APPLIED STA all_chapters_test_bank_for_exam

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
having a sample of students at a particular school take the exam. The quality of the students at
the chosen school varies widely and the school administrators are allowed to choose who gets to
take the exam. The administrators have a strong incentive for the school to do well on the exam.
Do you think the results will represent the true ability of the students at school? What kind of
bias, if any, do you think will be present? Explain.


Answer: Results will be biased because of selection bias.
Explanation: The administrators will probably systematically choose the good students to take
the test. This is a systematic exclusion of the bad students from the sample and will lead to a
biased score. This is called selection bias.

Results will be biased as Its selection bias , the administrator are systematically choosing the good
students and they excluded the bad students ..


137) The campaign manager for a candidate for governor in Arizona wants to conduct a poll to
better understand his candidate’s chances for the upcoming election.
a.What is the population of interest?
b.Why may the poll be biased if a simple random sample of voters in the last gubernatorial
election (four years prior) is taken?
Answer:
a. Voters in the upcoming election
b. Selection bias
Explanation: The campaign manager wants to know how voters in the upcoming election view
his candidate. That is the population of interest. Voters from the last election may not vote in the
coming election. Voters who were too young to vote four years ago are not included using this
sampling method. Additionally, migration into and out of Arizona will change the voter over a
four-year period.

,138) It is known that college students at a local community college study 12 hours per week
with a standard deviation of five hours. What are the expected value and variance for a sample of
nine students?


Expected value is 12 hrs

variance = (standard deviation )2 / n

= ( 5)* = 2.7779



Answer: Expected value equals 12, and variance equals 2.7779.
Explanation: The expected value of is the same as the expected value of individual
observation, that is, E( ) = E(X) = μ. The variance of the sample mean is computed as


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




139) A fast-food restaurant uses an average of 110 grams of meat per burger patty. Suppose
the amount of meat in a burger patty is normally distributed with a standard deviation of 20
grams. What is the probability that the average amount of meat in four randomly selected burgers
is less than 105 grams?


Answer: 0.3085
Explanation: If is normal, we can transform it into a standard normal random variable as


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


Z= . Compute P( < 105). Use z table.
The appropriate Excel function is =NORM.DIST(105,110,20/SQRT(4),TRUE) = 0.3085

,140) Suppose residents in a well-to-do neighborhood pay an average overall tax rate of 25%
with a standard deviation of 8%. Assume tax rates are normally distributed. What is the
probability that the mean tax rate of 16 randomly selected residents is between 20% and 30%.


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

Answer: 0.9876
Explanation: If is normal, we can transform it into a standard normal random variable as


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


Z= . Compute P(20% ≤ ≤ 30%).
Note that P(z1 ≤ Z ≤ z2) = P(Z ≤ z2) – P(Z ≤ z1). Use z table.

The appropriate Excel function is =NORM.DIST(0.3,0.25,0.08/SQRT(16),TRUE)-
NORM.DIST(0.2,0.25,0.08/SQRT(16),TRUE) = 0.9876



141) Suppose the average casino patron in Las Vegas loses $110 per day, with a standard
deviation of $700. Assume winnings/losses are normally distributed.
a.What is the probability that a random group of nine people averages more than $500 in
winnings on their one-day trip to Las Vegas?
b.What is the probability that a random group of nine people averages more than $500 in
losses on their one-day trip to Las Vegas?


Answer:
a. 0.0045
b. 0.0473
Explanation:
a. The standard deviation of is calculated as the positive square root of the variance. We call it


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


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

, corresponding value z on Z given by Z = .
Compute P( > 500). Note that P(Z > z) = 1 – P(Z ≤ z). Use z table.

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

b. Compute P( < −500). Use z table. The appropriate Excel function is =NORM.DIST(-500,-
110,700/SQRT(9),TRUE) = 0.0473




142) A ski resort gets an average of 2000 customers per weekday with a standard deviation of
800 customers. Assume the underlying distribution is normal. What is the probability a ski resort
averages between 1500 customers and 3000 customers per weekday over the course of four
weekdays?

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



Answer: 0.8881
Explanation: If is normal, we can transform it into a standard normal random variable as


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


Z= . Compute P(1,500 ≤ ≤ 3,000).
Note that P(z1 ≤ Z ≤ z2) = P(Z ≤ z2) – P(Z ≤ z1). Use z table.