March, 1999
To members of AP Statistics ListServe:
In the interest of sharing...This is a copy of a document I will share with my AP Statistics students. It may or may not be use
or interest to others. Comments, suggestions, criticisms etc., are welcome.
Sanderson M. Smith
==============================================================
SOME THOUGHTS RELATING TO THE
ADVANCED PLACEMENT STATISTICS EXAMINATION

Relax, and think. Remember that everyone else taking the exam is in a situation identical to yours. Realize that the
problems will probably look detailed compared to problems you have encountered in other math courses.

Make sure your calculator is functioning properly. Insert new batteries a day or so before the exam, and make sure all systems
are "go." Bring a backup calculator, if possible.

Read problems carefully. Bring colored pencil to highlight key words and phrases as you read the questions.

Don't confuse median and mean. They are both measures of center, but, for a given data set, they may differ by a considerable
amount.
* mean > median <===> distribution skewed right
* mean < median <===> distribution skewed left

Don't confuse coefficient of correlation and slope of leastsquares regression line.
* A slope close to 1 or 1 doesn't mean strong correlation.
* An r value close to 1 or 1 doesn't mean slope of least
squares regression line is close to 1 or 1.
* Relation between b (slope of regression line) and r (coefficient
of correlation) is b = r(Sy/Sx). This is on formula sheet
provided for the exam.
* Remember that r^2 > 0 doesn't mean r > 0. For instance,
if r^2 = 0.81, then r = 0.9 or r = 0.9.

Remember that the leastsquares regression line contains the point
(mean x, mean y), where mean x is the mean of the xvalues, and mean y is the mean of the yvalues.

A coefficient of correlation near 0 doesn't necessarily mean there are no meaningful relationships to be observed between the
two data sets. Consider x: 2 3 4 5 6 7 8 9 10 11 12
y: 6 30 8 50 10 70 12 90 14 110 16
In this case, r = .38, but a scatterplot displays something quite interesting. Moral of story: Whenever possible, look at the
"shape" of the data.

Be careful with the concept of simple random sample (SRS) For instance, if each individual in a group has an equal
probability of being chosen in a sample, it doesn't follow that the sample is an SRS. Consider a class of 6 boys and 6 girls.
I want to randomly pick a committee of two students from this group. I decide to flip a coin. If "heads," I will choose two
girls by a random process. If "tails," I will choose two boys by a random process. Now, each student has an equal probability
(1/6) of being chosen for the committee. However, the chosen two students do not represent an SRS of size two picked from
members of the class, for the selection process does not allow for a committee consisting of one boy and one girl. To have an
SRS of size two from this class of 6 boys and 6 girls, each committee of two students would have to have an equal probability
of being chosen.

Look at graphs and displays carefully. For graphs, note carefully what is represented on the axes, and be aware of number
scale. Some questions the provide tables of numbers and graphs relating to the numbers can be answered simply by "reading" the
graphs.

Don't confuse standard deviation and variance. Remember that standard deviation units are the same as the data units, while
variance is measured in "square units."

Be aware of numerical statistical changes when transformations are made on a data set, W.
* Adding the same number to each number in W simply shifts the data.
This doesn't change standard deviation and variance.
* Multiplying all numbers in W by a constant does change standard
deviation and variance. For instance, if all members of W are
multiplied by 4, then the new set has a standard deviation that
is 4 times larger than the standard deviation of W, and a variance
that is 16 times the variance of W.
Simple examples:
Mean StDev. Variance Range
Consider data set S:
1,2,3,4,5 3 1.4142 2 4
Add 7 to each element of S:
8,9,10,11,12 10 1.4142 2 4
Multiply elements of S by 4:
4,8,12,16,20 12 5.6569 32 16
Multiply elements of S by 4,
then add 7:
11,15,19,23,27 19 5.6569 32 16

Be aware of, but be careful with statements (a) and (b) since they represent simplified versions of sophisticated concepts.
(a) When combining two sets by addition,
means add;
standard deviations do not add;
variances add.
(b) When combining two sets by subtraction,
means subtract;
standard deviations do not subtract;
variances add.
Simple examples:
Let S = {5, 9} and T = {1,3}.
Then set S+T = {5+1,5+3,9+1,9+3} = {6,8,10,12}, and
set ST = {51,53,91,93} = {2,4,6,8}
S T S+T ST
Mean 7 2 9 5
St. Dev. 2 1 2.2361 2.2361
Variance 4 1 5 5
Note that: mean(S+T) = mean(S) + mean(T)
mean(ST) = mean(S)  mean(T)
variance(S+T) = variance(ST) = variance(S) + variance(T)

Recognize a binomial distribution situation when it arises. Thinking in terms of slots, if you have a set number of slots,
and the probability of getting a "success" in each slot is constant, then you have a binomial setting. Consider, for instance,
rolling a die ten times. There are ten slots to be filled, and the probability of filling each slot is 1/6.
__ __ __ __ __ __ __ __ __ __
Using the TI83, the probability of getting exactly three 6's is
(10C3)*(1/6)^3*(5/6)^7
= binompdf(10,1/6,3) = 0.155045, or about 15.5%.
The probability of getting less than four 6's is
binomcdf(10,1/6,3) = 0.93027, or about 93%.
The probability of getting four or more 6's in 10 rolls of a single die is about 7%.
If x is the number of 6's obtained when ten dice are rolled, then
mean(x) = 10(1/6) = 1.6667, and
st.dev(x) = sqrt[10(1/6)(5/6)] = 1.1785
Another example:
Assume a large population is 32% Hispanic. If a random sample of 15 people is chosen, this can be represented by a binomial
model with 15 slots. The probability of "success" for each slot is 0.32.
__ __ __ __ __ __ __ __ __ __ __ __ __ __ __
The probability that this sample would contain at least 5 Hispanics is
1  binomcdf(15,.32,4) = 1  0.4477 = 0.5523, or about 55%.
If x represents the number of Hispanics in a random sample of size 15, then
mean(x) = 15(.32) = 4.8, and
st.dev(x) = sqrt[15(.32)(.68)] = 1.8067

Binomial distribution > normal distribution as number of trials increases. If N is the number of trials in a binomial
setting, and if p represents the probability of "success" in each trial, then a general rule of thumb states that a normal
distribution can be used to approximate the binomial distribution if Np ? 5 and N(1p) ? 5.

Recognize a discrete random variable situation when it arises (and don't confuse it with a binomial situation.)
Let x = the number of heads obtained when five coins are tossed.
xvalue probability
0 1/32 = .03125
1 5/32 = .15625
2 10/32 = .3125
3 10/32 = .3125
4 5/32 = .15625
5 1/32 = .03125
mean(x) = 0(.03125) + 1(.15625) +2 (.3125)
+ 3(.3125) +4 (.15625) + 5(.03125) = 2.5
var(x) = .03125(02.5)^2 + .15625(12.5)^2 + .3125(22.5)^2
+.3125(32.5)^2 + .15625(42.5)^2 + .03125(52.5)^2 = 1.25
st.dev(x) = sqrt[var(x)] = sqrt(1.25) = 1.118

Simpson's Paradox:
This usually involves percentages.
Example:
Win Total % Win Win Total % Win
First Period A: 80 100 80% B: 78 100 78%
Second Period A: 20 40 50% B: 2 5 40%
Totals A: 100 140 71.4% B: 80 105 76.2%
In this example, A's winning percentage exceeds B's for both of two periods, but B has a better overall winning percentage.

Realize that logarithmic transformations can be practical and useful. Among other things, taking logs cuts down the magnitude
of numbers. Also, if {(x,y)} has an exponential pattern, then {(x,log y)} has a linear pattern.
Example:
x y log y
1 24 1.3802
2 192 2.2833
3 1,536 3.1864
4 12,188 4.0859
7 6,290,000 6.7987
8 49,900,000 7.6981
An exponential fit to (x,y) on the TI83 yields y = 3(8^x),
with r = .9999. If we attempt to extrapolate and predict a value for y when x = 9, we get y = 3(8^9) = 402,653,184.
A linear fit to (x,log y) on the TI83 yields log y = .9027286x + 0.477395, with r = .9999. If x = 9, then log y =
.9027286(9)+0.477395 = 8.6019524. Hence.
y = 10^8.6019524 = 399,900,917.

Type I error: Rejecting a null hypothesis when it is true.
Type II error: Accepting a null hypothesis when it is false.
Power of a test: Probability of correctly rejecting a null hypothesis.
Simple example:
Population #1: A A A A A A A A B B
Population #2: A B B B B B B B B B
Without knowing which of the populations is represented, an element is randomly chosen. After viewing the element, the chooser
must guess the population from which it came.
Null hypothesis (Ho): The element came from population #1.
Alternate hypothesis (Ha): The element came from population #2.
Test decision: Accept Ho if the element is A;
otherwise reject Ho and accept Ha.
Here is a probability chart:
Ho True Ho False
Accept Ho 80% 10% < probability of Type II error
Reject Ho 20% 90% < power of the test

probability of Type I error

In hypothesis testing, the level of significance is the probability of making a Type I error.

Thoughts on multiple choice statistics questions.
* Relate to the question. What topic is being referenced?
* Read carefully. Bring colored pencil to highlight key words and
phrases. After deciding on an answer choice, glance at
the highlighted words and phrases to make sure you haven't made a
careless mistake or an incorrect assumption.
* Realize scoring is (Number Right)  (1/4)(Number Wrong). Careless
mistakes hurt.
* You don't have to answer all of the questions to get a good overall
score.
* If an answer is "obvious," think about it. If it's so obvious to
you, it's probably obvious to others... and the chances are good
that it is not the correct response. For example, suppose one set
of test scores has a mean of 80, and another set of scores on the
same test has a mean of 90. If the two sets are combined, what is
the mean of the combined scores. The "obvious" answer is 85 (and
will certainly appear among the answer choices), but you, as an
intelligent statistics student, realize that 85 is not necessarily
the correct response.
* If a question and/or answer choice set appears to be detailed and
you need to do a lot of reading to reach a conclusion, most of the
answer choices will probably be obviously incorrect. Don't be
frightened off by questions and/or answer set choices that seem to
be wordy. Just read carefully, and use the highlighting technique
previously mentioned.
* If you can eliminate one or more of the answer choices, you should
respond, even if you have to guess from the remaining choices.

Thoughts on free response questions.
* Read carefully, sentence by sentence, and use colored pencil to
highlight key words or phrases.
* Relate to the problem. Decide what statistical concept/idea is
involved. This will allow you to make an intelligent approach to
questions asked. If you get started on an intelligent path, you
will probably get some points even if you make some mistakes along
the way.
* Be neat, Make it clear to the reader what you are attempting to do.
However, don't write too much. Overkill can waste valuable time.
* Questions may well look very detailed. You may be given much more
information than you actually need. This is likely to be true if you
are shown a computer printout. Don't get flustered by the way
a problem "looks" when you first glance at it. The 1997 AP Exam
provides good examples of problems that look scary, but which are
really quite reasonable if you remain levelheaded.
* Some questions may give you considerable leeway in choosing an
approach to a solution. Consider your options carefully and take
the one that requires the least amount of time.
* Don't be calculatorinefficient. It is certainly possible to waste
time punching numbers into a calculator. Entering lists of numbers
into a calculator can be timeconsuming, and certainly doesn't
represent a display of statistical intelligence. If, upon reading
an AP question, you think you will have to enter many numbers into a
calculator, you are probably overlooking something. Reread the
problem, and look for a quicker path to a solution.
