Optimizing Confidence Intervals in Statistics: A Study Guide

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___ Hide question 1 feedback Z-Critical Value = NORM.SINV(.995) = 2.575 n = n = Question 2 0 / 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 must be surveyed? Answer: (Round up your answer to nearest whole number) ___19___Incorrect Response( 20) Hide question 2 feedback Z-Critical Value =NORM.S.INV(.96) = 1.750686 n = n = Question 3 1 / 1 point

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: ___ 87___ Hide question 3 feedback Z-Critical Value = NORM.SINV(.95) = 1.645 n = n = Question 4 0 / 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 number) ___46___Incorrect Response( 50) Hide question 4 feedback Z-Critical Value = NORM.S.INV(.9875) = 2.241403 n = n = Question 5 1 / 1 point

The population standard deviation for the height of college baseball players is 3.2 inches. If we want to estimate 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: ___ 57___ Hide question 5 feedback Z-Critical Value = NORM.S.INV(.95) = 1.645 n = n = Question 6 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___ Hide question 6 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 n = n = Question 7 1 / 1 point

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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) ___11___ Hide question 7 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 n = n = Question 8 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 ( ___0.274___(50 %)___0.526___(50 %) ) (round to 3 decimal places) Hide question 8 feedback Z-Critical Value = NORM.S.INV(.995) = 2.575 LL = 0.4 - 2.575* UL = 0.4 -+2.575* Question 9 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? 0.309 to 0.391 0.305 to 0.395 0.321 to 0.379 0.316 to 0.384 Hide question 9 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 LL = 0.35 - 1.96* UL = 0.35 +1.96* Question 10 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

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. 94% of customers click on ads on their smartphones. We are 94% confident that the true proportion of customers who click on ads on their smartphones is between 0.56 and 0.62. We are 94% confident that the true proportion of customers who click on ads on their smartphones is 0.59. Question 11 0.5 / 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___(50 %) < p < ___0.507___Incorrect Response( 0.506, .506) (round to 3 decimal places) Hide question 11 feedback Z-Critical Value = NORM.S.INV(.96) = 1.750686 LL = 0.42 - 1.750686 * UL = 0.42 + 1.750686 *

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Question 12 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 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: 46.9 Incorrect Response( 46.8) Answer for blank # 2: 50.1 Incorrect Response( 50.2) Hide question 12 feedback T-Critical Value = T.INV.2T(.12,50) = 1.581805 LL = 48.5 - 1.581805 * UL = 48.5 +1.581805 *

Question 13 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: increases decreases remains the same Question 14 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 93% 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, 98.4 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: 201.8 Incorrect Response( 222.5) Answer for blank # 2: 414.5 Incorrect Response( 393.9) Hide question 14 feedback T-Critical Value = T.INV.2T(.07,11) = 2.006663 LL = 308.1667 - 2.006663* UL = 308.1667 +2.006663* Question 15 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.___

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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: 110.6 Incorrect Response( 231.5) Answer for blank # 2: 229.5 Incorrect Response( 384.9) Hide question 15 feedback T-Critical Value = T.INV.2T(.10,11) = 1.795885 LL = 308.1667 - 1.795885 * UL = 308.1667 +1.795885 * Question 16 1 / 1 point A sample of 9 production managers with over 15 years of experience has an average salary of $71,000 and a sample standard deviation of $18,000. Assuming that the salaries of production managers with over 15 years experience are normally distributed, you can be 95% confident that the mean salary for all production managers with at least 15 years of experience is between what two numbers. Place your LOWER limit, rounded to a whole number, in the first blank. Do not use a dollar sign, a comma, or any other stray mark. For example, 54321 would be a legitimate entry.___. Place your UPPER limit, rounded to a whole number, in the second blank. Do not use a dollar sign, a comma, or any other stray mark. For example, 65432 would be a legitimate entry.___

___ Answer for blank # 1: 57164 (50 %) Answer for blank # 2: 84836 (50 %) Hide question 16 feedback T-Critical Value = T.INV.2T(.05,8) = 2.306004 LL = 71000 - 2.306004* UL = 71000 +2.306004* Question 17 0 / 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 99% confidence interval for the mean germination time. (–3.250, 3.250) Incorrect Response (13.063, 18.537)

(12.550, 19.050) Correct Answer (12.347, 19.253) (13.396, 18.204) Hide question 17 feedback T-Critical Value = T.INV.2T(.01,9) = 3.249836 LL = 15.8 - 3.249836* UL = 15.8 + 3.249836* 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

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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 Hide question 18 feedback T- Critical Value = T.INV.2T(.05,13) = 2.160369 LL = 43 - 2.160369* UL = 43 + 2.160369* Question 19 1 / 1 point A researcher finds a 95% confidence interval for the average commute time in minutes using public transit is (15.75, 28.25). What is the correct interpretation of this interval?

There is a 95% chance commute time in minutes using public transit is between 15.75 and 28.25 minutes. We are 95% confident that all commute time in minutes for the population using public transit is between 15.75 and 28.25 minutes. We are 95% confident that the interval between 15.75 and 28.25 minutes contains the population mean commute time in minutes using public transportation. We are 95% confident that the interval between 15.75 and 28.25 minutes contains the sample mean commute time in minutes using public transportation. Question 20 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 (17.884, 40.806) (27.512, 31.178) (13.582, 45.108)

(16.572, 42.118) (–1.833, 1.833) Hide question 20 feedback T-Critical Value = T.INV.2T(.10,9) = 1.833113 LL = 29.345 - 1.833113* UL = 29.345 + 1.833113*

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MATH303 Week 5 QUIZ 4 Attempt

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