Updated Comprehensive Statistics All Chapters Test Bank Complete Exam Questions and Answers with Step-by-Step Solutions, Detailed Explanations, Data Analysis Methods, Probability Concepts, Hypothesis Testing, Inferential Statistics Applications, and Struc

Chapter 7 :
SHORT ANSWER. Write the word or phrase that best completes each statement or
answers the question.
136) The California Department of Education wants to gauge the difficulty of a new exam by
F F F F F F F F F F F F F F F



having a sample of students at a particular school take the exam. The quality of the students at the
F F F F F F F F F F F F F F F F F F F



chosen school varies widely and the school administrators are allowed to choose who gets to take
F F F F F F F F F F F F F F F F



the exam. The administrators have a strong incentive for the school to do well on the exam. Do
F F F F F F F F F F F F F F F F F F



you think the results will represent the true ability of the students at school? What kind of bias, if
F F F F F F F F F F F F F F F F F F F



any, do you think will be present? Explain.
F F F F F F F




Answer: Results will be biased because of selection bias.
FF F F F F F F F



Explanation: The administrators will probably systematically choose the good students to take
FF F F F F F F F F F F F



the test. This is a systematic exclusion of the bad students from the sample and will lead to a biased
F F F F F F F F F F F F F F F F F F F F



score. This is called selection bias.
F F F F F




Results will be biased as Its selection bias , the administrator are systematically choosing the good
F F F F F F F F F F F F F F F F



students and they excluded the bad students ..
F FF F F F F F F




137) The campaign manager for a candidate for governor in Arizona wants to conduct a poll to
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better understand his candidate’s chances for the upcoming election.
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a.What is the population of interest?
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b.Why may the poll be biased if a simple random sample of voters in the last gubernatorial
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election (four years prior) is taken?
F F F F F




Answer: FF



a. Voters in the upcoming election
F F F F F F



b. Selection bias
F F



Explanation: The campaign manager wants to know how voters in the upcoming election view
FF F F F F F F F F F F F F F



his candidate. That is the population of interest. Voters from the last election may not vote in the
F F F F F F F F F F F F F F F F F F



coming election. Voters who were too young to vote four years ago are not included using this
F F F F F F F F F F F F F F F F F



sampling method. Additionally, migration into and out of Arizona will change the voter over a
F F F F F F F F F F F F F F F



four-year period. F

,138) It is known that college students at a local community college study 12 hours per week
F F F F F F F F F F F F F F F F



with a standard deviation of five hours. What are the expected value and variance for a sample of
F F F F F F F F F F F F F F F F F F



nine students? F




Expected value is 12 hrs F F F F




F variance = (standard deviation )2 / n
F FF F F F F F




F F FF F FF FF FF F FF FF FF = ( 5)* = 2.7779
F F F F F F F



F




Answer: Expected value equals 12, and variance equals 2.7779.
FF F F F F F F F




Explanation: The expected value of is the same as the expected value of individual
FF F F F F F F F F F F F F F F




observation, that is, E( ) = E(X) = μ. The variance of the sample mean is computed as
F F F F F F F F F F F F F F F F F




Var( ) = F F FF =5^2/9 = 2.7779. F F




139) A fast-food restaurant uses an average of 110 grams of meat per burger patty. Suppose the
F F F F F F F F F F F F F F F F



amount of meat in a burger patty is normally distributed with a standard deviation of 20 grams.
F F F F F F F F F F F F F F F F F



What is the probability that the average amount of meat in four randomly selected burgers is less
F F F F F F F F F F F F F F F F F



than 105 grams? F F




Answer: 0.3085 FF




Explanation: If is normal, we can transform it into a standard normal random variable as
FF F F F F F F F F F F F F F F F




Z= F F , and any value of on
F F F F F F F F has a corresponding value z on Z given by
F F F F F F F F F




Z= F F . Compute P( < 105). Use z table.
F F F F F F F




The appropriate Excel function is =NORM.DIST(105,110,20/SQRT(4),TRUE) = 0.3085
F F F F F F F

,140) Suppose residents in a well-to-do neighborhood pay an average overall tax rate of 25%
F F F F F F F F F F F F F F



with a standard deviation of 8%. Assume tax rates are normally distributed. What is the
F F F F F F F F F F F F F F F



probability that the mean tax rate of 16 randomly selected residents is between 20% and 30%.
F F F F F F F F F F F F F F F




=NORM.DIST(30%,25%,8%/SQRT(16),TRUE)-NORM.DIST(20%,25%,8%/SQRT(16),TRUE)

Answer: 0.9876 FF




Explanation: If is normal, we can transform it into a standard normal random variable as
FF F F F F F F F F F F F F F F F




Z= F F , and any value of on
F F F F F F F F has a corresponding value z on Z given by
F F F F F F F F F




Z= F F . Compute P(20% ≤
F F F F F ≤ 30%). F F




Note that P(z1 ≤ Z ≤ z2) = P(Z ≤ z2) – P(Z ≤ z1). Use z table.
F F F F F F F F F F F F F F F F F




The appropriate Excel function is =NORM.DIST(0.3,0.25,0.08/SQRT(16),TRUE)-
F F F F F



NORM.DIST(0.2,0.25,0.08/SQRT(16),TRUE) = 0.9876 F F




141) Suppose the average casino patron in Las Vegas loses $110 per day, with a standard
F F F F F F F F F F F F F F F



deviation of $700. Assume winnings/losses are normally distributed.
F F F F F F F



a.What is the probability that a random group of nine people averages more than $500 in
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winnings on their one-day trip to Las Vegas?
F F F F F F F



b.What is the probability that a random group of nine people averages more than $500 in
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losses on their one-day trip to Las Vegas?
F F F F F F F

FFF




Answer: FF



a. 0.0045
F F



b. 0.0473 F



Explanation: FF




a. The standard deviation of
F F F F F F is calculated as the positive square root of the variance. We call it
F F F F F F F F F F F F F




the standard error of the sample mean, and it is computed as se( ) =
F F F F F F F F F F F F F F . If F F F is normal, we can
F F F F




transform it into a standard normal random variable as Z =
F F F F F F F F F F F , and any value of on
F F F F F F F F has a F F




corresponding value z on Z given by Z = F F F F F F F F F . F




Compute P( > 500). Note that P(Z > z) = 1 – P(Z ≤ z). Use z table.
F F F F F F F F F F F F F F F F F F




The appropriate Excel function is =1-NORM.DIST(500,-110,700/SQRT(9),TRUE) = 0.0045
F F F F F F F F

, b. Compute P( < −500). Use z table. The appropriate Excel function is =NORM.DIST(-500,-
F F FF F F F F F F F F F F




110,700/SQRT(9),TRUE) = 0.0473 F F




142) A ski resort gets an average of 2000 customers per weekday with a standard deviation of
F F F F F F F F F F F F F F F F



800 customers. Assume the underlying distribution is normal. What is the probability a ski resort
F F F F F F F F F F F F F F F



averages between 1500 customers and 3000 customers per weekday over the course of four
F F F F F F F F F F F F F F



weekdays?

=NORM.DIST(3000,2000,800/SQRT(4),TRUE)-NORM.DIST(1500,2000,800/SQRT(4),TRUE) = 0.8881 FF F




Answer: 0.8881 FF




Explanation: If is normal, we can transform it into a standard normal random variable as
FF F F F F F F F F F F F F F F F




Z= F F , and any value of on
F F F F F F F F has a corresponding value z on Z given by
F F F F F F F F F




Z= F F . Compute P(1,500 ≤
F F F F F ≤ 3,000).
F F




Note that P(z1 ≤ Z ≤ z2) = P(Z ≤ z2) – P(Z ≤ z1). Use z table.
F F F F F F F F F F F F F F F F F




143) A mining company made some changes to their mining process in an attempt to save fuel.
F F F F F F F F F F F F F F F F



Before the changes were made, it took an average of 20 gallons of diesel fuel to mine 1,000
F F F F F F F F F F F F F F F F F F



pounds of copper. Suppose the standard deviation of fuel used per 1,000 pounds of copper mined
F F F F F F F F F F F F F F F F



is 6 gallons. After the changes were made, the company only used an average of 18 gallons of
F F F F F F F F F F F F F F F F F F



diesel for the next 30,000 pounds of copper mined.
F F F F F F F F



a.How unusual would it be to get a sample average of 18 gallons or less for 30,000 pounds of
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copper mined if the changes to the mining process had no effect?
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b.Do you think the changes in the mining process actually lowered the fuel used? Explain.
FFFF F F F F F F F F F F F F F F

FFF




a) F F FF FF F F




=NORM.DIST(18,20,6/SQRT(30),TRUE) = 0.033945 F F FF F




b. Yes, it's very unlikely to average 18 gallons or less for 30,000 pounds of copper mined
F F F F F F F F F F F F F F F F F