Week 5 Homework: Confidence Intervals & Sample Size Calculations

Week 5 Knowledge Check Homework Practice Questions - Re… Attempt 1 of 4 Written Apr 3, 2024 2:31 PM - Apr 3, 2024 4:33 PM Attempt Score 14 / 20 - 70 % Overall Grade (Highest Attempt) 15.5 / 20 - 77.5 % Question 1 1 / 1 point ___ 231___ Hide ques±on 1 feedback 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 = n =

Question 2 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. Question 3 1 / 1 point ___ 20___ Hide ques±on 3 feedback 94% of college basketball players have height between 65.6 and 67.1 inches. There is a 94% chance 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 between 65.6 and 67.1 inches. We are 94% confident that the population mean height of college basketball players is 66.35 inches. 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 =

Question 4 0 / 1 point ___ 74___ (8) Hide ques±on 4 feedback Question 5 1 / 1 point 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) Z-Critical Value =NORM.S.INV(.975) = 1.96 n = n = 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 0.03 ppm of mercury with 95% confidence. Assume the population standard deviation is 0.138 ppm

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

___ 82___ Hide ques±on 5 feedback Question 6 0 / 1 point ___ 49___ (50) Hide ques±on 6 feedback of mercury. What sample size is needed? Round up to the nearest integer. Answer: Z-Critical Value = NORM.S.INV(.975) = 1.96 n = n = 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 number) Z-Critical Value = NORM.S.INV(.9875) = 2.241403 n =

Question 7 1 / 1 point ___ 87___ 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: n = The population standard deviation for the height of college hockey players is 3.4 inches. If we want to estimate 90% 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.SINV(.95) = 1.645 n = n =

Hide ques±on 8 feedback Question 9 1 / 1 point ___ .274___ (50 %) ___ .526___ (50 %) Hide ques±on 9 feedback 0.94 30 0.60 0.06 30/500 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*

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

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? Hide ques±on 10 feedback Question 11 1 / 1 point 0.305 to 0.395 0.316 to 0.384 0.309 to 0.391 0.321 to 0.379 Z-Critical Value = NORM.S.INV(.975) = 1.96 LL = 0.35 - 1.96* UL = 0.35 +1.96*

___ .334___ (50 %) ___ .506___ (50 %) Hide ques±on 11 feedback Question 12 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 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. < 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 *

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: 231.2 (231.5) Answer for blank # 2: 385.1 (384.9) Hide ques±on 12 feedback Question 13 1 / 1 point T-Critical Value = T.INV.2T(.10,11) = 1.795885 LL = 308.1667 - 1.795885 * UL = 308.1667 +1.795885 *

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

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 13 feedback Question 14 0 / 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 )? Place your answer, as a whole number, in the blank. For example 567 would be a legitimate entry. ___ ___ Answer: 768 (767) Get larger Nothing Get smaller As the Sample Size increases, the Margin of Error decreases. Making the interval smaller.

Hide ques±on 14 feedback Question 15 0 / 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: 49.47 (49.45) Answer for blank # 2: 53.13 (53.15) Z-Critical Value = NORM.S.INV(.975) = 1.96 n = n =

Hide ques±on 15 feedback 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: 2.492 (2.797) Hide ques±on 16 feedback T-Critical Value =T.INV.2T(.05,39) = 2.022691 LL = 51.3 - 2.022691* UL = 51.3 + 2.022691* In Excel, =T.INV.2T(0.01,24)

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Question 17 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 17 feedback (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*

Question 18 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 18 feedback 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* UL = 43 + 2.160369*

Question 19 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 99% confidence interval using the sample data. See Attached Excel for Data miles data.xlsx Hide ques±on 19 feedback Question 20 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 (27.144, 33.356) (27.256, 33.244) (26.746, 33.754) (–3.106, 3.106) (26.025, 34.475) T-Critical Value = T.INV.2T(.01,11) = 3.105807

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

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.

WK 5(1)

Related Documents