Homework 5 Attempt 1

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American Military University *

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302

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Statistics

Date

Jan 9, 2024

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docx

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Uploaded by DeanHippopotamus5891

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 number) ___ 11 ___ Hide question 1 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 n = � ∗ � ∗ � 2 �� 2 n = .5 ∗ .5 ∗ 1.962.32 Question 2 1 / 1 point 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 ___ 140 _ __ Hide question 2 feedback

Z-Critical Value = NORM.S.INV(.975) = 1.96 n = .0147 ∗ .9853 ∗ 1.962.022 Question 3 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 must be surveyed? Answer: (Round up your answer to nearest whole number) ___ 20 __ _ Hide question 3 feedback Z-Critical Value =NORM.S.INV(.96) = 1.750686 n = � ∗ � ∗ � 2 �� 2 n = .5 ∗ .5 ∗ 1.7506862.22 Question 4 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 5 0 / 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: Increases Decreases Remains the same Question 6 0 / 1 point 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: ___231 ___ ( 118) Hide question 6 feedback Z-Critical Value = NORM.S.INV(.985) = 2.17009

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n = �� 2 ∗ � 2 �� 2 n = 32 ∗ 2.170092.62 Question 7 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 7 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 n = �� 2 ∗ � 2 �� 2 n = 3.42 ∗ 1.962.52 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? 0.309 to 0.391 0.316 to 0.384 0.321 to 0.379

0.305 to 0.395 Hide question 8 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 LL = 0.35 - 1.96*.35 ∗ .65750 UL = 0.35 +1.96*.35 ∗ .65750 Question 9 1 / 1 point In a random sample of 80 people, 52 consider themselves as baseball fans. Compute a 95% confidence interval for the true proportion of people consider themselves as baseball fans and fill in the blanks appropriately. We are 95% confident that the true proportion of people consider themselves as baseball fans is between _ __ 0.545 _ __ (50 %) and ___ 0.755 ___ (50 %) . (round to 3 decimal places) Hide question 9 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96

LL = 0.65 - 1.96*.65 ∗ .3580 UL = 0.65 + 1.96*.65 ∗ .3580 Question 10 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. 0.6101< � ^<0.7399 0.6704< � ^<0.6796 0.325< � ^<0.675 –1.96< � ^<1.96 Hide question 10 feedback Z - Critical Value =NORM.S.INV(.975) = 1.96 LL = .675 - 1.96*.675 ∗ .325200 UL = .675 + 1.96*.675 ∗ .325200 Question 11 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

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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 11 feedback Z-Critical Value = NORM.S.INV(.995) = 2.575 LL = 0.4 - 2.575*.40 ∗ .60100 UL = 0.4 -+2.575*.40 ∗ .60100 Question 12 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 �^=0.0875)?

Place your answer, as a whole number, in the blank. For example 567 would be a legitimate entry. ___ ___ Answer: 767 Hide question 12 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 n = � ∗ � ∗ � 2 �� 2 n = .0875 ∗ .9125 ∗ 1.962.022 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. ___ Answer: 4000 Hide question 13 feedback Use

n = p*p*( �� )2 1000= .5*.5 *( � .05)2 4000 = ( � .05)2 4000= � .05 4000 ∗ .05= � 3.16227 = Z Now solve for n, using Z n = .5*.5 *(3.16227.025)2 Question 14 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. ___ Answer for blank # 1: 0.285 (50 %) Answer for blank # 2: 0.562 (50 %) Hide question 14 feedback

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Z-Critical Value =NORM.S.INV(.995) = 2.575 LL = .4235 - 2.575*.4235 ∗ .576585 UL = .4235 + 2.575*.4235 ∗ .576585 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.___ 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: 229.5 (231.5) Answer for blank # 2: 110.6 (384.9) Hide question 15 feedback T-Critical Value = T.INV.2T(.10,11) = 1.795885 LL = 308.1667 - 1.795885 *147.92812 UL = 308.1667 +1.795885 *147.92812 Question 16 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 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: 222.5 (50 %) Answer for blank # 2: 393.9 (50 %) Hide question 16 feedback T-Critical Value = T.INV.2T(.07,11) = 2.006663 LL = 308.1667 - 2.006663*147.92812 UL = 308.1667 +2.006663*147.92812 Question 17 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 (27.144, 33.356) (27.256, 33.244) (26.746, 33.754)

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(–3.106, 3.106) (26.025, 34.475) Hide question 17 feedback T-Critical Value = T.INV.2T(.01,11) = 3.105807 �� =30.25−3.105807 ∗ 4.7121712 �� =30.25+3.105807 ∗ 4.7121712 Question 18 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. 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. Question 19 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 93% confidence interval for the population mean height of college basketball players based on this sample and fill in the blanks appropriately. ___ 63.846 _ __ (50 %) < μ < ___ 65.841 ___ (50 %) (round to 3 decimal places) Hide question 19 feedback T-Critical Value =T.INV.2T(.07,31) = 1.876701 �� =64.84375−1.876701 ∗ 3.00654232 �� =64.84375+1.876701 ∗ 3.00654232 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. 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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Homework 5 Attempt 1

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