week 5 homework questions

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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 bradymcdede

Attempt Score 19.5 / 20 - 97.5 % Overall Grade (Highest Attempt) 19.5 / 20 - 97.5 % stion 1 1 / 1 p 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 ___ uestion 1 feedback cal Value = NORM.S.INV(.975) = 1.96 47 ∗ .9853 ∗ 1.962.022 on 2 1 / 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 ___ uestion 2 feedback cal Value =NORM.S.INV(.96) = 1.750686 ∗ � ∗ � 2 �� 2

.5 ∗ 1.7506862.22 on 3 1 / 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 4 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 ___ uestion 4 feedback cal Value = NORM.S.INV(.975) = 1.96 ∗ � ∗ � 2 �� 2 .5 ∗ 1.962.32 on 5 1 / 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: ___118 ___ uestion 5 feedback cal Value = NORM.S.INV(.985) = 2.17009 � 2 ∗ � 2 �� 2 ∗ 2.170092.62 on 6 1 / The population standard deviation for the height of college football players is 3.3 inches. If we want to estimate a 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: ___61 ___ uestion 6 feedback cal Value =NORM.S.INV(.95) = 1.645 � 2 ∗ � 2 �� 2 2 ∗ 1.6452.72 on 7 1 / 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:

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___87 ___ uestion 7 feedback cal Value = NORM.SINV(.95) = 1.645 � 2 ∗ � 2 �� 2 2 ∗ 1.6452.62 on 8 1 / 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 8 feedback Z - Critical Value =NORM.S.INV(.975) = 1.96 LL = .675 - 1.96*.675 ∗ .325200 UL = .675 + 1.96*.675 ∗ .325200 n 9 1 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 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 0.59. We are 94% confident that the true proportion of customers who click on ads on their smartphones is between 0.56 and 0.62. 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. Question 10 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: 0.06 0.60 30 0.94 Hide question 10 feedback 30/500 n 11 1 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 < ___0.506 ___ (50 %) (round to 3 decimal places) uestion 11 feedback

cal Value = NORM.S.INV(.96) = 1.750686 .42 - 1.750686 *.42 ∗ .58100 .42 + 1.750686 *.42 ∗ .58100 on 12 1 / 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 12 feedback T-Critical Value = T.INV.2T(.05,8) = 2.306004 LL = 71000 - 2.306004*180009 UL = 71000 +2.306004*180009 n 13 1

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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 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 question 14 feedback T-Critical Value =T.INV.2T(.05,39) = 2.022691 LL = 51.3 - 2.022691*5.840 UL = 51.3 + 2.022691*5.840 n 15 1

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 15 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 n = � ∗ � ∗ � 2 �� 2 n = .0875 ∗ .9125 ∗ 1.962.022 n 16 0.5 You are told that a random sample of 150 people from Iowa has been given cholesterol tests, and 60 of these people had levels over the "safe" count of 200. Construct a 95% confidence interval for the population proportion of people in Iowa with cholesterol levels over 200. Place your LOWER limit, rounded to 3 decimal places, in the first blank___. For example, 0.678 would be a legitimate entry. Place your UPPER limit, rounded to 3 decimal places, in the second blank ___. For example, 0.789 would be a legitimate entry. Make sure you include the 0 before the decimal. ___

Answer for blank # 1: 0.332 (0.322, .322) Answer for blank # 2: 0.478 (50 %) Hide question 16 feedback Z-Critical Value =NORM.S.INV(.975) = 1.96 LL = .4 - 1.96*.4 ∗ .6150 UL = .4 +1.96*.4 ∗ .6150 n 17 1 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) uestion 17 feedback cal Value =T.INV.2T(.07,31) = 1.876701 64.84375−1.876701 ∗ 3.00654232 64.84375+1.876701 ∗ 3.00654232 on 18 1 / In a certain town, a random sample of executives have the following personal incomes (in thousands); Assume the

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population of incomes is normally distributed. Find the 95% confidence interval for the mean income. See Attached Excel for Data. income data.xlsx 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*13.626914 UL = 43 + 2.160369*13.626914 n 19 1 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 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. Done

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