AP Statistics 2017

(Marvins-Underground-K-12) #1
I   .       Let μ = the true    mean    score   for all students    taking  the exam.

H (^) 0: μ = 520 (or, H 0 : μ ≤ 520)
H (^) A : μ > 520
II . We will use a one-sample t- test. We consider the 175 students to be a random sample of
students taking the exam. The sample size is large, so the conditions for inference are
present. (Note : Due to the large sample size, it is reasonable that the sample standard
deviation is a good estimate of the population standard deviation. This means that you
would receive credit on the AP exam for doing this problem as a z -test although a t -test is
preferable.)
III . df = 175 – 1 = 174 0.05 < P -
value < 0.10 (from Table B, with df = 100—always round down).
(Using the TI-83/84, P -value = tcdf(1.52,100,174)=0.065 . If we had used s = 96 as an
estimate of s based on a large sample size, and used a z -test rather than a t -test, the P -value
would have been 0.064. This is very close to the P -value determined using t . Remember
that this is a t -test, but that the numerical outcome using a z -test is almost identical for large
samples.)
IV . The P- value, which is greater than 0.05, is reasonably low but not low enough to provide
strong evidence that the current class is superior in math ability as measured by the SAT.


5.

I   .       Let μ = the true    average number  of  pages   in  the books   Booker  reads.
H 0 : μ ≤ 375.

H (^) A : μ > 375.
II . We are going to use a one-sample t -test test at α = 0.05. The problem states that the sample
is a random sample. A dotplot of the data shows good symmetry and no significant
departures from normality (although the data do spread out quite a bit from the mean, there
are no outliers):
The conditions for the t -test are present.
III .
(from Table B). (tcdf(1.48,100,12 ) gives P -value = 0.082, or you can use STAT TESTS

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