Statistics 103:
Answers to Practice Problems for Final Exam
1. False. The mean is roughly half way between the median and 75 percentile, so
that about 37.5 percent of people have taxes greater than the mean.
2. 1000
3. True.
The mean should increase when the zeros are removed.
4. False:
5. Around 16%. Any value between 12 and 20 gets credit.
6.
married, divorced, single.
7.
False
8.
False
9. Single.
line.
Use the box plot.
The median is close to zero, as evidence by the lack of a median
10. True. The sample size for divorced people is substantially smaller, leading
to increased SE.
11. Yes. The sample sizes are large in each group, and there are no very
serious outliers. Hence, the Central Limit Theorem should kick in.
12. Since the degrees of freedom equal 400, we can use 1.97 as the approximate
multiplier from the t-table. (996.086  754.942)  1.97
which simplifies to (5.77, 476.43).
1507.10 2 / 257  1162.05 2 / 242 ,
13. It appears that the average property tax for male household heads in the
population is higher than the average property tax for female household heads in
the population. We are 95% confident that the amount of the difference is
between $5.77 and $476.43. If we wanted to narrow this range, we’d need to
collect more data.
14. Check all of the following that are true:
_X_ If we took another random sample of 500, then another, then another, and so
on, we’d expect 95% of the formed confidence intervals to contain the population
difference in average property taxes.
15. Test the null hypothesis that there is no difference in average property
taxes between male and female household heads. State your null and alternative
hypotheses, the test statistic, the p-value, and your conclusions. Consider a
p-value near 0.05 to be small.
Let
Let
 1 be
 2 be
the population average property tax for men.
the population average property tax for women.
The null hypothesis is:
Ho:
1   2 .
The alternative hypothesis is
Ha:
1not   2
The value of the test statistic equals:
t
996.086  754.942
1507.10 2 / 257  1162.05 2 / 242
 2.0
Since the degrees of freedom is 400, the p-value associated with this test
statistic equals approximately 0.045. Hence, there is a 4.5% chance of seeing
such a difference in the sample averages when in fact the two population
averages are equal. This is a fairly small chance, so that we reject the null
hypothesis. There does appear to be evidence that the population average
property taxes for men and women differ.
16.
Check all of the following that are true.
_X_ It may be the case that the results are due to chance, and our conclusion
from the hypothesis test is wrong.
_X_ The chance of getting a value of the test statistic as or more extreme than
what was observed, assuming the null hypothesis is true, equals the p-value.
17.
0
18. _X_
The slope of the line would be positive.
19.
i)Let
p1 be the population percentage of people who get colds under placebo.
Let p 2 be the population percentage of people who get colds under Vitamin C.
p1  p2 .
The alternative hypothesis is Ha: p1  p2
The null hypothesis is:
Ho:
The value of the test statistic equals:
z
31 / 140  17 / 139
(31 / 140)(1  31 / 140) / 140  (17 / 139)(1  17 / 139) / 139
 2.2
The p-value associated with this test statistic equals 0.014. Hence, there is a
1.4% chance of seeing such a difference in the sample percentages when in fact
the two population percentages are equal. This is a fairly small chance, so
that we reject the null hypothesis. There does appear to be evidence that the
population incidence rates of colds when taking Vitamin C or placebo differ.
ii) You should not grant the request. The placebo ensures that any effects due
to the way the drug is administered are equally present in the Vitamin C and
control groups. For example, people may feel better because they are taking a
pill, regardless of whether it is Vitamin C or not. A control group that does
not take a pill would not have this effect
iii)
The SE would be approximately,
(31 / 140)(1  31 / 140) / 500  (17 / 139)(1  17 / 139) / 500
iv) Because the treatments were assigned randomly to the skiers, the background
characteristics in the two groups should be similar. Hence, valid causal
conclusions can be drawn for these people: Vitamin C appears to work for skiers.
However, I am reluctant to generalize these conclusions to other populations
because skiers may react differently than the general public.
20.
i) True.
ii) False.
iii) False.
pattern.
Larger sample size means smaller SE, which means narrower CI.
A nonrandom pattern is a violation.
We hope to find a random
iv) False. With a sample size of 4, it is hard to reject the null hypothesis in
favor of the alternative. The SE is too large. Hence, we cannot conclude much
at all from this study.
v) False. There was no control group over the same time period, so that we have
no way to tell if it is the program or something else that caused their scores
to increase.
vi) False. Pick the exam with the smaller SD so that scores will be closer to
the average of 75.
vii) False. Management and sex are not independent, as can be seen in the
conditional probabilities of being ion management.
21.
i) True. The expected value for the sample percentage is .10, and the standard
error for the sample percentage is .03. Translating to numbers out of 100, we
have 10 give or take 3.
ii) False. Since the SE = .03, there is a 68% chance it will be between 7% and
13%, or 7 and 13.
iii) False. The population has 10% minority members. There is no “give or
take.”
iv) False. There is roughly a 2.5% chance.
22. i)
The colleague is correct because
E (Y )  E (C  I  G  X )  E (C )  E ( I )  E (G )  E ( X ) .
(ii) The colleague is not correct because he or she did not add in the
covariance terms when computing the variance. We need to add
2Cov(C,I)+2Cov(C,G)+2Cov(C,X)+2Cov(I,G)+2Cov(I,X)+2Cov(G,X).
23. i) Let A be the event that you get an A.
Let S be the event that you study hard.
We want Pr(A|S).
We know that Pr(S|A) = .75, and that Pr(S| not A) = .20.
Pr(A) = .40.
Hence, we can find that Pr(A|S) = Pr(A and S)/Pr(S)
Also, we have that
= Pr(S|A)Pr(A) / Pr(S)
= (.75)(.40) / [(.75)(.40) + (.20)(.60)]
= .3/.42
= .714.
ii) We want Pr(A | not S).
Pr(A| not S) =
=
=
=
Pr(A and not S) / Pr(not S)
Pr(not S|A) Pr(A) / ( 1- Pr(S))
(1-.75)(.40) / (1 – .42)
.172.
2
24. i)
Pr(Y<2) =
1
1
1
 ln( 3) 4  y dy  ln( 3) ( ln( 4  y)) |
2
1
1
2.5
(ii) Pr(1.5<Y<2.5) =
1
1
1
Pr(Y<2.5 | Y>2) =
3
E (Y )   y
(iv)
1
ln( 3)  ln( 2)
 .37.
ln( 3)
 ln( 3) 4  y dy  ln( 3) ( ln( 4  y)) |
2
(iii)

2.5
2

ln( 2)  ln( 1.5)
 .262.
ln( 3)
Pr(Y  2, Y  2.5) Pr( 2  Y  2.5) .262


 .416.
Pr(Y  2)
1  Pr(Y  2)
.63
1
1
dy .
ln( 3) 4  y
25. i) Pr( X  1)  .4.
ii) E (Y )  0(.4)  1(.5)  2(.2)  .9.
iii) E (Y | X  1)  0(.15 / .35)  1(.2 / .35)  2(0)  0.57.
iv)
SD(Y )  Var (Y ) .
Var (Y )  E (Y 2 )  E (Y ) 2  0(.4)  1(.5)  4(.2)  .9 2  1.3  .81  .49.
So, SD(Y) = .7.
v)
Cov( X , Y )  E ( XY )  E ( X ) E (Y )  (0)(0)(. 1)  (0)(1)(. 3)  (0)( 2)(. 2)  (1)(0)(. 15)  (1)(1)(. 2)  (1)( 2)(0)
 (2)(0)(. 05)  (1)( 2)(0)  (2)( 2)(0)  E ( X ) E (Y )  .2  (.45)(. 90)  .205.
vi) E (T )  E (1.5Y  .2 X )  1.5E (Y )  .2 E ( X )  1.5(.9)  .2(.45)  1.26.
Var (T )  1.5 2 Var (Y )  .2 2 Var ( X )  2(1.5)( .2)Cov( X , Y )  2.25(.49)  .04(.3475)  .6(.205)  1.24.
26. Let X be the random variable for your earnings. The sample space of X is
found by determining all possible outcomes (in terms of dollars) of the three
rolls. Note that each roll is independent, and you have a 1/3 chance of winning
on any roll. Below, W means win and L means lose.
Rolls
x
Pr(X=x)
WWW
30
WWL
14
WLW
16
LWW
WLL
18
0
(1 / 3) 3
(1 / 3) 2 (2 / 3)
(1 / 3) 2 (2 / 3)
(1 / 3) 2 (2 / 3)
( 2 / 3)
LWL
2
LLW
4
LLL
-12
(ii)
2
(1 / 3)
(2 / 3) 2 (1 / 3)
(2 / 3) 2 (1 / 3)
(2 / 3) 3
E ( X )  30(1 / 3) 3  (14  16  18)(1 / 3) 2 (2 / 3)  (0  2  4)( 2 / 3) 2 (1 / 3)  12(2 / 3) 3  2.
Var ( X )  30 2 (1 / 3) 3  14 2 (1 / 3) 2 (2 / 3)  16 2 (1 / 3) 2 (2 / 3)  18 2 (1 / 3) 2 (2 / 3)  0 2 (2 / 3) 2 (1 / 3)  2 2 (2 / 3) 2 (1 / 3) 
4 2 (2 / 3) 2 (1 / 3)  12 2 (2 / 3) 3  2 2  132.44.
27.
Last problem
(i) False. The data are not 0-1 data, so the standard error formula for
percentages does not apply. The data actually are continuous, even though the
values lie between zero and one. People tip at different rates. (This is like
the problem on the practice midterm with the alumni donation percentages.)
(ii) False. While the calculation of the z-statistics and area under the normal
curve are correct, we cannot use the central limit theorem for a sample
percentage when there are only 5 observations. Recall that we need np>10 and
n(1-p)>10, both of which fail when n=5 and p=.30.
The way to solve this problem is to use the fact that
Pr(less than 40% blue) = Pr(0 blue or 1 blue) = Pr(0 blue) + Pr(1 blue).
Pr(0 blue) = (.7)(.7)(.7)(.7)(.7).
Pr(1 blue) = 5 (.7)(.7)(.7)(.7)(.3).
get one blue in five M&Ms.
28. Invest your money with me.
Yeah, I’ll take care of it.
a)
1
9 x 2 x3 1
 |.2  .5  .004  .496.
6
2
.2
Pr( X  .20)  
b)
1
E( X ) 
c)
9 x3 9 x 4 1
1 6  24 |1  0
The extra 5 comes from the five ways to
1
Var ( X )  E ( X 2 )  E ( X ) 2 
9x4
9 5
1 6 dx  0  30 x  .60.
d%
Var (.75Y  .25 X )  .752Var (Y )  .252Var ( X )  2(.75)(.25)Cov( X , Y )
Cov( X , Y )  Cor ( X , Y ) SD( X ) SD(Y ) , we have Cov(X,Y) = .45(.77)(.10)=.0348.
Since
Plugging in, we get Var(.75Y+.25X) = .056
So, the SD = .237
e) Pr(X>0) = 0.5, based on integrating f(x).
Pr(at least one month positive) = 1 – Pr(no months positive) = 1 - .5^12.
f) There’s a 50% chance that any month is positive. We want the chance of
getting at least 70/120 = 58.33% of the 120 months with positive return. Since
100 is a large sample size, we can use the central limit theorem to figure out
this chance.
Pr( Z 
.5833  .50
.0833
)  Pr( Z 
)  .034.
.0456
.5(.5) / 120
29. Weather predictions
a) Pr(Sun | Pred says Sun ) = Pr (Sun, Pred says Sun) / Pr(Sun)
= (.80)(.40) / ((.80)(.40) + (.10)(.30) + (.33)(.30)) = .71
b) Pr(Cloudy or Sunny | Pred says rain) = 1 – Pr(Rain | Pred says rain)
Pr(Rain | Pred says rain) = Pr(Rain, Pred says rain) / Pr(Pred says rain)
= (.5)(.3)/((.5)(.3)+(.33)(.3)+(.05)(.4) = .557.
So, Pr(cloudy or sunny | Pred says rain)
30.
= 1 - .557 = .443.
Cov( X , Y )  E (( X  E ( X ))(Y  E (Y )))
 E ( XY  XE (Y )  YE ( X )  E ( X ) E (Y ))
 E ( XY )  E (Y ) E ( X )  E ( X ) E (Y )  E ( X ) E (Y )  E ( XY )  E ( X ) E (Y )
The key step is that in E(XE(Y)), we can pull out E(Y) because it is a constant,
thus leaving E(Y)E(X).
31.
Var ( X 1  X 2 )  Var ( X 1 )  Var ( X 2 ) 
To prove that Var ( X 1 ) 
 12
n1
 12
n1

, we have that
 22
n2
.
Var ( X 1 )  Var (

1
n1
2
1
1
( X 1  X 2  ...  X n ))  2 Var ( X 1  X 2  ...  X n )
n1
n1
Var ( X 1 )  Var ( X 2 )  ...  Var ( X n ) 
A similar proof is used for
1
n1
2

2
1

  1  ...   1 
2
2
n1 1
n1
2
2

 12
n1
Var ( X 2 ) .
n
1 n
1
1 n
1 n
1
X i )  E ( X i )   E ( X i )     (n )   .

n i 1
n i 1
n i 1
n i 1
n
1
1
1
1
E (V )  E (( X 1  X 2 ) / 2)  E ( X 1 )  E ( X 2 )       .
2
2
2
2
1
1
1
1
b) Var (V )  Var (( X 1  X 2 ) / 2)  2 Var ( X 1 )  2 Var ( X 2 )  (1000 / 25)  (10000 / n 2 )
4
4
2
2
32.
E (W )  E ( X 1 )  E (
To make Var(V)<Var(W), we solve
1
1
(1000 / 25)  (10000 / n2 )  (1000 / 25)
4
4
for
n 2 to obtain 83.333. So, we need at 84
sampled individuals from group 2 to ensure that V is more efficient than W.
33. Drinking and Driving
i) The problem states that the measurement of breathalyzer in percentages for
someone with blood alcohol level .095 follows a normal curve with mean .095 and
standard deviation .004.
We want the probability of getting more than .10 as a measurement. Let's
standardize .10 by subtracting the mean and dividing by the standard deviation:
z = (.10-.095)/.004 = 1.25
We want the area under the normal curve to the right of 1.25, which is .1056.
ii) The measurement of breathalyzer in percentages for someone with blood
alcohol level .15 follows a normal curve with mean .15 and standard deviation of
.004.
We want the chance that the measurement for this person is less than .10.
Standardizing, we get
z = (.10 - .15) / .004 = -12.5.
The chance of observing a z-value less than (to the left of) 12.5 is very, very
small (and not even on the table).
iii) For people with true levels of .10, the chance that any one individual will
be booked is .50 (the median value for these people is .10). Hence, the
probability that any one individual will not be booked is also 0.50.
Now, want Pr(at least one individual is booked). Since the only possible
outcomes are "at least one person is booked" and "no one is booked", it is true
that: Pr(at least one is booked) + Pr(no one is booked) = 1.
So, Pr(at least
one is booked) = 1 - Pr(no one is booked). Now, Pr(no one is booked) = Pr(first
person not booked, and second person not booked, and... etc..., and eighth
person not booked, and ninth person not booked). Since the test is done
separately on each person, we can assume that readings are independent.
Hence, Pr(first person not booked, and second person not booked, and...,
etc..., and ninth person not booked) = Pr(first person not booked) * Pr(second
person not booked) * ... * Pr(ninth person not booked)
= 0.5 * 0.5 * ... * 0.5
(nine of these 0.5s) = 0.5^9.
Finally, we have Pr(at least one person booked) = 1 - 0.5^9.
34. Sex of children
i) Because the babies' sexes are independent, Pr ( M, F, M, F, M, F) = Pr(M)
* Pr(F) * Pr(M) * Pr(F) * Pr(M) * Pr(F)
= .4986 * .5014 * .4986 * .5014 *
.4986 * .5014
ii) Because the babies' sexes are independent, Pr ( F, F, F, F, F, F) = Pr(F)
* Pr(F) * Pr(F) * Pr(F) * Pr(F) * Pr(F)
= .4986 * .4986 * .4986 * .4986 *
.4986 * .4986
iii) Because the babies' sexes are independent, Pr ( F | M, M, M ) = Pr (F) =
.5014
35.
To find the MLE, we first need the joint distribution of the data.
Pr(Y1  0, Y2  2, Y3  1, Y4  4)  Pr(Y1  0) Pr(Y2  2) Pr(Y3  1) Pr(Y4  4)
 0 e 
 
 0!
 2 e 

 2!
 1e 

 1!
 4 e 

 4!
 7 e 4
 
48

The likelihood function equals this probability, conceived as a function of the
unknown  . We then find the value of  that maximizes the likelihood
function. This is done by setting the derivative of the function to equal zero,
which results in:
 7 e 4 
d 
 48
d


  ( 7  6 e  4   4 7 e  4  )  0
After cancellations, we get that
  7/ 4
is the maximum likelihood estimate.
36. a) I would use the chi-squared analysis (analysis ii). These data are
counts, not continuous data. The t-test assumes that the data are continuous.
Plus, if you think about it, the sample average frequency for the digits has to
equal 50, so that the t-test is completely meaningless.
b) Because the p-value is so large, there is a good chance of seeing these
results when the null hypothesis is true. Therefore, we cannot reject the null
hypothesis. The data are consistent with the hypothesis that each digit has a 1
in 10 chance of being selected.
c) We just need to make one frequency really small and the other really large.
For example, we could make the frequency for eight to be 1, and the frequency
for nine to be 99. This would mean that eight happens much less than 10% of the
time, and nine happens much more than 10% of the time, thereby rejecting the
null hypothesis. Note that the two frequencies should add to 100 since we are
not changing the frequencies for other digits, and the original sum of the
frequencies was 100.
37. a)
Favorite color
Red
Blue
Green
Total
b)
Favorite color
Red
Blue
Green
Total
Male
10
10
20
40
Male
1
1
38
40
Female
20
20
40
80
Female
78
1
1
80
You want
of males
who like
who like
to make a table so that the %
who like red equals the % of males
blue, and also equals the % of males
green. An example is shown here.
You want to make a table so that the % of
males who like red is very different from
either the % of males who like green or the %
of males who like blue. Example shown here.