Forecasting Sales Revenue for Anita Limited in 2021

This question involves using a linear trend equation to forecast the sales revenue of Anita Limited for the year 2021 based on their annual sales revenue from 2015 to 2020.

Student Number: (enter on the line below)
Student Name: (enter on the line below)
HI6007
STATISTICS FOR BUSINESS DECISIONS
FINAL
TRIMESTER 1, 2021
Assessment Weight: 50 total marks
Instructions:
 All questions must be answered by using the answer boxes provided in this paper.
 Completed answers must be submitted to Blackboard by the published due date
and time.
Submission instructions are at the end of this paper.
Purpose:
This assessment consists of six (6) questions and is designed to assess your level of
knowledge of the key topics covered in this unit
HI6007 Final T1 2021

Question 1 ( 7
marks)
a. Holmes institute Students evaluation of the course they follow, askes following questions
from students. Identify the type of data and measurement scale for each with relevant
justifications.
i. How many interactive tutorials did you attend in this semester?
ii. What was your group assignment grade (HD, D, C, P, F)?
iii. Rate the lecturer (very effective, effective, not too effective and not at all effective)
iv. Which campus you are registered in (Melbourne, Sydney, Brisbane or Gold coast)
(2 marks)
ANSWER: ** Answer box will enlarge as you type
I
The data undertaken here is the quantitative data because the answer provided to this will be in
form of numbers only. Hence, this involves the use of the ordinal scale and due to the numbers this
scale has been used.
Ii
Here the type of data being used is the qualitative and the measurement scale is the nominal scale
only. This is particularly because of the reason that it only involves the qualitative data.
Iii
This the ordinal type of data and the measurement scale used in this is the ordinal scale and the
reason behind this use is that this can be ordered in some sequence like the very effective, effective,
not too effective and not at all effective
Iv
This come under the nominal type of data and uses the nominal scale of measurement. This is
particularly being selected because of the reason that this is a qualitative data and it is undertaken
under the qualitative type of data.
b. An investor recorded the following annual returns of one of his investments. You are
required to calculate and comment on;
i. Mean return.
ii. Variance and standard deviation of the return.
iii. Geometric return.
Year 2016 2017 2018 2019 2020
Return 15% 17% 19% 10% -5%
(5 marks)
ANSWER: ** Answer box will enlarge as you type
I
The mean return= total of the return divided by the total number of returns.
= 56/ 5
= 11.2%
Ii
Variance= 0.007456
Standard deviation= 0.086348
Iii
Geometric mean= 0.148362
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Question 2 (11 marks)
Nature lovers’ association of Australia, launched a campaign to encourage paper less communication
and/or recycling of used papers to save the trees to reduce global warming. Hence, many small
businesses have scaled up their business such as new forms of online document exchanges and
collecting used papers and cardboards from households and companies.
Abita Recycling Ltd is one such company which is operates in Melbourne. The financial analysist of
the company has estimated that the firm would make a profit if the mean weekly collection of
papers and cardboards from each household exceeded 1KG. To examine the feasibility of a recycling
plant, a random sample of 100 households was selected and sample mean and standard deviations
are 1.1KG and 0.35KG respectively.
Following the 6-step process of hypothesis testing, you are required to examine do this information
provide sufficient evidence at 99% confidence to allow the analyst to conclude that a recycling plant
would be profitable?
ANSWER:
Step 1- setting up the hypothesis
H0- there is no significant relation between feasibility of recycling and mean and standard deviation
H1- there is significant relation between feasibility of recycling and mean and standard deviation
H1= μ>1.1 H0= μ< 1.1
Step 2 is to set the level of significance which in present case if 99 % of 0.01
Step 3 involves calculation of test statistics by the formula
The step 4 involves the comparison of the test statistics and the critical value and marking it on the
graph
Further the step 5 is to state the conclusion fir formally and then less formally.
Question 3 (11 marks)
Edex limited is a renowned agricultural chemical manufacturer in Australia. They conduct many
research and development in the field of Agri and Horticulture. Company wanted to examine the
effect of temperature on farming of their selected range of products.
Company has produced following results based on their data gathering.
15° C 35 24 36 39 32
25° C 30 31 34 23 27
35° C 23 28 28 30 31
You are required to answer following questions;
a. State the null and alternative hypothesis for single factor ANOVA to test for any significant
difference in the perception among three groups. (1 marks)
ANSWER:
H0- there is no significant effect of temperature over farming of the selected range of products.
H1- there is significant effect of temperature over farming of the selected range of products.
b. State the decision rule at 5% significance level. (2 marks)
ANSWER:
The decision rule of 5 % significance level states that in case the significance value is more than 5 %
or 0.05 then the null hypothesis will be selected. On the other hand, in case if the significance value
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will be less than 0.05 then the alternate hypothesis will be selected rejecting the alternate
hypothesis.
c. Calculate the test statistic. (6 marks)
ANSWER:
ANOVA
Source of
Variation SS df MS F P-value F crit
Between Groups
76.1333
3 2
38.0666
7
1.91289
8
0.19006
2
3.88529
4
Within Groups 238.8 12 19.9
Total
314.933
3 14
d. Based on the calculated test statistics, decide whether there are any significant differences
between the yield based on the given temperature levels. (2 marks)
ANSWER:
With the above evaluation of the data it is clear that the significance value id 0.19 which is
greater which means that there is no significant effects of the temperature over the selected
range of the products.
Note: No excel ANOVA output allowed in question3. Students need to show all the steps in
calculations.
Question 4 (7 marks)
Melbourne Uni Lodge has decided to provide cup of cold or hot drinks for their tenants to attract
them after the Covid pandemic. They have determined that mean number of cups of drinks per day
is 2.00 with the standard deviation of 0.6. There will be 125 new tenants in the upcoming months.
What is the probability that the new tenants will consume more than 240 cups of drinks per day?
ANSWER:
Z= x-μ (mean) / Standard deviation
= (125 – 2) / 0.6
= 205
Question 5 (7 marks)
Yummy Lunch Restaurant needs to decide the most profitable location for their business expansion.
Marketing manager plans to use a multiple regression model to achieve their target. His model
considers yearly revenue as the dependent variable. He found that number of people within 2KM
(People), Mean household income(income), no of competitors and price as explanatory variables of
company yearly revenue.
The following is the descriptive statistics and regression output from Excel.
Revenue People Income
Competitor
s Price
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Mean 343965.68 5970.26 41522.96 2.8 5.68
Standard Error 5307.89863 139.0845281 582.1376385 0.142857 0.051030203
Median 345166.5 6032 41339.5 3 5.75
Mode #N/A 5917 #N/A 3 6
Standard Deviation
37532.5111
5 983.47613 4116.334718 1.010153 0.360838027
Sample Variance 1408689393 967225.2984 16944211.51 1.020408 0.130204082
Sum 17198284 298513 2076148 140 284
Count 50 50 50 50 50
SUMMARY
OUTPUT
Regression Statistics
Multiple R 0.77
R Square A
Adjusted R Square B
Standard Error 25139.79
Observations 50.00
ANOVA
df SS MS F
Significance
F
Regression C 40585376295 F H 3.0831E-08
Residual D 28440403984 G
Total E 69025780279
Coefficients Standard Error t Stat P-value
Intercept -68363.1524 78524.7251 -0.8706 0.3886
People 6.4394 3.7051 I 0.0891
Income 7.2723 0.9358 J 0.0000
Competitors -6709.4320 3818.5426 K 0.0857
Price 15968.7648 10219.0263 L 0.1251
You are required to;
a. Complete the missing entries from A to L in this output (2 marks)
ANSWER:
A = R2 = 1 – RSS / TSS = 1 – 28440403984 / 69025780279 = 1 – 0.412 = 0.588
B = 1 – (1-R2)*n-1 / n – k -1 = 1 – (1-0.588)*49 / 45 = 1-0.4486 = 0.5514
C= k -1 = 4-1 = 3
D = n – k = 50 – 4 = 46
E = n – 1 = 50 – 1 = 49
F = MSTR = SSTR / k-1 = 13528458765
G = MSE = SSE / n-k = 618269651
H= MSTR / MSE = 21.88
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s = √ SSE / n-2 = 28440406984 / 48 = √592508416.33 = 24341.49
sb1 = s / √SS = 24341.49 / √69025780279 = 24341.49/ 262727.58 = 0.0926
I = b1 / sb1 = 6.4394 / 0.0926 = 69.53
J = b2 / sb1 = 7.2723 / 0.0926 = 78.53
K = b3 / sb1 = -6709.4320 / 0.0926 = -724560.69
L = b4 / sb1 = 15968.7648/0.0926 = 172448.86
b. Derive the regression model (1
mark)
ANSWER:
Y = income (dependent variable)
a = intercept
b1 = slope (people)
b2 = slope (competitors)
b3 = slope (price)
x1 = people
x2 = competitors
x3 = price
y= a + b1*x1 + b2* x2 + b3 * x3
y = -68363.1524 + 6.4394x1 – 6709.4320x2 + 15968.7648x3
c. What does the standard error of estimate tell you about the model? (1
mark)
ANSWER:
The standard error of the estimate tells about the model that it is a measure of accuracy of
predictions. the regression line is that LINE which minimises the sum of the squared deviations of
reproduction which is called as the sum of squared error. This is standard error under the regression
identify the average distance which the observed value has from the regression line. Hence in the
present case the standard error of 25 139.7 9 states that the average distance from the regression
line of all the variables is 2513 9.79.
d. Assess the independent variables signiicance at 5% level (develop hypothesis if necessary in
the analysis)? (3 marks)
ANSWER:
In the present case the standard volume of the independent factor that is the people is 0.08. this
simply means that at the 5% significance level the p-value of people that is 0.08 is more. In this
simply means that null hypothesis is being selected which means that there is no significant
relationship between all the variables.
Question 6 (7 marks)
Anita Limited has shared their annual sales revenue over the last 6 financial years from 2015 to 2020.
Year Sales ($ 000)
2015 4500
2016 5100
2017 4900
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2018 5400
/2019 5670
2020 6000
You are required to;
a. Using linear trend equation forecast the sales revenue of Anita Limited for 2021. (5
marks)
ANSWER:
a)
2014 2015 2016 2017 2018 2019 2020 2021
0
1000
2000
3000
4000
5000
6000
7000
f(x) = 277.428571428571 x − 554450.476190476
Sales
The derived trend equation is Y = 277.43x – 554450 where y indicates sales (dependent variable) and
x indicates year (independent variable)
For the year 2021, where x = 2021
The forecasted value of sales would be y = (277.43 * 2021) – 554450 = 6236.03
b. Calculate the forecasted sales difference if you use 3-period weighted moving average
designed with the following weights: 2018 (0.1), 2019 (0.3) and 2020(0.6).
(2 marks)
ANSWER:
Forecasted sales on the basis of 3- period weighted moving average
= 5400 * 0.1 + 5670 * 0.3 + 6000 * 0.6 = 5841
Difference in forecasted sales value = 6236.03 – 5841 = 395.03.
Note: See the formula sheet on the next page.
FORMULA SHEET
K = 1 + 3.3 log10 n
Summary Measures (n – sample size; N – Population size)
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μ=
∑
i=1
N
Xi
N X =
∑
i=1
n
Xi
n
^p= X
n
s2= 1
n−1 ∑
i=1
n
( xi−x )2 Or s2= 1
n−1 [ (∑
i=1
n
xi
2
)−n x2
]
Or s2= 1
n−1 [ (∑
i=1
n
xi
2
)−
(∑
i=1
n
xi )2
n ]
σ 2= 1
N ∑
i=1
N
( xi−μ )2 Or σ 2= 1
N [ ( ∑
i=1
N
xi
2
)−n μ2
]
s Range
4 CV = σ
μ cv = s
x
Location of the pth percentile:
Lp= p
100 (n+1)
IQR = Q3 – Q1
Expected value of a discrete random variable
E ( x ) =μ=∑ x∗f (x)
Variance of a discrete random variable
Var ( x )=∑ ( x−μ )2 f ( x )
Z and t formulas:
Z= x −μ
σ
Z= x −μ
σ
√ n
Z= ^p− p
√ pq
n
t= x−μ
s
√ n
Confidence intervals
Mean: x ± tα/ 2
s
√ n
Proportion:
^p ± z α
2 √ ^p ^q
n
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x ± zα/ 2
σ
√ n

n= zα /2
2 p q
B2
Time Series Regression
b1=
∑
t =1
n
[ ( t−t ) ( yt − y ) ]
∑
t =1
n
( t −t )2
b0=Y −b1 t
T t=b0 +b1 t
ANOVA:
SSE=∑
j=1
k
( nj −1 ) s j
2
Simple Linear Regression:
SSE = ∑ ( yi− ^yi )2 SST = ∑ ( yi− y ) 2
SSR= ∑ ( ^yi− y )2
Coefficient of determination
Correlation coefficient
r = ∑ ( x −x ) ( y− y )
√ (∑ ( x−x )2
) (∑ ( y− y )2
) or r =
∑ XY −∑ X ∑ Y
N
√ ( ∑ X2− ( ∑ X )
2
N )( ∑ Y 2− ( ∑ Y )
2
N )
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MSTR= SSTR
k−1
SSTR=∑
j=1
k
n j ( x j− ´x ) 2
MSE= SSE
nT −k
F = MSTR / MSE
SST =∑
j =1
k
∑
i=1
n j
( xij−´x )2
^y=b0 +b1 x
b1=∑ ( xi −x ) ( yi − y )
∑ ( xi−x )2
b0= y−b1 x
SST = SSR + SSE
R2= SSR/SST

R2¿ ¿
Testing for Significance
Confidence Interval for β1
b1 ± tα /2 sb1
Multiple Regression:
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r xy=(sign of b1 ) √ Coefficient of Determination
s 2 = MSE = SSE/(
n  2) s = √ MSE= √ SSE
n−2
t= b1
sb1
sb1
= s
√∑ ( xi−x )2 F = MSTR / MSE
y =
0 +
1x1 +
2x2 + . . . +
pxp +

^y= b0 + b1x1 + b2x2 + . . . + bpxp
Ra
2=1−(1−R2) n−1
n− p−1
R2 = SSR/SST
MSR = SSR/k-1 MSE = SSE/n-k

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F distribution
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Submission Directions:
The assignment will be submitted via Blackboard. Each student will be permitted only ONE
submission to Blackboard. You need to ensure that the document submitted is the correct one.
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Academic Integrity
Holmes Institute is committed to ensuring and upholding Academic Integrity, as Academic
Integrity is integral to maintaining academic quality and the reputation of Holmes’ graduates.
Accordingly, all assessment tasks need to comply with academic integrity guidelines. Table 1
identifies the six categories of Academic Integrity breaches. If you have any questions about
Academic Integrity issues related to your assessment tasks, please consult your lecturer or tutor
for relevant referencing guidelines and support resources. Many of these resources can also be
found through the Study Skills link on Blackboard.
Academic Integrity breaches are a serious offence punishable by penalties that may range from
deduction of marks, failure of the assessment task or unit involved, suspension of course
enrolment, or cancellation of course enrolment.
Table 1: Six categories of Academic Integrity breaches
Plagiarism Reproducing the work of someone else without attribution. When
a student submits their own work on multiple occasions this is
known as self-plagiarism.
Collusion Working with one or more other individuals to complete an
assignment, in a way that is not authorised.
Copying Reproducing and submitting the work of another student, with or
without their knowledge. If a student fails to take reasonable
precautions to prevent their own original work from being copied,
this may also be considered an offence.
Impersonation Falsely presenting oneself, or engaging someone else to present as
oneself, in an in-person examination.
Contract cheating Contracting a third party to complete an assessment task,
generally in exchange for money or other manner of payment.
Data fabrication and
falsification
Manipulating or inventing data with the intent of supporting false
conclusions, including manipulating images.
Source: INQAAHE, 2020
If any words or ideas used the assignment submission do not represent your original words or
ideas, you must cite all relevant sources and make clear the extent to which such sources were
used.
In addition, written assignments that are similar or identical to those of another student is also a
violation of the Holmes Institute’s Academic Conduct and Integrity policy. The consequence for a
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violation of this policy can incur a range of penalties varying from a 50% penalty through
suspension of enrolment. The penalty would be dependent on the extent of academic
misconduct and your history of academic misconduct issues.
All assessments will be automatically submitted to Self-Assign to assess their originality.
Further Information:
For further information and additional learning resources please refer to your Discussion Board
for the unit.
END OF TUTORIAL ASSIGNMENT
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