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Transcript
Alex Weller
348
Problems 16-20
Ten students were randomly selected from a high school to take part in a program
designed to raise their reading comprehension. Each student took a test before and after
completing the program. The mean of the differences between the score after the program
and the score before the program is 16. It was decided that all students in the school
would take part in this program during the next school year. Let  A denote the mean
score after the program and B denote the mean score before the program for all students
in the school. The 95 percent confidence interval estimate of the true mean difference for
all students is (9, 23). Which of the following statements is a correct interpretation of this
confidence interval?
A)
B)
C)
D)
E)
 A > B with probability 0.95.
 A < B with probability 0.95.
 A is around 23 and B is around 9.
For any  A and B with (  A - B )  14, the sample result is quite likely.
For any  A and B with 9 < (  A - B ) < 23, the sample result is quite likely.
E. A confidence interval gives two numbers that are meant to encompass the mean. So if
a confidence interval was done on the differences, then any future means obtained by
calculating and averaging the differences between the before and after scores should fall
within the interval.
Gina’s doctor told her that the standardized score (z-score) for her systolic blood
pressure, as compared to the blood pressure of other women her age, is 1.50. Which of
the following is the best interpretation of this standardized score?
A) Gina’s systolic blood pressure is 150.
B) Gina’s systolic blood pressure is 1.50 standard deviations above the average
systolic blood pressure of women her age.
C) Gina’s systolic blood pressure is 1.50 above the average systolic blood pressure of
women her age.
D) Gina’s systolic blood pressure is 1.50 times the average systolic blood pressure of
women her age.
E) Only 1.5% of women Gina’s age have a higher systolic blood pressure than she
does.
B. The z-score for a number is the number of standard deviations away from the mean
that number is; thus if Gina’s z-score is 1.50, then B is the correct answer.
The physician’s Health Study, a large medical experiment involving 22,000 male
physicians, attempted to determine whether aspirin could help prevent heart attacks. In
this study, one group of about 11,000 physicians took an aspirin every other day, while a
control group took a placebo. After several years, it was determined that the physicians in
the group that took aspirin had significantly fewer heart attacks than the physicians in the
control group. Which of the following statements explains why it would not be
appropriate to say that everyone should take an aspirin every other day?
I.
II.
III.
A)
B)
C)
D)
E)
The study included only physicians, and different results may occur in
individuals in other occupations.
The study included only males and there may be different results for
females.
Although taking aspirin may be helpful in preventing heart attacks, it may
be harmful to some other aspects of health.
I only
II only
III only
II and III only
I, II, and III
E. I is true because physicians probably do take better care of their health than other
people, so that could be a big factor in their frequency of heart attacks. II is true because
the study was only done on males, and females could have completely different statistics
for their frequency of heart attacks. III is true because the study did not check to see if
there were any adverse effects of the aspirin, they looked only at heart attacks. They
should have looked to see if there were any other common ailments among the test
subjects.
Questions 19-20 refer to the following information.
Every Thursday, Matt and Dave’s Video Venture has “roll-the-dice” day. A
customer may choose to roll two fair dice and rent a second movie for an amount (in
cents) equal to the numbers uppermost on the dice, with the larger number first. For
example, if the customer rolls a two and a four, a second movie may be rented for $0.42.
If a two and a two are rolled, a second movie may be rented for $0.22. Let X represent the
amount paid for a second movie on roll=the-dice day. The expected value of X is $0.47
and the standard deviation of X is $0.15.
If a customer rolls the dice and rents a second movie every Thursday for 20
consecutive weeks, what is the total amount that the customer would expect to pay for
these second movies?
A)
B)
C)
D)
E)
$0.45
$0.47
$0.67
$3.00
$9.40
E. Multiply the expected amount ($0.47) times the number of times the customer rolled
that amount (20). So .47*20 = 9.4
If a customer rolls the dice and rents a second movie every Thursday for 30
consecutive weeks, what is the approximate probability that the total amount paid for
these second movies will exceed $15.00?
A)
B)
C)
D)
E)
0
0.09
0.14
0.86
0.91
C. Multiply 30 by .47, which equals 14.1. Then take the square root of 30 times the
variance, or the squared standard deviation. The whole thing should look like this:
30  0.152  0.8216 To get the final probability, plug the numbers into a Normal Cdf
in a TI-89 as follows. The lower value is 15, the amount you are looking for; the upper
value is infinite. The mean is 14.1 and the standard deviation is 0.8216. Press enter twice
and the Cdf pops up, displaying 0.1367. Rounding it off, you get .14, which is an answer
choice. The screens below show the test on the calculator.