MAST20005-2020
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20005
Subject
Statistics
Date
Apr 3, 2024
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Semester 2 Assessment, 2020
School of Mathematics and Statistics
MAST20005 Statistics
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MAST20005 Statistics
Semester 2, 2020
Question 1 (11 marks)
We will fit the following regression model:
Y
i
=
α
+
βx
i
+
i
,
i
∼
N(0
, σ
2
),
i
= 1
, . . . ,
10.
Consider the following R output:
> summary(x)
Min. 1st Qu.
Median
Mean 3rd Qu.
Max.
6.18
29.21
60.10
55.15
83.90
94.47
> sd(x)
[1] 31.56569
> sort(x)
[1]
6.18 20.17 26.55 37.21 57.29 62.91 66.08 89.84 90.82 94.47
> f = lm(y ~ x)
> summary(f)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
1.45528
3.05901
0.476
0.647
x
4.01616
0.04874
82.394 5.25e-13 ***
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 4.616 on 8 degrees of freedom
Multiple R-squared:
0.9988,
Adjusted R-squared:
0.9987
F-statistic:
6789 on 1 and 8 DF,
p-value: 5.251e-13
(a) For each of the following quantities, state or calculate its value if possible, or otherwise
explain why it is not possible.
(i)
x
(4
.
5)
(ii) ¯
x
(iii)
α
(iv) ˆ
σ
2
(v) Sample correlation coefficient,
r
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MAST20005 Statistics
Semester 2, 2020
(b) For each of the following pairs of hypotheses, carry out the test if it is possible, using a
5% significance level, or otherwise explain what further information you need in order
to do it.
(i)
H
0
:
α
= 0 versus
H
1
:
α
6
= 0
(ii)
H
0
:
α
= 0 versus
H
1
:
α >
0
(iii)
H
0
:
α
= 4 versus
H
1
:
α
6
= 4
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MAST20005 Statistics
Semester 2, 2020
Question 2 (8 marks)
The School of Mathematics and Statistics conducted a survey about the time spent commuting
by students each day. A total of 40 students were surveyed and the responses are summarised
in the table below.
0–1 hours
1–2 hours
2–3 hours
3+ hours
Number of students
10
16
10
4
(a) Let
p
be the proportion of students taking 1–2 hours to commute.
Calculate a 95%
confidence interval for
p
.
(b) Let
p
1
, p
2
, p
3
, p
4
be the proportion of all students (not just those surveyed) that would
be in each of the groups defined in the table above.
Using the survey data and a
10% significance level, test whether the commuting time groups follow the distribution
p
1
=
p
2
=
p
3
=
p
4
= 1
/
4.
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MAST20005 Statistics
Semester 2, 2020
(c) Using a 5% significance level, test whether the commuting time follows the exponential
distribution with the mean equal to 2.
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MAST20005 Statistics
Semester 2, 2020
Question 3 (11 marks)
Let
X
be the length of a tomato seed (in millimeters). Ten randomly selected seeds are mea-
sured, giving the following measurements:
6.2
8.4 11.2
5.7
7.8
8.3
9.4
7.5 12.0
7.7
For these data, we have ¯
x
= 8
.
42 and
s
= 1
.
97. You may assume that
X
∼
N(
μ, σ
2
).
(a) Calculate a 95% confidence interval for
μ
.
(b) Calculate a 95% confidence interval for
σ
2
.
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MAST20005 Statistics
Semester 2, 2020
(c) Let
X
*
be the length of a new seed. Assuming
σ
= 2 is known, calculate a 95% prediction
interval for
X
*
.
(d) Let
X
*
be the length of a new seed. If
σ
is unknown, calculate a 95% prediction interval
for
X
*
.
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MAST20005 Statistics
Semester 2, 2020
Question 4 (16 marks)
Let
X
1
, . . . , X
n
be a random sample from the following distribution with parameter
λ
and pdf:
f
(
x
|
λ
) =
λe
-
λ
(
x
-
2)
,
x
>
2
.
(a) Determine a sufficient statistic for
λ
.
(b) Find the maximum likelihood estimator (MLE) of
λ
.
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MAST20005 Statistics
Semester 2, 2020
(c) Is this MLE unbiased?
(d) Find the Cram´
er–Rao lower bound for unbiased estimators of
λ
.
(e) Find the
p
quantile,
π
p
.
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MAST20005 Statistics
Semester 2, 2020
(f) Find the pdf of
X
(2)
.
(g) Let
n
= 10 and
λ
= 1
/
8, find Pr(
X
(2)
>
4).
Page 10 of 19 pages
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MAST20005 Statistics
Semester 2, 2020
Question 5 (13 marks)
Tingjin wants to compare three types of fertiliser. Each type of fertiliser is applied to five plants,
and the growth of each plant (in centimetres) is measured after one month.
Plant
Statistics (per-fertiliser)
Fertiliser
1
2
3
4
5
Mean
Standard deviation
Type A
4.5
2.7
5.7
7.5
4.0
4.88
1.82
Type B
3.7
0.4
4.5
3.1
0.9
2.52
1.79
Type C
8.5
2.6
6.3
4.3
3.8
5.10
2.32
(a) Assuming a normal distribution and equal variances for the the plant growth, test
whether the growth with Type A is larger than that with Type B (
α
= 0
.
05).
(b) Use the sign test with
α
= 0
.
05 to test if the median plant growth with Type A is larger
than 3 centimetres.
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MAST20005 Statistics
Semester 2, 2020
(c) Consider a one-way ANOVA model, please complete the following ANOVA table:
Source
df
SS
MS
F
Treatment (fertiliser type)
Error
Total
(d) Is there evidence of any differences in the average plant growth between the three types
of fertiliser?
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MAST20005 Statistics
Semester 2, 2020
Question 6 (7 marks)
Let
X
1
, . . . , X
n
∼
N(0
,
1
/θ
) be a random sample.
(a) Let
f
(
θ
) =
e
-
θ
with
θ
>
0 be the prior distribution for
θ
.
Derive the posterior distribution for
θ
.
(b) Find the posterior mean for
θ
.
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MAST20005 Statistics
Semester 2, 2020
(c) Find a conjugate prior distribution for
θ
.
(d) Let
f
(
θ
) = 1
/θ
with
θ
>
0 be the prior distribution for
θ
.
Explain why
f
(
θ
) is called an improper prior.
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MAST20005 Statistics
Semester 2, 2020
Question 7 (14 marks)
Consider the following regression model without an intercept term:
Y
i
=
βx
i
+
i
,
i
= 1
, . . . , n,
where
1
, . . . ,
n
∼
N(0
, σ
2
) are independent random variables.
(a) Find the distribution of
Y
i
.
(b) Find the estimator of
β
using least squares estimation.
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MAST20005 Statistics
Semester 2, 2020
(c) Show that this estimator (from part (b)) is unbiased.
(d) Find the joint density function of
Y
1
, . . . , Y
n
.
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MAST20005 Statistics
Semester 2, 2020
(e) Assuming
β
is known, find the maximum likelihood estimator of
σ
2
.
(f) Is this estimator (from part (e)) unbiased? Justify your answer.
End of Exam — Total Available Marks = 80
Turn the page for appended material
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MAST20005 Statistics
Semester 2, 2020
Appendix
Distributions
•
The pdf of
X
∼
N(
μ, σ
2
) is,
f
(
x
) =
1
√
2
πσ
exp
-
(
x
-
μ
)
2
2
σ
2
.
•
The pdf of an exponential distribution with mean equal to 1
/λ
is,
f
(
x
) =
λ
exp
{-
λx
}
,
(
x
>
0)
.
•
The pdf of
X
∼
Gamma(
α, β
) is,
f
(
x
) =
β
α
Γ(
α
)
x
α
-
1
e
-
xβ
,
(
x
>
0)
and it has mean
E
(
X
) =
α/β
and var(
X
) =
α/β
2
.
•
The pdf of
X
∼
Beta(
α, β
) is,
f
(
x
) =
Γ(
α
+
β
)
Γ(
α
)Γ(
β
)
x
α
-
1
(1
-
x
)
β
-
1
,
(0
6
x
6
1)
and it has mean
E
(
X
) =
α/
(
α
+
β
) and var(
X
) =
αβ/
{
(
α
+
β
)
2
(
α
+
β
+ 1)
}
.
R output
> p1 <- c(0.01, 0.025, 0.05, 0.1, 0.9, 0.95, 0.975, 0.99)
> qnorm(p1)
[1] -2.326 -1.960 -1.645 -1.282
1.282
1.645
1.960
2.326
> qt(p1, df = 7)
[1] -2.998 -2.365 -1.895 -1.415
1.415
1.895
2.365
2.998
> qt(p1, df = 8)
[1] -2.896 -2.306 -1.860 -1.397
1.397
1.860
2.306
2.896
> qt(p1, df = 9)
[1] -2.821 -2.262 -1.833 -1.383
1.383
1.833
2.262
2.821
> qt(p1, df = 10)
[1] -2.764 -2.228 -1.812 -1.372
1.372
1.812
2.228
2.764
> qt(p1, df = 11)
[1] -2.718 -2.201 -1.796 -1.363
1.363
1.796
2.201
2.718
> qt(p1, df = 12)
[1] -2.681 -2.179 -1.782 -1.356
1.356
1.782
2.179
2.681
> qt(p1, df = 13)
[1] -2.650 -2.160 -1.771 -1.350
1.350
1.771
2.160
2.650
> qt(p1, df = 14)
[1] -2.624 -2.145 -1.761 -1.345
1.345
1.761
2.145
2.624
> qt(p1, df = 26)
[1] -2.479 -2.056 -1.706 -1.315
1.315
1.706
2.056
2.479
> qt(p1, df = 27)
[1] -2.473 -2.052 -1.703 -1.314
1.314
1.703
2.052
2.473
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MAST20005 Statistics
Semester 2, 2020
> qchisq(p1, df = 1)
[1] 0.0001571 0.0009821 0.0039321 0.0157908 2.7055435 3.8414588 5.0238862 6.6348966
> qchisq(p1, df = 2)
[1] 0.02010 0.05064 0.10259 0.21072 4.60517 5.99146 7.37776 9.21034
> qchisq(p1, df = 3)
[1]
0.1148
0.2158
0.3518
0.5844
6.2514
7.8147
9.3484 11.3449
> qchisq(p1, df = 7)
[1]
1.239
1.690
2.167
2.833 12.017 14.067 16.013 18.475
> qchisq(p1, df = 8)
[1]
1.646
2.180
2.733
3.490 13.362 15.507 17.535 20.090
> qchisq(p1, df = 9)
[1]
2.088
2.700
3.325
4.168 14.684 16.919 19.023 21.666
> qchisq(p1, df = 10)
[1]
2.558
3.247
3.940
4.865 15.987 18.307 20.483 23.209
> qf(p1, 1, 13)
[1] 0.0001632 0.0010206 0.0040868 0.0164196 3.1362051 4.6671927 6.4142543 9.0738057
> qf(p1, 1, 14)
[1] 0.0001628 0.0010178 0.0040756 0.0163739 3.1022134 4.6001099 6.2979386 8.8615927
> qf(p1, 2, 12)
[1] 0.01006 0.02537 0.05151 0.10629 2.80680 3.88529 5.09587 6.92661
> qf(p1, 2, 13)
[1] 0.01006 0.02537 0.05150 0.10622 2.76317 3.80557 4.96527 6.70096
> qf(p1, 3, 11)
[1] 0.03686 0.06957 0.11411 0.19148 2.66023 3.58743 4.63002 6.21673
> qf(p1, 3, 12)
[1] 0.03697 0.06975 0.11436 0.19173 2.60552 3.49029 4.47418 5.95254
> qf(p1, 3, 13)
[1] 0.03706 0.06991 0.11456 0.19195 2.56027 3.41053 4.34718 5.73938
> pbinom(0:5, 5, 0.5)
[1] 0.03125 0.18750 0.50000 0.81250 0.96875 1.00000
> pbinom(0:5, 10, 0.5)
[1] 0.0009766 0.0107422 0.0546875 0.1718750 0.3769531 0.6230469
> pbinom(0:5, 9, 0.5)
[1] 0.001953 0.019531 0.089844 0.253906 0.500000 0.746094
> pbinom(0:10, 10, 0.25)
[1] 0.056 0.244 0.526 0.776 0.922 0.980 0.996 1.000 1.000 1.000 1.000
> qgamma(p1, 8, 1)
[1]
2.906
3.454
3.981
4.656 11.771 13.148 14.423 16.000
> qgamma(p1, 9, 1)
[1]
3.507
4.115
4.695
5.432 12.995 14.435 15.763 17.403
> qgamma(p1, 10, 1)
[1]
4.130
4.795
5.425
6.221 14.206 15.705 17.085 18.783
Page 19 of 19 pages
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The area of the whole pattern is 1600 cm2. They need to know the length of the pennant base.
80 cm
How long is the base of each pennant?
cm
2 of 4 . Oo0
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1st blank: 0.6267, 0.5006, 33, 0.7843
2nd blank: 1.44, 2.25, 1.15, 1.80
3rd blank: 0.7843, 0.6267, 0.5006, 0.4
4th blank: 2.25, 1.80, 1.15, 1.44
5th blank: larger or smaller
6th blank: 0.2689, 0.1293, 0.0145, 0.0378
7th blank: 0.0378, 0.1293, 0.0100, 0.2612
8th blank: flatter and more spread out or more like the z distribution
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