Statistics Assignment: Data Analysis, Probability & Interpretation

This statistics assignment solution includes detailed answers to questions covering a range of statistical concepts. The assignment begins with an analysis of stock prices for WBC and NAB, including calculations of frequency, relative frequency, and graphical representations of market capitalization. It then moves on to analyze new motor vehicle sales data by state, calculating descriptive statistics such as mean, standard deviation, minimum, maximum, and quartiles. The assignment further explores probability concepts, including calculating the probability of household internet access across different states and years. Finally, it addresses problems involving normal and uniform distributions, as well as hypothesis testing, providing step-by-step calculations and interpretations. The solution also uses Excel to assist in statistical calculations.

Running head: Statistics 1
Statistics
By:
Student ID:
Course No:
Tutor:
Date:

Statistics 2
Question 1
a)
Prices ($)
2009 WBC
NA
B
Jan 17 19
April 19 19
July 19 21
October 20 21
2010 Jan 20 22
April 21 22
July 21 22
October 22 23
2011 Jan 22 24
April 22 24
July 23 24
October 24 24
2012 Jan 25 24
April 25 25
July 26 25
October 26 26
2013 Jan 28 26
April 28 26
July 29 28
October 30 28
2014 Jan 30 29
April 30 29
July 30 29
October 31 29
2015 Jan 32 30
April 32 31
July 32 31
October 32 31
2016 Jan 32 32
April 33 32
July 33 32
October 33 33
2017 Jan 34 33
April 34 33
July 35 34
October 39 37
(ASX, n.d).

Statistics 3
WB
C
NA
B
9 9 7 1 9 9
9 8 8 6 6 5 5 4 3 2 2 2 1 1 0 0 2 1 1 2 2 2 3 4 4 4 4 4 5 5 6 6 6 8 8 9 9 9 9
9 5 4 4 3 3 3 2 2 2 2 2 1 0 0 0 0 3 0 1 1 1 2 2 2 3 3 3 4 7
b)
Prices ($)
WBC NAB
17 19
19 19
19 21
20 21
20 22
21 22
21 22
22 23
22 24
22 24
23 24
24 24
25 24
25 25
26 25
26 26
28 26
28 26
29 28
30 28
30 29
30 29
30 29
31 29
32 30
32 31
32 31
32 31

Statistics 4
32 32
33 32
33 32
33 33
34 33
34 33
35 34
39 37
Class
Interv
al
(NAB
&
WBC
prices) Lower Upper
Midpoi
nt
Frequenc
y (NAB)
Relative
Frequenc
y (NAB)
Frequenc
y (WBC)
Relative
Frequenc
y (WBC)
14-16 14 16 15 0 0.00 0 0.00
16-18 16 18 17 0 0.00 1 0.03
18-20 18 20 19 2 0.06 3 0.08
20-22 20 22 21 3 0.08 6 0.17
22-24 22 24 23 6 0.17 1 0.03
24-26 24 26 25 5 0.14 5 0.14
26-28 26 28 27 4 0.11 2 0.06
28-30 28 30 29 5 0.14 3 0.08
30-32 30 32 31 5 0.14 4 0.11
32-34 32 34 33 5 0.14 8 0.22
34-36 34 36 35 0 0.00 2 0.06
36-38 36 38 37 1 0.03 0 0.00
38-40 38 40 39 0 0.00 1 0.03
Totals 36 1.0 36 1.00

Statistics 5
15 17 19 21 23 25 27 29 31 33 35 37 39
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
8
9
NAB & WBC Share price
Frequency
Share prices
c)
Companie
s
Market Capital (Million
$Aus)
CBA 129,320.00
ANZ 75,150.00
HGH 742,200.00
AFG 263,150.00
BEN 5,500.00
CYB 4,990.00

Statistics 6
CBA ANZ HGH AFG BEN CYB
-
100,000.00
200,000.00
300,000.00
400,000.00
500,000.00
600,000.00
700,000.00
800,000.00
129,320.00
75,150.00
742,200.00
263,150.00
5,500.00 4,990.00
Companies
Market capital in Mill $Aus
d)
NAB has a P/E ratio of 12.370 whereas WBC has a P/E ratio of 11.010. An investor is
recommended to invest in WBC with low P/E ratio. This is because a low P/E value is an
indication that the growth potential of that company is still not known to the market hence, a
good opportunity to venture into it (Hirsa, & Neftci, 2013; Satchell, & Knight, 2011). On the
other hand, a relatively high P/E value is a warning of an over-stocked price, hence not
advisable for investment (Baker & Powell, 2009; Hjalmarsson, 2010).
QUESTION 2
a) , b)
Month State
NSW VIC QLD SA WA TAS NT ACT
Jan-16 28392 22426 17193 5208 7861 1298 626 1509
Feb-16 32633 26782 18711 5988 8725 1319 851 1528
Mar-
16
35963 27716 21470 6087 9054 1515 1023 1550
Apr- 29391 23963 17753 5174 7601 1313 969 1563

Statistics 7
16
May-
16
32867 26125 19565 5949 7935 1503 1180 1560
Jun-16 43449 34262 27270 7613 10749 1853 1240 1542
Jul-16 30219 25517 18445 5210 8234 1389 831 1524
Aug-
16
31788 27096 18062 5785 8144 1743 881 1519
Sep-16 34424 28766 19702 6551 8668 2074 880 1531
Oct-16 31850 27546 16870 5450 7580 1937 713 1552
Nov-
16
32955 28991 18733 6259 7657 1903 808 1570
Dec-
16
33632 27065 19252 6464 8026 1918 734 1573
Mean
33,130.2
5
27,187.9
2
19,418.8
3
5,978.1
7
8,352.83 1,647.0
8
894.67 1,543.4
2
Std
3,874.43 2,913.56 2,762.81 705.48 889.27 286.47 182.79 21.12
Min
28,392.0
0
22,426.0
0
16,870.0
0
5,174.0
0
7,580.00 1,298.0
0
626.00 1,509.0
0
Q1
31,395.7
5
25,973.0
0
17,984.7
5
5,390.0
0
7,810.00 1,371.5
0
789.50 1,527.0
0
Media
n 32,750.0
0
27,080.5
0
18,722.0
0
5,968.5
0
8,085.00 1,629.0
0
865.50 1,546.0
0
Q3
33,830.0
0
27,978.5
0
19,599.2
5
6,310.2
5
8,431.63 1,906.7
5
982.50 1,560.7
5
Max
43,449.0
0
34,262.0
0
27,270.0
0
7,613.0
0
10,749.0
0
2,074.0
0
1,240.0
0
1,573.0
0
(ABS, 2017)
c.

Statistics 8
NSW VIC QLD SA WA TAS NT ACT
Min 28392 22426 16870 5174 7580 1298 626 1509
Q1
3003.7
5 3547
1114.7
5 216 230 73.5 163.5 18
Media
n
1354.2
5 1107.5 737.25 578.5 275 257.5 76 19
Q3 1080 898 877.25 341.75 346.625 277.75 117 14.75
Max 9619 6283.5
7670.7
5
1302.7
5
2317.37
5 167.25 257.5 12.25
1 2 3 4 5 6 7 8
0
1000
2000
3000
4000
5000
6000
New MotorVehicle sales By state
State
Motor Vehicle sales
d. NSW had the highest sales of new motor vehicles (43,449), followed by VIC (24, 262) and
NT recorded the least sales (1,240). There was a substantial decline in the trend of the
number of cars sold between NSW, VIC, and QLD states. The trend of decline between SA
and WA was relatively minimal, and the trend among the remaining (TAS, NT, and ACT)
was very minimal and relatively insignificant.
Question 3
a.
Households with internet access
2008-
09
2010-
11
2012-
13
2014-
15
2016-
17
000 000 000 000 000
State or territory Total
New South Wales 1890.2 2163.6 2274.5 2407.9 2439.9 11176.

Statistics 9
1
Victoria 1468.6 1683.5 1830.6 1934.2 2008.2 8925.1
Queensland 1182.1 1347.3 1517 1552.4 1591.9 7190.7
South Australia 431.8 496.5 554 565.1 575.5 2622.9
Western Australia 626.1 714 814.8 843.6 859.7 3858.2
Tasmania 125.7 146.2 164.6 172 177.7 786.2
Northern Territory 45.5 50.7 58.8 58.1 57.6 270.7
Australian Capital
Territory
108.2 121.8 128.9 137.2 140.1 636.2
Total 5878.2 6723.6 7343.2 7670.5 7850.6 35466.
1
(ABS, 2018)
Probability that a randomly selected household lives in Victoria
= 8925.1/ 35466.1
= 0.251652
=0.252
b. probability that a randomly selected household lives in Tasmania and has an internet
access in the year 2010-11
P (household lives in Tasmania) x (has internet access in 2010-11)
= (786.2/ 35466.1) x (146.2/ 6723.6)
= 0.0221676 x 0.0217443
= 0.043912
= 0.044
c. household has an internet access in 2012-13, what is the probability that a randomly
selected household lives in New South Wales?
P (household has internet access in 2012-13) + P (household selected lives in NSW)
= (7343.2/ 35466.1) + (11176.1/ 35466.1)
= 0.2070484 + 0.3151206
= 0.5221691
= 0.522

Statistics
10
d. probability that a randomly selected household has an internet access in 2010-11 or
2012-13?
P (household has internet in 2010-11) + P (household has internet in 2012-13)
= (6723.6/ 35466.1) + (7343.2/ 35466.1)
= 0.1895782 + 0.2070484
= 0.3966266
=0.397
Question 4
a.
Mean = 73 mins
sd = 8 mins
upper 5% = ?
X ∼ N (73, 82)
Upper 5% = 3.65
Z =(76.65-73)/8
= 0.456
b.
let X = number of degrees Celsius, a = 12, b = 25, X ∼ U (12, 25).
Probability density function, f (X) = 1 /25-12= 1/ 13 for 12 ≤ X ≤ 25.
Find P (X < 21)
= (21-12) x 1/13
= 0.692
c.
i. probability that the sample percentage of correct identifications is greater than 65
P ( X>0.65) = (base) (height) = 0.35 x 1
= 0.35
ii. probability that the sample will have between 50% and 60% of the identifications correct
Let X represent the correct sample tested, X ∼ U (0.5, 0.6)

Statistics
11
Probability that the selected sample falls between 50% and 60%
f (X) = 1 /0.6 - 0.5 = 1/ 0.1 for 0.5 ≤ X ≤ 0.6
= 0.1 x 1/0.1
= 1
(Spiegel, Schiller, Srinivasan, & LeVan, 2009)
Question 5
a.
let X = successful trials, a =0.16, b= 0.71
X ∼ U (0.16, 0.71)
Probability density function, f (X) = 1 /0.71-0.16= 1/ 0.55 for 0.16 ≤ X ≤ 0.71.
Find P (X >= 0.7)
= (0.7 +0.16) x 1/0.55
0.86/0.55
= 1.564
b. cancelling flight after booking
210/500 = 0.42
let X cancelled bookings, a =0.42, b= 0.5
X ∼ U (0.42, 0.5)
Find P (0.42 <=X >= 0.5)
= 0.08
c.
X
x-
101.22
2
(x-
101.22
2)
110
8.7777
78
77.049
38
123
21.777
78
474.27
16
80 -
21.222
450.38
27

Statistics
12
2
131
29.777
78
886.71
6
105
3.7777
78
14.271
6
65
-
36.222
2
1312.0
49
84
-
17.222
2
296.60
49
118
16.777
78
281.49
38
95
-
6.2222
2
38.716
05
total 911
3831.5
56
mean
101.22
22
3831.556/n-1
3831.556/8
= 478.9444
= sqrt (478.9444)
= 21.8848
Therefore,
Mean = 101.22
Sd = 21.88
Df = n-1
= 9-1
=8
Alpha level
= (1-0.95)/2
= 0.025
From t-distribution table, df =8, alpha = 0.025
= 2.306

Statistics
13
Dividing sd by the square root of sample size
21.88/√(9)
= 7.205
Multipliying by t-value
7.295 x 2.306 = 16.822
Lower end = 101.22 -16.822
= 84.398
Upper end = 101.22 +16.822
=118.042
CI (84.398, 118.042)

Statistics
14
References
Australian Bureau of Statistics (ABS), (2017). New motor vehicles sales by State between
Jan
2016 and Dec 2017. Retrieved from
http://www.abs.gov.au/AUSSTATS/abs@.nsf/DetailsPage/9314.0December
%202017?OpenDocument
Australian Bureau of Statistics (ABS), (2018). Household Use of Information Technology,
Australia, 2016-17. Retrieved from
http://www.abs.gov.au/ausstats/abs@.nsf/mf/8146.0
Australian Security Exchange (ASX). (n.d). Prices and Research. Retrieved from
https://www.asx.com.au/asx/share-price-research/company/NAB
Baker, H. K., & Powell, G. (2009). Understanding financial management: A practical guide.
John Wiley & Sons.
Hirsa, A., & Neftci, S. N. (2013). An introduction to the mathematics of financial derivatives.
Academic Press.
Hjalmarsson, E. (2010). Predicting global stock returns. Journal of Financial and
Quantitative Analysis, 45(1), 49-80.
Satchell, S., & Knight, J. (2011). Forecasting volatility in the financial markets. Elsevier.
Spiegel, M. R., Schiller, J. J., Srinivasan, R. A., & LeVan, M. (2009). Probability and
statistics (Vol. 2). New York: Mcgraw-hill.

Statistics
15