM.SINV(.995) = 2.575 n = n = SD 2 Z 2 ME 2
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Question 4 1 / 1 point ___ 11___ Hide ques ! on 4 feedback Question 5 1 / 1 point 3 2 2.575 2 .5 2 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 = p q Z 2 ME 2 .5 .5 1.96 2 .3 2
___ 178___ Hide ques ! on 5 feedback Question 6 1 / 1 point ___ 8___ 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 = SD 2 Z 2 ME 2 3.4 2 1.96 2 .5 2 There is no prior information about the proportion of Americans who support gun control in 2018. If we want to estimate 95% confidence interval for the true proportion of Americans who support gun control in 2018 with a 0.36 margin of error, how many randomly selected Americans must be surveyed? Answer: (Round up your answer to nearest whole number)
Question 7 1 / 1 point ___ 140___ Hide ques ! on 7 feedback Z-Critical Value =NORM.S.INV(.975) = 1.96 n = n = p q Z 2 ME 2 .5 .5 1.96 2 .36 2 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 95% 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 2%? Round up to the nearest integer. Answer Z-Critical Value = NORM.S.INV(.975) = 1.96 n =
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Question 8 1 / 1 point 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 8 feedback .0147 .9853 1.96 2 .02 2 0.321 to 0.379 0.305 to 0.395 0.316 to 0.384 0.309 to 0.391 Z-Critical Value = NORM.S.INV(.975) = 1.96 LL = 0.35 - 1.96* UL = 0.35 +1.96* .35 .65 750 .35 .65 750
Question 9 1 / 1 point ___ 0.274___ (50 %) ___ 0.526___ (50 %) Hide ques ! on 9 feedback Question 10 1 / 1 point A random sample found that forty percent of 100 Americans were satisfied with the gun control laws in 2017. Compute a 99% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2017. Fill in the blanks appropriately. A 99% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2017 is ( ) (round to 3 decimal places) Z-Critical Value = NORM.S.INV(.995) = 2.575 LL = 0.4 - 2.575* UL = 0.4 -+2.575* .40 .60 100 .40 .60 100 Suppose a marketing company wants to determine the current proportion of customers who click on ads on their smartphones. It was estimated that
___ 0.334___ (50 %) ___ 0.506___ (50 %) Hide ques ! on 10 feedback Question 11 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. the current proportion of customers who click on ads on their smartphones is 0.42 based on a random sample of 100 customers. Compute a 92% confidence interval for the true proportion of customers who click on ads on their smartphones and fill in the blanks appropriately. < p < (round to 3 decimal places) Z-Critical Value = NORM.S.INV(.96) = 1.750686 LL = 0.42 - 1.750686 * UL = 0.42 + 1.750686 * .42 .58 100 .42 .58 100 (–1.645, 1.645) (0.0764, 0.1636)
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Hide ques ! on 11 feedback Question 12 1 / 1 point A survey of 85 families showed that 36 owned at least one DVD player. Find the 99% confidence interval estimate of the true proportion of families who own at least one DVD player. Place your limits, rounded to 3 decimal places , in the blanks. Place the lower limit in the first blank___ and the upper limit in the second blank___ When entering your answer do not use any labels or symbols other than the decimal point. Simply provide the numerical values. For example, 0.123 would be a legitimate entry. Make sure you put a 0 before the decimal. (.12, .88) (0.0436, 0.1164) (0.068, 0.172) Z-Critical Value = NORM.S.INV(.95) = 1.645 LL = .12 - 1.645* UL = .12 + 1.645* .12 .88 150 .12 .88 150
___ Answer for blank # 1: 0.285 (50 %) Answer for blank # 2: 0.562 (50 %) Hide ques ! on 12 feedback Question 13 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. Z-Critical Value =NORM.S.INV(.995) = 2.575 LL = .4235 - 2.575* UL = .4235 + 2.575* .4235 .5765 85 .4235 .5765 85 μ
___ Answer: 2.797 Hide ques ! on 13 feedback Question 14 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 95% 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 %) Hide ques ! on 14 feedback In Excel, =T.INV.2T(0.01,24) T-Critical Value =T.INV.2T(.05,39) = 2.022691 LL = 51.3 - 2.022691*
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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 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.___ ___ UL = 51.3 + 2.022691* 5.8 40 5.8 40
Answer for blank # 1: 231.5 (50 %) Answer for blank # 2: 384.9 (50 %) Hide ques ! on 15 feedback Question 16 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. How many units would have to be sampled to be 95% confident that you can estimate the fraction of defective parts within 2% (using the information from today's sample--that is using the result that T-Critical Value = T.INV.2T(.10,11) = 1.795885 LL = 308.1667 - 1.795885 * UL = 308.1667 +1.795885 * 147.928 12 147.928 12
)? Place your answer, as a whole number, in the blank. For example 567 would be a legitimate entry. ___ ___ Answer: 767 Hide ques ! on 16 feedback Question 17 1 / 1 point ^ p = 0.0875 Z-Critical Value = NORM.S.INV(.975) = 1.96 n = n = p q Z 2 ME 2 .0875 .9125 1.96 2 .02 2 A randomly selected sample of college basketball players has the following
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___ 63.846___ (50 %) ___ 65.841___ (50 %) Hide ques ! on 17 feedback 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 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 LL = 64.84375 1.876701 3.006542 32 UL = 64.84375 + 1.876701 3.006542 32
Hide ques ! on 18 feedback Question 19 1 / 1 point A random sample of college football players had an average height of 66.35 inches. Based on this sample, (65.6, 67.1) found to be a 90% confidence interval for the population mean height of college football players. Select the correct answer to interpret this interval. (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* 22.03462 10 22.03462 10
Question 20 1 / 1 point 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. Done We are 90% confident that the population mean height of college football players is between 65.6 and 67.1 inches. There is a 90% chance that the population mean height of college football players is between 65.6 and 67.1 inches. We are 90% confident that the population mean height of college football palyers is 66.35 inches. A 90% of college football players have height between 65.6 and 67.1 inches. 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.
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