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Variable - Applications of Statistics - Exam Key, Exams of Statistics

This is the Exam Key of Applications of Statistics which includes Variable, Researcher, Type of Variable, Observational Unit, Sample Size, Dendritic Branches, Emanating, Humans and Animals, Disease etc. Key important points are: Variable, Researcher, Type of Variable, Observational Unit, Sample Size, Dendritic Branches, Emanating, Humans and Animals, Disease, Blood

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2012/2013

Uploaded on 02/26/2013

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STAT 205 Name:___ANSWER KEY______
Fall 2007
Exam 1
s2 = )1(
)( 2
1
=
n
yyi
n
i
P{E1 U E2} = P{E1} + P{E2} – P{E1 E2}
P{E1 E2} = P{E1}P{E2| E1}
µY = yi P{Y = yi}
σY
2 = (yi – µY)2
P{Y = yi}
= E(Y2) – (E(Y))2
P{Y = j} = nCj pj (1 – p)n-j
µY = np
σY
2 = np(1 – p)
Z =
σ
µ
)(
Y
pf3
pf4
pf5

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STAT 205 Name:___ANSWER KEY______

Fall 2007

Exam 1

s

2

2

1

=

n

y (^) i y

n

i

P{E 1 U E 2 } = P{E 1} + P{E2} – P{E 1 ∩ E 2 }

P{E 1 ∩ E 2 } = P{E 1}P{E2| E1}

μ Y = ∑y i P{Y = y i }

σY

2

= ∑(y i – μ Y)

2

P{Y = y i }

= E(Y

2

) – (E(Y))

2

P{Y = j} = n Cj p

j

(1 – p)

n-j

μ Y = np

σY

2

= np(1 – p)

Z =

( Y −μ)

Part I: Answer eight of the following nine questions. If you complete more than eight, I

will grade only the first eight. Five points each.

  1. Fill in the blank. A researcher counts the number of dendritic branches emanating

from a nerve cell taken from the brains of 36 newborn guinea pigs.

Variable __ number of dendritic braches ______________________

Type of Variable__quantitative - discrete_____________________

Observational Unit_nerve cell from brain of newborn guinea pig_

Sample Size__36__________________________________________

  1. Trypanosomes are parasites which cause disease in humans and animals. In an early

study of trypanosome morphology, researchers measured the lengths (in μm) of 500

individual trypanosomes taken from the blood of a rat. These data were used to create the

following histogram.

Using the histogram, describe the shape of the

distribution of these trypanosome lengths.

Asymmetric and bimodal

  1. Recall the empirical rule which allows us to estimate the percent of observations falling

within 1 SD (68%), 2 SDs (95%), and 3 SDs (99.7%) of the mean. Could the empirical

rule be used for the trypanosome data from question (2)? Why or why not?

The empirical rule requires that the distribution be unimodal and not too skewed. The

histogram of trypanosome length shows these requirements are not met.

  1. Suppose a disease is inherited via a sex-linked mode of inheritance, so that a male

offspring has a 50% chance of inheriting the disease, but a female offspring has no chance

of inheriting the disease. Further suppose that 51.3% of births are male. What is the

probability that a randomly chosen child will have the disease? Hint: Drawing a tree

diagram might be useful here.

Pr{having the disease} = (0.513)(0.5) + (0.487)(0) = 0.2565 or 25.65%

  1. Find Z (^) 0.
  1. (Circle the correct answer) Suppose the weights of USC females follow a normal

distribution. A randomly selected female is told that her Z-score is -2. What information

does this give her about her weight in this distribution?

Her weight is 2 pounds above the mean.

Her weight is 2 pounds below the mean.

Her weight is 2 standard deviations above the mean.

Her weight is 2 standard deviations below the mean.

Part II: Answer every part of the next three problems. Read each question

carefully, and show your work for full credit.

  1. (24 pts.) According to the prostate cancer foundation, prostate cancer will

strike 1 in 6 American men. Then the probability that an American man will

develop prostate cancer is 1/6 = 0.167.

a) If a random sample of 30 men were taken, find the probability that at most

one of the men will develop prostate cancer.

By hand

P{at most one of the men}

= P{Y ≤ 1}

= P{Y = 0} + P{Y = 1}

= 30 C 0 (0.167)

0 (0.833)

30

  • 30 C 1 (0.167)

1 (0.833)

29

= (0.833)

30

  • (30)(0.167)

1 (0.833)

29

= 0.00416246 + 0.

= 0.

Using TI-

Binomcdf(30, 0.167, 1) = 0.

b) Find the probability that exactly five men from the sample will develop

prostate cancer.

By hand

P{exactly 5}

= P{Y = 5}

= 30 C 5 (0.167)

5 (0.833)

25

= 142506 (0.167)

5 (0.833)

25

= 142506 (0.000129892)(0.010378272)

= 0.

Using TI-

Binompdf(30, 0.167, 5) = 0.

  1. (12 pts.) A classical method of researching the nature vs. nurture question is using

identical twins to conduct research. A book by Susan Farber ( Identical Twins Reared Apart ,

New York: Basic Books, 1981) contains a chronicle and reanalysis of identical twins reared

apart. One question of interest is whether there are significant differences in the IQ scores of

identical twins where one member of the pair is reared by natural parents and the other is

not.

A random sample of 32 sets of twins under the above circumstances was taken. The

researcher is hoping to use a statistical test which requires the assumption of the data

coming from a normal population. Using the QQplot below, discuss whether this

assumption appears to be met or violated and why.

When answering this question, keep in mind that I care more about how you reach your

decision, than the actual decision itself. Two to four sentences should suffice…

Overall, this plot looks fairly linear. I do see that the points go above the line at

the high end of the distribution (i.e. the observations are higher than what we

would expect if they came from a normal distribution) and go below the line at the

low end of the distribution (similarly, are lower than we’d expect from a normal).

But the deviation is slight and the points end up going back to the line. Hence, I

see little evidence that these points do not come from a normal distribution.