Business Statistics 11 Assignment: Statistical Analysis and Report
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Homework Assignment
AI Summary
This business statistics assignment analyzes data from the telecommunication and retail industries. It includes tables and plots comparing quarterly opening prices, market share, and turnover trends. The assignment covers statistical concepts such as probability calculations, hypothesis testing, and z-scores, with detailed explanations and formulas. The student demonstrates statistical analysis using Excel and provides interpretations of the results, including return trends and market capitalization. The assignment addresses various statistical problems, including calculating probabilities, and interpreting results related to urban school enrollment and brand identification, providing a comprehensive overview of statistical techniques in a business context.

Running head: BUSINESS STATISTICS
BUSINESS STATISTICS
Name of the Student
Name of the University
Author Note
Course ID:
BUSINESS STATISTICS
Name of the Student
Name of the University
Author Note
Course ID:
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1BUSINESS STATISTICS
Table of Contents
Answer to 1:...............................................................................................................................2
Answer to 2:...............................................................................................................................4
Answer to 3:...............................................................................................................................6
Answer to 4:...............................................................................................................................6
Answer to 5:...............................................................................................................................8
References................................................................................................................................11
Table of Contents
Answer to 1:...............................................................................................................................2
Answer to 2:...............................................................................................................................4
Answer to 3:...............................................................................................................................6
Answer to 4:...............................................................................................................................6
Answer to 5:...............................................................................................................................8
References................................................................................................................................11

2BUSINESS STATISTICS
Answer 1:
a) Table 1: Quarterly opening prices of the telecommunication industry TLS and TPM
The table above portrays the quarterly opening and unadjusted prices of TPM and TLS for a
period of 8 years.
Table 2: Steam and Leaf plot display table
Answer 1:
a) Table 1: Quarterly opening prices of the telecommunication industry TLS and TPM
The table above portrays the quarterly opening and unadjusted prices of TPM and TLS for a
period of 8 years.
Table 2: Steam and Leaf plot display table
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3BUSINESS STATISTICS
The chart above displays the steam and leaf plot of TLS and TPM companies of Australia.
With leaf of TPM in the left side and TLS leaf on the right.
b)
Figure 1: Comparison between prices of TPM and TLS Telecommunication Company
c)
Table 3: List of market share of telecommunication companies listed in ASX-200
Figure 2: Market capitalisation of each telecommunication industry in Australia
d)
The chart above displays the steam and leaf plot of TLS and TPM companies of Australia.
With leaf of TPM in the left side and TLS leaf on the right.
b)
Figure 1: Comparison between prices of TPM and TLS Telecommunication Company
c)
Table 3: List of market share of telecommunication companies listed in ASX-200
Figure 2: Market capitalisation of each telecommunication industry in Australia
d)
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Figure 3: Return trend of the companies
The above trend as well as the risk and return table suggests that it is liable and worthy to
invest in TPM. This is because TPM generates higher return in the telecommunication
industry.
Answer to 2:
a)
Table 4: Mean and SD of the retail group
b)
Table 5: Calculation for statistical variables
Figure 3: Return trend of the companies
The above trend as well as the risk and return table suggests that it is liable and worthy to
invest in TPM. This is because TPM generates higher return in the telecommunication
industry.
Answer to 2:
a)
Table 4: Mean and SD of the retail group
b)
Table 5: Calculation for statistical variables

5BUSINESS STATISTICS
c)
Figure 4: Box Plot and Whiskers plot for retailing group
d)
Figure 5: Turnover trend of retail companies
The trend line indicates that basic goods, cafes and miscellaneous trading generates greater
monthly turnover than the departmental stores. Additionally, clothing and footwear generates
relatively higher revenue than the departmental chain.
c)
Figure 4: Box Plot and Whiskers plot for retailing group
d)
Figure 5: Turnover trend of retail companies
The trend line indicates that basic goods, cafes and miscellaneous trading generates greater
monthly turnover than the departmental stores. Additionally, clothing and footwear generates
relatively higher revenue than the departmental chain.
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Answer to 3:
a) Probability of a person randomly selected person living in Queensland is 0.28.
b) Probability of a randomly selected person living in Tasmania will have access to internet
for social networking is 0.21.
c) Probability of a pe4rson living in Victoria will have access to internet for educational
purpose is 0.08.
d) Probability of having a chance to enjoy both banking and entertainment facilities is 0.41.
Answer to 4:
a) Provided, mean (x) = 51 minutes and, standard deviation (SD) = 4 minutes.
The value of Z for upper 5% is -1.96
The formula is
Therefore, the value of population mean is
μ=58.84
Answer to 3:
a) Probability of a person randomly selected person living in Queensland is 0.28.
b) Probability of a randomly selected person living in Tasmania will have access to internet
for social networking is 0.21.
c) Probability of a pe4rson living in Victoria will have access to internet for educational
purpose is 0.08.
d) Probability of having a chance to enjoy both banking and entertainment facilities is 0.41.
Answer to 4:
a) Provided, mean (x) = 51 minutes and, standard deviation (SD) = 4 minutes.
The value of Z for upper 5% is -1.96
The formula is
Therefore, the value of population mean is
μ=58.84
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b) Given, the height varies between 110 cm to 135 cm.
Therefore, the value of p is 0.44
c) The brand is selected by the students in an equally likely way.
i. Total Correct identification, p0 =50% = 0.5
Amount of correct identification in sample, p = 65% = 0.65
The formula is
Therefore, Z = 4.75
Therefore, P (p>0.65) = P (z > 4.75)
Therefore, P equals 0.0001
ii. Probability of the sample lying between the correct identification ranges of 60 to 70
percent is P (0.6 < p < 0.7)
b) Given, the height varies between 110 cm to 135 cm.
Therefore, the value of p is 0.44
c) The brand is selected by the students in an equally likely way.
i. Total Correct identification, p0 =50% = 0.5
Amount of correct identification in sample, p = 65% = 0.65
The formula is
Therefore, Z = 4.75
Therefore, P (p>0.65) = P (z > 4.75)
Therefore, P equals 0.0001
ii. Probability of the sample lying between the correct identification ranges of 60 to 70
percent is P (0.6 < p < 0.7)

8BUSINESS STATISTICS
P (z<6.33) - P(z<3.16)
= 0.9997-0.988
= .0117
Answer to 5:
a) Given, about 62 percent of students attends school in an urban district. If 290 out of 500
students enrols in such schools, the probability of enrolment is
The formula for z is
P (z<6.33) - P(z<3.16)
= 0.9997-0.988
= .0117
Answer to 5:
a) Given, about 62 percent of students attends school in an urban district. If 290 out of 500
students enrols in such schools, the probability of enrolment is
The formula for z is
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9BUSINESS STATISTICS
Thus, P ( z ≥−1.84 ) = 0.967
b) Given, p0=0.5
p= 185
500 =0.37
Let
Therefore,
As the p-value is, lower than 0.01 we reject null-hypothesis and accept the alternative.
c) Mean of the given values = (210 +125 + 80 +131 + 105 +65 +84+118+95)/ 9
=112.56
Thus, P ( z ≥−1.84 ) = 0.967
b) Given, p0=0.5
p= 185
500 =0.37
Let
Therefore,
As the p-value is, lower than 0.01 we reject null-hypothesis and accept the alternative.
c) Mean of the given values = (210 +125 + 80 +131 + 105 +65 +84+118+95)/ 9
=112.56
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SD2 = (210-112.56)2 + (125-112.56) 2 + (80-112.56) 2 + (131-112.56) 2 + (105-112.56) 2 + (65-
112.56) 2 + (84-112.56) 2 + (118-112.56) 2 + (95-112.56) 2/ (9-1)
= (9494.55+154.75+1060.15+340.03+57.15+2261.95+815.67+29.59+308.35)/8
=1815.27
=√1815.27
SD= 42.6,
n=9
SD2 = (210-112.56)2 + (125-112.56) 2 + (80-112.56) 2 + (131-112.56) 2 + (105-112.56) 2 + (65-
112.56) 2 + (84-112.56) 2 + (118-112.56) 2 + (95-112.56) 2/ (9-1)
= (9494.55+154.75+1060.15+340.03+57.15+2261.95+815.67+29.59+308.35)/8
=1815.27
=√1815.27
SD= 42.6,
n=9

11BUSINESS STATISTICS
References
Black, K. U. (2019). Business statistics: for contemporary decision making. Wiley.
David, M. (2017). Statistics for Managers, Using Microsoft Excel. Pearson Education India.
Jaggia, S., Kelly, A., Salzman, S., Olaru, D., Sriananthakumar, S., Beg, R., & Leighton, C.
(2016). Essentials of Business Statistics: communicating with numbers. McGrawhill
Education.
Siegel, A. (2016). Practical business statistics. Academic Press.
References
Black, K. U. (2019). Business statistics: for contemporary decision making. Wiley.
David, M. (2017). Statistics for Managers, Using Microsoft Excel. Pearson Education India.
Jaggia, S., Kelly, A., Salzman, S., Olaru, D., Sriananthakumar, S., Beg, R., & Leighton, C.
(2016). Essentials of Business Statistics: communicating with numbers. McGrawhill
Education.
Siegel, A. (2016). Practical business statistics. Academic Press.
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