Homework 5 Attempt 2

.docx

School

American Military University *

*We aren’t endorsed by this school

Course

302

Subject

Statistics

Date

Jan 9, 2024

Type

docx

Pages

14

Uploaded by DeanHippopotamus5891

Report
Question 1 0 / 1 point 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: ___22 ___ ( 61) Hide question 1 feedback Z-Critical Value =NORM.S.INV(.95) = 1.645 n = �� 2 ∗ � 2 �� 2 n = 3.32 ∗ 1.6452.72 Question 2 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 2 feedback
Z-Critical Value = NORM.S.INV(.95) = 1.645 n = �� 2 ∗ � 2 �� 2 n = 3.22 ∗ 1.6452.72 Question 3 1 / 1 point 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 of mercury. What sample size is needed? Round up to the nearest integer. Answer: ___ 82 ___ Hide question 3 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 n = �� 2 ∗ � 2 �� 2 n = .1382 ∗ 1.962.032 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 ___ Hide question 4 feedback Z-Critical Value = NORM.S.INV(.985) = 2.17009 n = �� 2 ∗ � 2 �� 2 n = 3.52 ∗ 2.170092.52 Question 5 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 ___ Hide question 5 feedback Z-Critical Value = NORM.SINV(.995) = 2.575 n = �� 2 ∗ � 2 �� 2 n = 32 ∗ 2.5752.52
Question 6 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 6 feedback Z-Critical Value = NORM.SINV(.95) = 1.645 n = �� 2 ∗ � 2 �� 2 n = 3.42 ∗ 1.6452.62 Question 7 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. 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. 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 _ __ (50 %) < p < ___ 0.506 _ __ (50 %) (round to 3 decimal places) Hide question 8 feedback Z-Critical Value = NORM.S.INV(.96) = 1.750686 LL = 0.42 - 1.750686 *.42 ∗ .58100 UL = 0.42 + 1.750686 *.42 ∗ .58100 Question 9 1 / 1 point
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
Recommended textbooks for you
Text book image
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
Text book image
Big Ideas Math A Bridge To Success Algebra 1: Stu...
Algebra
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:Houghton Mifflin Harcourt
SEE MORE TEXTBOOKS
Recommended textbooks for you
Text book image
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
Text book image
Big Ideas Math A Bridge To Success Algebra 1: Stu...
Algebra
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:Houghton Mifflin Harcourt
SEE MORE TEXTBOOKS