2014 AP Stats Exam
Solutions
17 + 7 24
=
= 0.72
33
33
25 +12 37
=
= 0.55224
67
67
Residential Status appears to make a difference in
participation for this sample of students. On campus
students are more likely to participate in one or more
activities (72%) than off campus students (55%).
Because the percentages in the stacked bar charts are
not the same, we can see that there is an association
between residential status and level of participation
in activities. On campus students are slightly more
likely to participate in 2 or more activities (21% vs.
18%) and off campus students are more likely to not
participate in an activity (45% vs. 27%).
Since p=0.23 is greater than any reasonable
significance level (.01, .05, 0.10), the administrator
should fail to reject the null hypothesis. This means
that there is not significant evidence that residential
status and level of participation in activities at the
university have an association.
3 2 1 1
P(X = 3) = × × =
= 0.0119
9 8 7 84
Yes, since the probability is so small, it gives us reason
to doubt the manager’s claim. Only about 1% of the
time would we select all 3 women from this group if it
truly was a random sample. The probability of this
happening is not very likely!
No, this does not correctly simulate the selection. The
dice roles are independent. They simulate a situation
where 3 out of the 9 people are chosen WITH
replacement, so that the probability of picking a
woman remains constant at one-third. However, we
need to simulate choosing 3 WITHOUT replacement,
which would change the probabilities after each
selection as shown in part a. We should use a
different method, such as drawing names out of a hat.
We could do this without replacement and that would
correctly simulate the process.
P(X > 140)
Looking up -1.90 on the
140 -120
normal table, I get 0.0287.
z=
= 1.90
10.5
The probability of them losing
some state funding is 0.0287.
School A would be less likely to lose funding because it would be
less likely to show a mean greater than 140. This is because the
sampling distribution for samples of size 3 will have less variability
than for samples of size 1. It would be less likely to find an
AVERAGE of 3 be over 140 than one single observation be over 140.
P(X > 140)
z=
PROOF:
140 - 120
= 3.30 Looking up -3.30, we get a probability of
10.5
3
0.0003 which is definitely less likely than our answer in part (a).
3
8
æ 2ö
P(Monday or Friday) = ç ÷ =
= 0.064
è 5ø
125
Since income data is likely to be skewed to the right because of
many low incomes and only a few high outliers, the mean is likely to
be much higher than the median and most of the data. When data is
skewed, the median would be a more accurate representation of the
majority of the data.
Method 2 is better because Method 1 is biased. Method 1 is
a voluntary response sample. If only 600 (less than 10%)
answer as expected, those who answer may be different
from the others in some way. For instance, alumni might be
more likely to respond if their income is higher (because
they’re proud of it), which would cause an overestimation of
the average income. Method 2 avoids this nonresponse bias
by following up to make sure all randomly selected people
answer the question. A smaller unbiased random sample
will give better results than a larger biased sample. Method
1 will overestimate, however, Method 2 should have an
estimate close to the true average yearly income of the class
of 1988.
This is a matched pairs t test!!
ud = the true mean difference in price (woman – man)
H0 : ud = 0
Ha : u d > 0
Assumptions: Stated that they randomly selected 8
car models
Pop car models > 10 * 8
Normality: The plot of differences (given) is not
skewed (fairly symmetric), so normality can be
assumed.
Matched pairs t test:
t=
585 - 0
= 3.118
530.71
8
With 7 degrees of freedom, our p-value is 0.0084.
Let the significance level be 5%. Since .0084 < .05,
I would reject the null hypothesis and say that there is
significant evidence that in this county women pay
more than men on average for the same car!!
y = -1.5958 + 0.03726(175) = 4.9247
Residual = y - y = 5.88 - 4.9247 = 0.9553
Interpretation: The actual fuel consumption rate of
Car A is 0.9553 gallons per mile higher than the
predicted fuel consummation rate based on our LSRL.
Car B has a very small, positive residual, which means
that the predicted fuel consumption rate was very
close to the actual fuel consumption rate. (The actual
was only SLIGHTLY higher).
There is a moderate positive linear association
between the residuals of FCT vs. Length and the
Engine Size, shown on Graph II. Graph III shows
almost no linear association between Wheel Base and
the Residuals of FCR vs. Length. The points are
randomly scattered. In comparison, the residuals of
FCT has a stronger linear association with engine size
than with wheel base.
Jamal should use engine size to improve his
predictions, because it has a stronger association with
the fuel consumption rate. The lack of association on
Graph III tells us that using wheel base to predict FCR
will not give us better predictions than what we
already had with length alone. However, the positive
association on Graph II tells us that some of the
variability that has not yet been explained by length
could be explained by engine size.