WK 5 KC

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American Military University *

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302

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Statistics

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Jun 12, 2024

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Week 5 Knowledge Check Homework Practice Questions Your work has been saved and submitted Written Jun 4, 2024 12:21 AM - Jun 4, 2024 2:25 AMAttempt 1 of 4 Attempt Score 16.5 / 20 - 82.5% Overall Grade (Highest Attempt) 16.5 / 20 - 82.5 % Question 1 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 humber) 11 v

W Hide question 1 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 n = Question 2 1 /1 point Select the correct answer for the blank: If everything else stays the same, the required sample size as the confidence level increases to reach the same margin of error. Answer: v ) Increases ) Decreases ~ ) Remains the same 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. O 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 =

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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 v W Hide question 5 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 n =

Question 6 1 /1 point 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: 239 v W Hide question 6 feedback Z-Critical Value = NORM.SINV(.995) = 2.575 n = Question 7 1 /1 point There is no prior information about the proportion of Americans who support free trade in 2018. If we want to estimate a 97.5% confidence interval for the true proportion of Americans who support free trade in 2018 with a 0.16 margin of error, how many randomly selected

Americans must be surveyed? Answer: (Round up your answer to nearest whole humber) 50 v W Hide question 7 feedback Z-Critical Value = NORM.S.INV(.9875) = 2.241403 n = Question 8 1 /1 point Suppose a marketing company wants to determine the current proportion of customers who click on ads on their smartphones. It was estimated that 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. ___0.334 __ «v+(50%) <p < 0.506 v (50 %) (round to 3 decimal places)

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W Hide question 8 feedback Z-Critical Value = NORM.S.INV(.96) = 1.750686 LL = 0.42 -1.750686 * UL =0.42 + 1.750686 * Question 9 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 O 94% of customers click on ads on their smartphones. O 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. O We are 94% confident that the true proportion of customers who click on ads on their smartphones is - 0.59.

v 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 10 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? ~0.305 t0 0.395 - 0.321 t0 0.379 v ) 0.316t0 0.384 ~0.309 to 0.391 W Hide question 10 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 LL = 0.35 - 1.96* UL = 0.35 +1.96*

Question 11 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 11 feedback Z-Critical Value =NORM.S.INV(.995) =2.575 LL = .609756 - 2.575% UL = .609756 + 2.575* Question 12 0.5 / 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

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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. Answer for blank # 1: 0.286 3 (0.285, .285) Answer for blank # 2: 0.562 (50 %) W Hide question 12 feedback Z-Critical Value =NORM.S.INV(.995) = 2.575 LL = .4235 - 2.575*% UL = .4235 + 2.575% Question 13 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 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: 266.7 % (231.5) Answer for blank # 2: 349.7 i (384.9) W Hide question 13 feedback T-Critical Value = T.INV.2T(.10,11) = 1.795885 LL = 308.1667 - 1.795885 *

UL = 308.1667 +1.795885 * Question 14 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. Answer: 4000 W Hide question 14 feedback

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1000= .5*.5* 4000 = 3.16227 = Z Now solve for n, using Z n=.5*5*% 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 507? () Get larger . ) Nothing v ) Get smaller W Hide question 15 feedback As the Sample Size increases, the Margin of Error decreases. Making the interval smaller. Question 16 0/ 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: 2.172 3 (2.797) W Hide question 16 feedback

In Excel, =T.INV.2T(0.01,24) Question 17 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 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%) <u< 66.302 v (50 %) (round to 3 decimal places) W Hide question 17 feedback T-Critical Value =T.INV.2T(.01,31) = 2.744042 Question 18 1 / 1 point

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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) W Hide question 18 feedback T-Critical Value = T.INV.2T(.10,9) = 1.833113 LL = 29.345 - 1.833113* UL = 29.345 + 1.833113%

Question 19 0/ 1 point It is known that the number of hours a student sleeps per night has a normal distribution. The sleeping time in hours of a random sample of 8 students is given below. See Attached Excel for Data. sleep hours data.xlsx Compute a 92% confidence interval for the true mean time a student sleeps per night and fill in the blanks appropriately. We have 92 % confidence that the true mean time a student sleeps per night is between 6.067 % (5.847) and ___7.508 % (7.728) hours. (round to 3 decimal places) W Hide question 19 feedback T-Critical Value = T.INV.2T(.08,7) = 2.046011 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. 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. O There is a 92% chance that the population mean height of college football players is between 63.2 and ~ 65.9 inches. Done

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WK 5 KC

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