AP Statistics / Mr. Hansen

Name: __________KEY____________

1.

In order to demonstrate a cause-and-effect relationship, we must perform __an experiment_ , which means that we administer a __treatment__ while controlling as many of the ___lurking____ variables as we can reasonably afford to. The 3 key principles in designing a study of this type are __control__ (by which we mean __trying to make the treatment level, or presence or absence of treatment, be the ONLY variable that changes--all others are controlled __ ), __randomization__ (by which we mean __avoiding bias in selection or assignment of experimental units to groups__ ), and __replication__ (by which we mean __using a large enough sample to provide plausible evidence that the experimental result seen (or not seen) was not merely the result of chance __ ).

2.

The technical purpose of blocking in experimental design is to __reduce variability in the experimental units__ . This is a good thing to do (i.e., there is a real-world purpose or goal to performing blocking) because __the experimental effect, if any, will be more likely to stand out above the background "noise" caused by excessive variability__ .

3.

Any study in which the subjects know when they are being treated and when they are not being treated may be susceptible to the __placebo___ effect.

4.

Systematic discrepancies in the probability that certain members of the population are selected for a study, or selected for treatment, or selected for response, or treated in any way differently from other members of the population are all examples of __bias__ .

5.

Subjects in a certain imaginary experiment are randomly chosen to receive an all-expense-paid trip to Seattle for a mathematics conference. Their scores on a standardized achievement test measuring mathematical knowledge are computed before and after the trip. This design uses __matched pairs__ .

6.

Subjects in another imaginary experiment are randomly chosen to receive a low dose of aspirin daily or a high dose of aspirin daily. This design exhibits __multiple levels of treatment (but apparently, no control!)__ .

7.

Subjects in a third (and final, I promise) imaginary experiment paint the left side of their noses with one skin cream, the right side with another, and the bridge of their nose with a third. This design exhibits __matched triples__ .

8.

If heights of cornstalks in Farmer Bill’s field (in feet) are distributed according to N(6, 0.8) and if Farmer Irene’s are distributed according to N(6.5, 0.95), what is the probability that a randomly selected cornstalk from Bill’s field is taller than a randomly selected cornstalk from Irene’s field? For full credit, state any assumptions you are making.
Let B = ht. of a randomly selected stalk in Bill’s field, I = ht. of a randomly selected stalk in Irene’s field, and assume that B and I are independent. The answer is 34.4%. Note: WNFFC.

9.

What happens to the mean of a r.v. if all values are increased by 4? __mean will increase by 4__ What if all values are quadrupled? __mean will quadruple__

10.

What happens to the s.d. of a r.v. if all values are increased by 4? __s.d. stays the same (NO CHANGE)__ What if all values are quadrupled? __s.d. will quadruple__

11.

If temperatures on the Frizzlewatt scale in West Siberia are normally distributed with a mean of 15 and a s.d. of 14, what are the mean and standard deviation on the Fahrenheit scale? Assume that Frizzlewatt degrees = 4 · (Fahrenheit degrees) – 10.

Solving the given equation for degrees Fahrenheit, we have Fahr. = (Friz. + 10)/4. Thus the expected value (a.k.a. mean) is E(Fahr.) = E((Friz. + 10)/4) = E(Friz./4 + 10/4) = E(Friz./4) + 10/4 = E(Friz.)/4 + 10/4 = 15/4 + 10/4 = 25/4 = 6.25.

As for the variance, Var(Fahr.) = Var((Friz. + 10)/4) = Var(Friz./4 + 10/4) = Var(Friz./4) = Var(Friz.)/16 = 14²/16 = 12.25 Þ s.d. = Ö12.25 = 3.5. The method we used in class is considerably easier but is harder to notate in HTML.

12.

Classify the distributions of each of the following random variables as normal, approximately normal, binomial, approximately binomial, geometric, approximately geometric, or none of the above. If you choose any answer other than "none of the above," state the parameters of the distribution and answer the remaining questions.

(a)

X = # of spades received in a bridge hand (13 cards)

Type of distribution: __none of the above (actually called "hypergeometric")__
Note: Because the AP curriculum does not cover the hypergeometric distribution (and neither does your textbook), you may omit all of these computations except for the third one, which you should be able to do.
m_X = 3.25
s_X = 1.365
P(X = 3) = (₁₃C₃)(₃₉C₁₀)/(₅₂C₁₃) = 0.286 (you should be able to do this one)
P(X £ 3) = 0.585
P(X > 6) = 0.010
P(X ³ 6) = 0.052

(b)

X = # of free throws made in 40 attempts, where the probability of success per trial is initially 0.6 but improves slightly with practice

Type of distribution: __approx. binomial, B(40, 0.6)__
m_X = 24
s_X = 3.098
P(X = 3) = 4.03 · 10^–12
P(X £ 3) = 4.25 · 10^–12
P(X > 6) = 0.9999999941
P(X ³ 6) = 0.9999999993

(c)

X = # of double sixes received when Belinda rolls a pair of fair dice 120 times

Type of distribution: __exactly binomial, B(120, 1/36)__
m_X = 10/3 or approx. 3.333
s_X = 1.800
P(X = 3) = 0.223
P(X £ 3) = 0.572
P(X > 6) = 0.051
P(X ³ 6) = 0.118

(d)

X = # of phone calls made by Leisure Suit Larry to get a prom date, where the probability of success per call is approximately 0.2

Type of distribution: __approx. geometric with parameter p = 0.2__
m_X = 5 by the important formula in the box on p.441
s_X = 4.472 by the unimportant formula s_X = (Öq)/p that is not even in your textbook (hence not required for the test)
P(X = 3) = 0.128 by the important formula in the box on p.436
P(X £ 3) = 0.488 by calc. (geometcdf fcn.) or by using formula at bottom of p.442:
P(X £ 3) = 1 – P(X > 3) = 1 – q³ = 1 – 0.8³ = 0.488
P(X > 6) = 0.262
P(X ³ 6) = P(X > 5) = q⁵ = 0.8⁵ = 0.328

(e)

X = length (in cm) of the right big toe of American men, where the median is 5 and 68% of these men have toes measuring between 4.2 cm and 5.8 cm

Type of distribution: __approx. normal, N(5, 0.8)__
m_X = 5
s_X = 0.8
P(X = 3) = 0
P(X £ 3) = 0.006 by calc. (or by table with z = –2.5)
P(X > 6) = 0.106
P(X ³ 6) = 0.106

(f)

X = # of die rolls needed to get either a 3 or a 6

Type of distribution: __exactly geometric with parameter p = 1/3__
m_X = 3 by the important formula in the box on p.441
s_X = 2.449 (formula not in book; no need to do this one)
P(X = 3) = 0.148 by the important formula in the box on p.436
P(X £ 3) = 0.704
P(X > 6) = 0.088
P(X ³ 6) = 0.132

13.

For your own benefit, fill in the purpose and syntax of each of the following TI-83 functions. You are permitted to store "cheat hints" in the memory of your calculator if you wish. (Last year, Doug Bemis wrote a program to prompt for the needed values.)

Purpose of binompdf: __if X has a B(n, p) distrib., fcn. returns P(X = a)__
Syntax: binompdf( n, p, a )

Purpose of binomcdf: __ if X has a B(n, p) distrib., fcn. returns P(X £ a)__
Syntax: binomcdf( n, p, a )

Purpose of normalpdf: __not really needed for AP Statistics (see note below)__
Syntax: normalpdf( x, m, s)
Syntax note: If m and s are omitted, the calculator will assume you want the standard normal curve (m = 0, s = 1).

Purpose of normalcdf: __if X has a N(m, s) distrib., fcn. returns area under bell-shaped curve from lowerlimit to upperlimit__
Syntax: normalcdf( lowerlimit, upperlimit, m, s)
Syntax note: If m and s are omitted, the calculator will assume you want the standard normal curve (m = 0, s = 1).

Purpose of geometpdf: __ if X has a geometric distrib. with parameter p, fcn. returns P(X = a)__
Syntax: geometpdf( p, a )

Purpose of geometcdf: __ if X has a geometric distrib. with parameter p, fcn. returns P(X £ a)__
Syntax: geometcdf( p, a )