AP Statistics 2017

(Marvins-Underground-K-12) #1
The analysis    will    include a   significance    test    with    H   0   :   Heartaid    and the current medication  are

equally effective at preventing heart disease and H (^) A : Heartaid is more effective than the current
medication at preventing heart disease. Which of these would be a potential consequence of a Type
II error?
A. Patients will spend more money on Heartaid, even though it is actually not any more effective
than the current medication.
B. Doctors will begin to prescribe Heartaid to patients, even though it is actually not any more
effective than the current medication.
C. Patients will continue to use the current medication, even though Heartaid is actually more
effective.
D. Researchers will calculate the wrong P -value, making their advice to doctors invalid.
E. Researchers will calculate the correct P- value, but will misinterpret it in their conclusion.



  1. Researchers in the Southwest are studying tortoises—a species of animal that is affected by habitat
    loss due to human development of the desert. A total of 78 tortoises are being studied by
    researchers at several sites. The data on gender and species of all these tortoises is organized in the
    table shown below.


If  a   tortoise    from    this    study   is  to  be  selected    at  random, let A   =   the tortoise    is  female and  B   =   the
tortoise is a Morafka . Which of the following appropriately interprets the value of P(B|A)?
A. 31.3% is the probability that a randomly selected Morafka tortoise is a female tortoise.
B. 31.3% is the probability that a randomly selected female tortoise is a Morafka tortoise.
C. 19.2% is the probability that a randomly selected Morafka tortoise is a female tortoise.
D. 19.2% is the probability that a randomly selected female tortoise is a Morafka tortoise.
E. 31.3% is the probability that a randomly selected tortoise is a Morafka tortoise.



  1.          The main    purpose of  blocking    in  an  experiment  is  to:



A.          reduce  bias.
B. reduce confounding.
C. reduce variation within treatments.
D. reduce variation between treatments.
E. reduce the probability of a Type I error.



  1.          The midterm scores  for a   statistics  course  were    approximately   normally    distributed with    a   mean    of

    52 points and a standard deviation of 4 points. The final exam scores were approximately normally
    distributed with a mean of 112 points and a standard deviation of 10 points. One student had a score
    of 58 points on the midterm. If she had the same standardized score (z -score) on the final exam,
    what must her score have been?
    A. 15 points



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