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Research School of Finance, Actuarial Studies and Statistics
PAST FINAL EXAMINATION 1 SOLUTIONS
STAT7055 Introductory Statistics for Business and Finance
Writing Time: 180 minutes
Reading Time: 15 minutes
Exam Conditions:
Central examination.
Students must return the examination paper at the end of the examination.
This examination paper is not available to the ANU Library archives.
Materials Permitted in the Exam Venue:
(No electronic aids are permitted, e.g., laptops, phones).
Calculator (non-programmable).
Two A4 pages with notes on both sides.
Unannotated paper-based dictionary (no approval required).
Materials to be Supplied to Students:
Script book.
Scribble paper.
Instructions to Students:
Please write your student number in the space provided on the front of the script book.
Attempt all 5 questions.
Start your solution to each question on a new page and clearly label each solution with the corresponding
question number.
To ensure full marks show all the steps in working out your solutions. Marks may be deducted for failure
to show working or formulae.
Selected statistical tables are attached to the back of the examination paper.
If a required degree of freedom is not listed in a statistical table, please use the closest degree of freedom.
Unless otherwise stated, use a significance level of α = 5%.
Round all numeric answers to 4 decimal places.
Question: 1 2 3 4 5 Total
Marks: 19 17 17 9 41 103
Question 1 [19 marks]
(a) [4 marks] A student answers a multiple-choice examination question that offers
four possible answers. Suppose that the probability that he knows the answer to
the question is 0.8 and the probability that he guesses is 0.2. Assume that if the
student guesses, the probability of selecting the correct answer is 0.25. If the student
correctly answers a question, find the probability that he really knew the correct
answer.
Solution:
Let:
K = the student knows the answer.
KC = the student doesn’t know the answer (i.e., guesses).
C = the student answers correctly.
CC = the student answers incorrectly.
We know that P (K) = 0.8, P (KC) = 0.2, P (C|K) = 1 and P (C|KC) = 0.25.
We want to find P (K|C):
P (K|C) = P (K ∩ C)
P (C)
=
P (C|K)P (K)
P (C ∩K) + P (C ∩KC)
=
P (C|K)P (K)
P (C|K)P (K) + P (C|KC)P (KC)
=
1× 0.8
1× 0.8 + 0.25× 0.2
=
16
17
= 0.9412
(b) [5 marks] Suppose the student takes an examination with 3 multiple choice ques-
tions. Let X denote the number of questions he gets correct. Assuming that the
student answers each question independently, determine the probability distribution
of X and calculate the expected value of X.
Solution:
The probability of getting any given question correct is P (C) and was cal-
culated in part (a), P (C) = 0.85. Each question is an independent trial,
so the number of successes (or correct answers) has a binomial distribution,
X ～ Bin(n = 3, p = 0.85). Therefore, using the binomial formula, we know
that P (X = 0) = 0.003375, P (X = 1) = 0.057375, P (X = 2) = 0.325125 and
Past Final Examination 1 Page 2 of 16 STAT7055
P (X = 3) = 0.614125. Further, the expected value of X is:
E(X) = np = 3× 0.85 = 2.55
(c) [10 marks] Suppose the student takes an examination with 2 multiple choice ques-
tions. But this time, he gains confidence every time he knows the answer to a ques-
tion. Specifically, if he knows the answer to the current question, then for the next
question the probability that he knows the answer becomes 0.9 and the probability
that he guesses becomes 0.1. If he does not know the answer to the current ques-
tion, then for the next question the probability that he knows the answer and the
probability that he guesses remain 0.8 and 0.2, respectively. Again letting X denote
the number of questions he gets correct, determine the probability distribution of X
and calculate the expected value of X. We can assume the probabilities of knowing
the answer and guessing for the first question are 0.8 and 0.2, respectively.
Solution:
The key point to recognise is that the probabilities of knowing and not knowing
the answer to the second question change depending on whether or not he
knows the answer to the first question. Therefore, we can calculate the following
probabilities for each outcome:
Question 1 Question 2
Know Correct Know Correct x Probability
0 0 0 0 0 0.2× 0.75× 0.2× 0.75 = 0.0225
0 0 0 1 1 0.2× 0.75× 0.2× 0.25 = 0.0075
0 0 1 1 1 0.2× 0.75× 0.8× 1 = 0.12
0 1 0 0 1 0.2× 0.25× 0.2× 0.75 = 0.0075
0 1 0 1 2 0.2× 0.25× 0.2× 0.25 = 0.0025
0 1 1 1 2 0.2× 0.25× 0.8× 1 = 0.04
1 1 0 0 1 0.8× 1× 0.1× 0.75 = 0.06
1 1 0 1 2 0.8× 1× 0.1× 0.25 = 0.02
1 1 1 1 2 0.8× 1× 0.9× 1 = 0.72
We can combine the probabilities in the table above to obtain P (X = 0) =
0.0225, P (X = 1) = 0.0075 + 0.125 + 0.0075 + 0.06 = 0.195 and P (X = 2) =
0.0025 + 0.04 + 0.02 + 0.72 = 0.7825. Using these probabilities, the expected
value of X is:
E(X) =
∑
all x
(x× p(x)) =
∑
all x
(x× P (X = x)) = 1.76
Past Final Examination 1 Page 3 of 16 STAT7055
Question 2 [17 marks]
Suppose that X is a continuous random variable with the following probability density
function:
f(x) =
1
θ
, 0 ≤ x ≤ θ
Jimmy considers himself a budding statistician and he wants to investigate the true value
of θ.
(a) [2 marks] Jimmy thinks that the true value is actually θ = 2. Assuming Jimmy
is right, find each of the following probabilities:
P
(
0 < X < 1 3 ) , P ( 1 3 < X < 1 ) , P ( 1 < X < 7 4 ) , P ( 7 4 < X < 2 ) Solution: If θ = 2, X has a uniform distribution between 0 and 2. Therefore, for 0 < a < b < 2, the probability that a < X < b is given by: P (a < X < b) = (b a)× 1 2 Therefore we have: P ( 0 < X < 1 3 ) = 1 3 × 1 2 = 1 6 , P ( 1 3 < X < 1 ) = 2 3 × 1 2 = 1 3 P ( 1 < X < 7 4 ) = 3 4 × 1 2 = 3 8 , P ( 7 4 < X < 2 ) = 1 4 × 1 2 = 1 8 (b) [4 marks] In order to test whether θ = 2, Jimmy collects a sample of 500 of these X variables and records their values. He then tabulates how many variables fell into each range of part (a), and his results are summarised below. Based on this data, test whether θ = 2. Clearly state your hypotheses and use a significance level of α = 5%. Range of value ( 0 < X < 1 3 ) ( 1 3 < X < 1 ) ( 1 < X < 7 4 ) ( 7 4 < X < 2 ) Number of variables 101 165 191 43 Solution: We should use a Chi-squared goodness-of-fit test. The hypotheses are: H0 : p1 = 1 6 , p2 = 1 3 , p3 = 3 8 , p4 = 1 8 H1 : The population proportions do not match that given above. Past Final Examination 1 Page 4 of 16 STAT7055 The expected counts are calculated from H0 and are given below: e1 = 250 3 , e2 = 500 3 , e3 = 375 2 , e4 = 125 2 Therefore, the χ2-statistic is given by: χ2 = 4∑ i=1 (fi ei)2 ei = 9.9113 We need to compare this to a chi-squared distribution with 4 1 = 3 degrees of freedom and reject H0 when χ 2 > χ20.05,3 = 7.81. Since 9.9113 > 7.81, we reject
H0 and conclude that the population proportions are not as given in H0, i.e.,
θ 6= 2.
(c) [4 marks] Jimmy now wants to actually estimate θ. From the sample he collected
in part (b), he can calculate the sample mean, Xˉ =
∑500
i=1Xi
500
. Is Xˉ an unbiased
estimator of θ Why or why not If not, derive an unbiased estimator of θ.
Solution:
First find the expected value of Xˉ:
E(Xˉ) = E(X) =
θ
2
Therefore, since E(Xˉ) does not equal θ, Xˉ is not an unbiased estimator of θ.
An unbiased estimator of θ would be 2Xˉ, since:
E(2Xˉ) = 2E(X) = 2× θ
2
= θ
(d) [3 marks] Jimmy decides it might be a better idea to use an interval estimator
rather than a point estimator. Based on the sample mean of Xˉ = 0.9418 and
the population variance of σ2 = 0.3008, calculate a 95% confidence interval for
μ = E(X). Interpret this confidence interval.
Solution:
We apply the formula for a 95% confidence interval for μ when σ2 is known:(
Xˉ ± z0.025 σ√
n
)
= (0.8937, 0.9899)
The interpretation is, if we were to repeatedly construct confidence intervals in
this manner, we would expect 95% of them to contain μ.
Past Final Examination 1 Page 5 of 16 STAT7055
(e) [2 marks] Without actually performing the test, if you were to test H0 : μ = 1
against the two-tailed alternative at a significance level of α = 5%, would you reject
H0 Why or why not
Solution:
Yes, we would reject H0, because the value μ = 1 lies outside the confidence
interval constructed in part (d).
(f) [2 marks] Jimmy actually wanted an interval estimator for θ, not μ. Suggest a way
to convert the confidence interval for μ you constructed in part (d) to a confidence
interval for θ.
Solution:
Since θ = 2μ, we could simply multiply both endpoints of the confidence interval
in part (d) by 2:
(1.7875, 1.9797)
Question 3 [17 marks]
A survey was conducted to determine student, faculty, and administration attitudes on
a new university parking policy. The distribution of those favouring or opposing the
policy was as shown in the table below:
Student Faculty Administration
Favour 252 107 43
Oppose 139 81 40
(a) [4 marks] The parking office claims that the population proportion of people at
the university who are both students and in favour of the new parking policy is
greater than 34%. Test this claim at a significance level of α = 5%, clearly stating
your hypotheses.
Solution:
Let p denote the population proportion of people in the university who are both
students and in favour of the policy. The hypotheses are:
H0 : p = 0.34
H1 : p > 0.34
From the data we can calculate the sample proportion, p = 252
662
= 0.3807. The
test statistic is given by:
Z =
p p0√
p0(1 p0)
n
=
0.3807 0.34√
0.34×(1 0.34)
662
= 2.2087
Past Final Examination 1 Page 6 of 16 STAT7055
As this is a upper-tailed test, the rejection region is given by Z > z0.05 = 1.645.
Since 2.2087 > 1.645, we reject H0 and we conclude that p > 0.34, supporting
the parking office’s claim. Alternatively, the p-value equals 0.0136.
For parts (b) and (c), assume that the students, faculty and administration were sampled
independently.
(b) [4 marks] The parking office also claims that the population proportion who favour
the new parking policy is that same for both students and faculty. Test this claim
at a significance level of α = 5%, clearly stating your hypotheses.
Solution:
Let p1 denote the population proportion of students who favour the policy and
p2 denote the population proportion of faculty who favour the policy. The
hypotheses are:
H0 : p1 p2 = 0
H1 : p1 p2 6= 0
From the data we can calculate n1 = 391, n2 = 188, the sample proportions,
p 1 =
252
391
= 0.6445 and p 2 =
107
188
= 0.5691, and the combined proportion,
p = 252+107
391+188
= 0.6200. Since D0 = 0, the test statistic is given by:
Z =
(p 1 p 2) 0√
p (1 p )
(
1
n1
+ 1
n2
) = 0.6445 0.5691√
0.6200(1 0.6200) ( 1
391
+ 1
188
) = 1.7492
As this is a two-tailed test, the rejection region is given by Z > z0.025 = 1.96 or
Z < z0.025 = 1.96. Since < 1.96 < 1.7492 < 1.96, we fail to reject H0 and we conclude that p1 = p2, supporting the parking office’s claim. Alternatively, the p-value equals 0.0802. (c) [4 marks] The parking office also claims that the population proportion of students who favour the new policy minus the population proportion of administration who favour the new policy is equal to 0.1. Test this claim at a significance level of α = 5%, clearly stating your hypotheses. Solution: Let p1 denote the population proportion of students who favour the policy and p3 denote the population proportion of administration who favour the policy. Past Final Examination 1 Page 7 of 16 STAT7055 The hypotheses are: H0 : p1 p3 = 0.1 H1 : p1 p3 6= 0.1 From the data we can calculate n1 = 391, n3 = 83, the sample proportions, p 1 = 252 391 = 0.6445 and p 3 = 43 83 = 0.5181. The test statistic is given by: Z = (p 1 p 3) D0√ p 1(1 p 1) n1 + p 3(1 p 3) n3 = (0.6445 0.5181) 0.1√ 0.6445(1 0.6445) 391 + 0.5181(1 0.5181) 83 = 0.4408 As this is a two-tailed test, the rejection region is given by Z > z0.025 = 1.96 or
Z < z0.025 = 1.96. Since 1.96 < 0.4408 < 1.96, we fail to reject H0 and we conclude that p1 p3 = 0.1, supporting the parking office’s claim. Alternatively, the p-value equals 0.66. (d) [5 marks] The parking office now decides that it is too risky to make claims about population proportions. Instead, they now only claim that the attitudes regarding the new parking policy are independent of student, faculty or administration status. Test this claim at a significance level of α = 5%, clearly stating your hypotheses. Solution: Here we are doing a Chi-squared test of a contingency table. The hypotheses are: H0 : Attitude and status are independent. H1 : Attitude and status are not independent. The expected counts are given in the table below: Student Faculty Administration Favour 237.4350 114.1631 50.4018 Oppose 153.5650 73.8369 32.5982 The χ2-statistic is given by: χ2 = 2∑ i=1 3∑ j=1 (fij eij)2 eij = 6.1869 We need to compare this to a chi-squared distribution with (2 1)× (3 1) = 2 degrees of freedom and reject H0 when χ 2 > χ20.05,2 = 5.99. Since 6.1869 > 5.99,
we reject H0 and conclude that attitude is not independent to status.
Past Final Examination 1 Page 8 of 16 STAT7055
Question 4 [9 marks]
Suppose the manager of a manufacturing plant suspects that the output of a production
line (in number of units produced per shift) depends on two factors:
Which of two supervisors is in charge of the line.
Which of three shifts (day, swing, or night) is being measured.
Suppose that the two supervisors were each observed on three randomly selected days for
each of the three different shifts. The production outputs were recorded and summarised
below:
Shift
Day Swing Night
Supervisor
1 571 480 470
610 474 430
625 540 450
2 480 625 630
516 600 680
465 581 661
(a) [5 marks] Suppose the sample variances of production output for the two super-
visors are s21 = 5155.25 and s
2
2 = 6096.5. Test whether the mean production output
is the same for the two supervisors. Clearly state your hypotheses and use a signifi-
cance level of α = 5%. Assume that the population variances of production output
for the two supervisors are equal.
Solution:
Let μ1 and μ2 be the population mean production output for supervisors 1 and
2, respectively. The hypotheses are:
H0 : μ1 μ2 = 0
H1 : μ1 μ2 6= 0
From the data we can calculate Xˉ1 = 516.6667, Xˉ2 = 582 and s
2
p = 5625.875.
The test statistic is then equal to:
T =
(Xˉ1 Xˉ2) D0√
s2p
(
1
n1
+ 1
n2
) = (516.6667 582) 0√
5625.875
(
1
9
+ 1
9
) = 1.8478
We need to compare this to a t-distribution with n1 + n2 2 = 16 degrees of
freedom and reject H0 when T > t0.025,16 = 2.120 or T < t0.025,16 = 2.120. Past Final Examination 1 Page 9 of 16 STAT7055 Since 2.120 < 1.8478 < 2.120, we fail to reject H0 and we conclude that the population mean production output is the same for both supervisors. A two-way ANOVA has been conducted on this data and the results are summarised in the table below: Source Sum of squares Degrees of freedom Mean squares F Supervisor 19208 Shift 247 Interaction Error 8640 Total 109222 (b) [4 marks] Test whether there is an interaction between supervisor and shift. Clearly state your hypotheses and use a significance level of α = 5%. Solution: The hypotheses are: H0 : There is no interaction between supervisor and shift. H1 : There is an interaction. From the ANOVA table we know that SS(Int) = SS(Total) SS(Supervisor) SS(Shift) SSE = 81127 and we also know that df(Int) = (2 1)×(3 1) = 2 and df(E) = 18 2×3 = 12. So the test statistic is equal to: F = MS(Int) MSE = SS(Int) df(Int) SSE df(E) = 81127 2 8640 12 = 56.3382 We need to compare this to an F -distribution with 2 numerator and 12 denom- inator, degrees of freedom, and reject H0 when F > F0.05,2,12 = 3.89. Since
56.3382 > 3.89, we reject H0 and we conclude that there is an interaction be-
tween supervisor and shift.
Question 5 [41 marks]
It is generally thought that the weight of a turkey depends on its age. The following
data was collected on 13 turkeys that were raised on Australian farms:
Past Final Examination 1 Page 10 of 16 STAT7055
Age (weeks) Weight (pounds) Age (weeks) Weight (pounds)
28 13.3 26 11.8
20 8.9 21 11.5
32 15.1 27 14.2
22 10.4 29 15.4
29 13.1 23 13.1
27 12.4 25 13.8
28 13.2
Let Y denote Weight and X denote Age. The following summary statistics are provided:
sXY = 5.2404, s
2
X = 12.5769 and s
2
Y = 3.2847.
(a) [5 marks] Fit the regression model Y = β0 + β1X + . That is, calculate the
estimates β 0 and β 1.
Solution:
From the data we can calculate Xˉ = 25.9231 and Yˉ = 12.7846. Therefore we
have:
β 1 =
sXY
s2X
=
5.2404
12.5769
= 0.4167
β 0 = Yˉ β 1Xˉ = 12.7846 0.4167× 25.9231 = 1.9833
(b) [3 marks] We have been told that the sum of squares for regression is SSR =
26.202. Calculate the standard error of estimate, s.
Solution:
We can calculate the SSE as follows:
SSE = SS(Total) SSR = (n 1)×s2Y SSR = 12×3.2847 26.202 = 13.2144
Therefore,
s =
√
SSE
n 2 =
√
13.2144
11
= 1.0960
(c) [4 marks] Test whether there is a linear relationship between Weight and Age.
Clearly state your hypotheses and use a significance level of α = 5%.
Solution:
Past Final Examination 1 Page 11 of 16 STAT7055
The hypotheses are:
H0 : β1 = 0
H1 : β1 6= 0
We know that the standard error of β 1 is:
sβ 1 =
s√
(n 1)s2X
=
1.0960√
12× 12.5769 = 0.0892
Hence our test statistic becomes:
T =
β 1
sβ 1
=
0.4167
0.0892
= 4.6703
We need to compare this to a t-distribution with n 2 = 11 degrees of free-
dom and reject H0 if T > t0.025,11 = 2.201 or T < t0.025,11 = 2.201. Since 4.6703 > 2.201, we reject H0 and conclude that β1 6= 0, that is, there is a linear
relationship between Weight and Age.
(d) [5 marks] Calculate a 95% confidence interval for the expected weight of a turkey
that is 22 weeks old. Would this be wider or narrower than a prediction interval
for the weight of a turkey that is 22 weeks old Why or why not
Solution:
The 95% confidence interval for the expected weight of a turkey that is 22 weeks
old is given by: (
y g ± tα
2
,n 2 × s
√
1
n
+
(xg Xˉ)2
(n 1)s2X
)
where y g = 11.1500, t0.025,11 = 2.201, xg = 22, Xˉ = 25.9231, n = 13, s
2
X =
12.5769 and s = 1.0960. Plugging this all in we get:
(10.1296, 12.1704)
We would expect this confidence interval to be narrower than the corresponding
prediction interval because there is more variability associated with predicting
a particular weight than there is with estimating the expected weight.
Past Final Examination 1 Page 12 of 16 STAT7055
It turns out that the 13 turkeys were not all raised on the same farm. A regression
model with Weight as the dependent variable and Age, NSW and VIC as independent
variables was fitted to the data, where NSW is an indicator variable that equals 1 if the
turkey was raised in New South Wales and VIC is an indicator variable that equals 1 if
the turkey was raised in Victoria. Some regression output is displayed below:
Predictor Coef SE Coef T p-value
Intercept 1.4309 0.6574 2.18 0.0575
Age 0.4868 0.0257 18.91 0.0000
NSW 1.9184 0.2018 9.51 0.0000
VIC 2.1919 0.2114 10.37 0.0000
Analysis of Variance
Source DF SS MS F p-value
Regression 3 38.60575 12.86858
Residual Error 9 0.8112 0.0901
Total 12
(e) [1 mark] Write down the general form of the model.
Solution:
Letting Y and X denote Weight and Age, respectively, the general form of the
model is:
Y = β0 + β1X + β2NSW + β3VIC +
(f) [3 marks] Test the overall significance of the model. Clearly state your hypotheses
and use a significance level of α = 5%.
Solution:
The hypotheses are:
H0 : β1 = β2 = β3 = 0
H1 : Not all coefficient parameters are equal to 0.
The F -statistic is given by:
F =
MSR
MSE
=
12.86858
0.0901
= 142.8255
We need to compare this to an F -distribution with k = 3 numerator degrees of
freedom and n k 1 = 9 denominator degrees of freedom and reject H0 when
Past Final Examination 1 Page 13 of 16 STAT7055
F > F0.05,3,9 = 3.86. Since 142.8255 > 3.86, we reject H0 and we conclude that
the overall regression model is significant.
(g) [3 marks] Calculate the coefficient of determination, R2, and interpret the value.
Solution:
R2 =
SSR
SS(Total)
=
38.60575
38.60575 + 0.8112
= 0.9794
The interpretation is that 97.94% of the variation in Weight is explained by the
regression model.
(h) [4 marks] What is one of the drawbacks of using R2 Calculate the adjusted R2.
Solution:
One of the drawbacks of using R2 in multiple linear regression is that it will
always increase as we add more independent variables to the model, even if
they are not related to Y .
adj R2 = 1
SSE
n k 1
SS(Total)
n 1
= 1
0.8112
9
38.60575+0.8112
12
= 0.9726
(i) [3 marks] What do you conclude about the relationship between Weight and Age
Clearly state your hypotheses and use a significance level of α = 5%. Interpret the
coefficient parameter for Age.
Solution:
The hypotheses are:
H0 : β1 = 0
H1 : β1 6= 0
From the output, we see that the p-value for this test is 0.0000, which is much
less than 0.05, so we reject H0 and conclude that β1 6= 0. That is, we conclude
that once the other independent variables have been included in the model, there
is a significant linear relationship between Age and Weight. The interpretation
of the coefficient for Age is that it represents the expected change in Weight
arising from a one unit change in Age, when all other independent variables are
held fixed.
(j) [2 marks] Test whether a different intercept is needed for turkeys from NSW.
Clearly state your hypotheses and use a significance level of α = 5%.
Past Final Examination 1 Page 14 of 16 STAT7055
Solution:
To see if a different intercept is need for turkeys from NSW, we are testing:
H0 : β2 = 0
H1 : β2 6= 0
From the output, we see that the p-value for this test is 0.0000, which is much
less than 0.05, so we reject H0 and conclude that β2 6= 0. That is, a different
intercept is needed for turkeys from NSW.
(k) [2 marks] Test whether a different intercept is needed for turkeys from VIC.
Clearly state your hypotheses and use a significance level of α = 5%.
Solution:
To see if a different intercept is need for turkeys from VIC, we are testing:
H0 : β3 = 0
H1 : β3 6= 0
From the output, we see that the p-value for this test is 0.0000, which is much
less than 0.05, so we reject H0 and conclude that β3 6= 0. That is, a different
intercept is needed for turkeys from VIC.
(l) [5 marks] Suppose we want to test whether a different coefficient parameter for
Age is needed for turkeys from NSW and whether a different coefficient parameter
for Age is needed for turkeys from VIC. What additional variables do you need in
order to test these claims State clearly the general form of the model that you
would fit and the hypothesis tests you would conduct.
Solution:
We need to include as independent variables the interaction between Age and
NSW and the interaction between Age and VIC. The model that we would fit
is:
Y = β0 + β1X + β2NSW + β3VIC + β4(X × NSW) + β5(X × VIC) +
The two sets of hypotheses that we would need to test are:
H0 : β4 = 0
H1 : β4 6= 0
and H0 : β5 = 0
H1 : β5 6= 0
Past Final Examination 1 Page 15 of 16 STAT7055
(m) [1 mark] Based on your model from part (l), write down the form of the model
for turkeys from NSW.
Solution:
For turkeys from NSW, we have NSW = 1 and VIC = 0. Therefore,
Y = β0 + β1X + β2 × 1 + β3 × 0 + β4(X × 1) + β5(X × 0) +
= β0 + β2 + (β1 + β4)X +
END OF EXAMINATION
Past Final Examination 1 Page 16 of 16 STAT7055