INTRODUCTION
In the present time, it has been observed that statistical methods and research skills provide managers with detailed information about the current happenings and the latest trends and changes in different external and internal business scenarios (Berman and Wang, 2016). This supports making meaningful and sound decisions when uncertainties arise in the business situation.
In this report, different statistical techniques are used to resolve the business problem if any, and make proper decisions. Different kinds of graphs are used to make proper comparisons, and various kinds of useful tables and figures are prepared in this report. In addition, Estimation and testing significance level are used to determine valuable results.
QUESTION 1
A) Appropriate Graphical Technique to Compare the Amount
Australian food and fiber exports by state, 2010 and 2015 |
||
|
Exports ($million) |
|
State |
2010 |
2015 |
Victoria |
7,344 |
11,656 |
Queensland |
4,872 |
8,179 |
NSW |
4,959 |
6,979 |
WA |
4,219 |
6,350 |
SA |
3,391 |
5,255 |
Tasmania |
907 |
736 |
Others |
973 |
4,278 |
B) Appropriate Graphical Technique to Compare the Percentage Value of the Amount
State |
% change |
Victoria |
36.99 |
Queensland |
40.43 |
NSW |
28.94 |
WA |
33.56 |
SA |
35.47 |
Tasmania |
-23.23 |
Others |
77.26 |
C) Comments on Part (a) and (b)
The bar chart presented in Part A helps describe the actual amount of food and fiber exports in different states of Australia (Bickel and Lehmann, 2012). It shows that data on the export of food products in the years 2010 and 2015 help in the easy comparison of states' contributions to export figures in the respective countries. It is determined from the graph that in the year 2010, Victoria was the state that contributed the highest in exports, as total exports were 7344 $ million. Similarly, Tasmania was considered to be the weakest state in terms of exports, as the total amount was only 907 $ million in the respective year. In the year 2015, Tasmania again had the highest figures for exports, which were 11656 $million, and the figures for exports decreased in the respective year. Tasmania was the only state which showed a negative change in the figures of exports through these years. It is also figuring out that in 2010, exports from other states were lower, which is 973 million. However, the Australian government took several steps, such as the easy availability of financial resources, that increased production levels. This helped to increase the overall export level, as in the year 2015 the total amount reached 4278 $ million, which helped increase the overall profitability of the nation.
The line chart displayed above supports the actual percentage change in the figure for exports between 2010 and 2015. It is clear from the graph that Tasmania was the only state that showed a negative percentage change from this year, which is -23. 23%. The chart also shows the highest percentage of alteration in another state of Australia, which shares the highest amount of export in the year 2015. The main reason for the increase in exports is the fall in the inflation rate in 2010 (which was 2.9%), and in 2015 it remained at 1.5. This helps to raise the Australian international competitiveness and will likely improve and raise the export level in the respective year.
We Promise Exceptional Assignment Writing & No AI Shortcuts !
View Samples Order Now Assignment helpQUESTION 2
The numbers of weekly sales calls by a sample of 40 telemarketers are recorded and presented below table:
14 |
8 |
6 |
12 |
21 |
4 |
9 |
0 |
25 |
17 |
9 |
5 |
8 |
18 |
16 |
3 |
17 |
19 |
10 |
15 |
5 |
20 |
17 |
14 |
19 |
7 |
10 |
15 |
10 |
8 |
28 |
31 |
27 |
30 |
2 |
17 |
26 |
11 |
29 |
19 |
A)Frequency Distribution and a Relative Frequency Distribution.
Classes |
Tally Frequency |
Frequency (f) |
Relative Frequency (f/40) |
0 - 4 |
I |
4 |
0.1 |
5 - 9 |
IIIII IIII |
9 |
0.23 |
10 - 14 |
IIIII II |
7 |
0.18 |
15-19 |
IIIII IIIII I |
11 |
0.28 |
20- 24 |
II |
2 |
0.05 |
24 - 29 |
I |
5 |
0.13 |
30 - 34 |
II |
2 |
0.05 |
Total |
40 |
40 |
|
Frequency Distribution: This is a description that allows to display either in tabular or tabular format the amount of findings inside a given period (Dittmar and Duchin, 2015). It is used primarily in a mathematical sense related to normal distribution forecasting.
Relative Frequency: This is a kind of allocation that focuses on counting the overall population and frequency for a sub-group of people. Throughout this technique, there is a systematic experimental approach where experiments must be done over and over again to obtain the best result from such a trial as it is a test and it is feasible to get a specific frequency (Hutton, Jiang, and Kumar, 2014).
From the above table, it is determined that tally frequency is primarily used to calculate the frequency distribution. Class intervals 20 - 24 and 30 - 34 have the lowest frequency and class intervals 15 - 19 have the highest frequency of 11.
B) Cumulative Frequency Distribution:
Classes |
Frequency (f) |
Relative Frequency (f/40) |
Cumulative Frequency |
Cumulative Relative Frequency |
0 - 4 |
4 |
0.1 |
4 |
0.1 |
5 - 9 |
9 |
0.23 |
13 |
0.33 |
10 - 14 |
7 |
0.18 |
20 |
0.5 |
15-19 |
11 |
0.28 |
31 |
0.78 |
20- 24 |
2 |
0.05 |
33 |
0.83 |
24 - 29 |
5 |
0.13 |
38 |
0.95 |
30 - 34 |
2 |
0.05 |
40 |
1 |
Total |
40 |
|
|
|
As seen in the table above, the cumulative frequency could be described as a total of all corresponding frequencies up to the usual stage. It measures the accumulated frequency by summarizing the frequency at each point, whereas the relative accumulated frequency is determined by applying the relative frequency to the present time.
C) Relative Frequency Histogram
Classes |
Frequency |
Relative Frequency |
Cumulative Relative Frequency |
0 - 4 |
4 |
0.1 |
0.01 |
5 - 9 |
9 |
0.23 |
0.33 |
10 - 14 |
7 |
0.18 |
0.5 |
15-19 |
11 |
0.28 |
0.78 |
20- 24 |
2 |
0.05 |
0.83 |
24 - 29 |
5 |
0.13 |
0.95 |
30 - 34 |
2 |
0.05 |
1 |
Total |
40 |
|
|
D) An O-Give Curve for the Data
To create an O-give curve, transforming information to less than the form or more than the form is necessary initially (Meyr, Myers, and Pontious, 2014). To draw an O-give graph of relative frequency in the context of relevant information Less-than-form of the O-give graph was produced as shown below:
Classes |
Frequency |
Relative Frequency |
Less than type |
Cumulative Relative Frequency |
0 - 4 |
4 |
0.1 |
Less than 4 |
0.01 |
5 - 9 |
9 |
0.23 |
Less than 9 |
0.33 |
10 - 14 |
7 |
0.18 |
Less than 14 |
0.5 |
15-19 |
11 |
0.28 |
Less than 19 |
0.78 |
20- 24 |
2 |
0.05 |
Less than 24 |
0.83 |
24 - 29 |
5 |
0.13 |
Less than 29 |
0.95 |
30 - 34 |
2 |
0.05 |
Less than 34 |
1 |
Total |
40 |
|
|
|
The o-Give curve for Cumulative Frequency is developed below:
Classes |
Frequency (f) |
Less than type |
Cumulative Frequency |
0 - 4 |
4 |
Less than 4 |
4 |
5 - 9 |
9 |
Less than 9 |
13 |
10 - 14 |
7 |
Less than 14 |
20 |
15-19 |
11 |
Less than 19 |
31 |
20- 24 |
2 |
Less than 24 |
33 |
24 - 29 |
5 |
Less than 29 |
38 |
30 - 34 |
2 |
Less than 34 |
40 |
Total |
40 |
|
|
E) Proportion of Data Lower Than 20
From the table above, the data that is lower than 20 can be determined by using the following equation, such as:
Data less than 20 is 31
Thus, 31/40= 0.775
F) Proportion of Data Higher Than 24
In the above tabular discussion, it is easy to figure out the information related to the frequencies that are higher than 24. thus
Data more than 24 is 7
So, 7/40 = 0.292 (approx.)
Question 3
A) Appropriate Graphic Descriptive Measure
According to the case, it was evaluated that inflation has reduced Australia's currency buying power at the moment. Consequently, due to increased inflation and reduced spending power intensity, shareholders' aspirations also increased, as they guessed how much returns they could get on their investment prices.
Year |
Rate of inflation (%) |
All-Ordinaries index |
1995 |
1.9 |
2000.8 |
1996 |
4.6 |
2231.7 |
1997 |
2.6 |
2662.7 |
1998 |
0.3 |
2608.2 |
1999 |
1.3 |
2963 |
2000 |
2.4 |
3115.9 |
2001 |
5.9 |
3352.4 |
2002 |
2.9 |
3241.5 |
2003 |
3 |
3032 |
2004 |
2.4 |
3499.8 |
2005 |
2.4 |
4197.5 |
2006 |
2.9 |
4933.5 |
2007 |
2.5 |
6337.6 |
2008 |
4.3 |
5513.5 |
2009 |
2.4 |
4127.6 |
2010 |
2.9 |
4632.8 |
2011 |
3.3 |
4553.9 |
2012 |
1.6 |
4385.2 |
2013 |
2.5 |
5110.5 |
2014 |
2.9 |
5423.9 |
2015 |
1.5 |
5713.4 |
Graphical with Rate of inflation
Time graph of Ordinary index
B) Scatter Plot
C) Numerical Summary Plot
Year |
Rate of inflation (%) (x) |
All-Ordinaries index (y) |
(x)2 |
1995 |
1.9 |
2000.8 |
3.61 |
1996 |
4.6 |
2231.7 |
21.16 |
1997 |
2.6 |
2662.7 |
6.76 |
1998 |
0.3 |
2608.2 |
0.09 |
1999 |
1.3 |
2963.0 |
1.69 |
2000 |
2.4 |
3115.9 |
5.76 |
2001 |
5.9 |
3352.4 |
34.81 |
2002 |
2.9 |
3241.5 |
8.41 |
2003 |
3.0 |
3032.0 |
9 |
2004 |
2.4 |
3499.8 |
5.76 |
2005 |
2.4 |
4197.5 |
5.76 |
2006 |
2.9 |
4933.5 |
8.41 |
2007 |
2.5 |
6337.6 |
6.25 |
2008 |
4.3 |
5513.5 |
18.49 |
2009 |
2.4 |
4127.6 |
5.76 |
2010 |
2.9 |
4632.8 |
8.41 |
2011 |
3.3 |
4553.9 |
10.89 |
2012 |
1.6 |
4385.2 |
2.56 |
2013 |
2.5 |
5110.5 |
6.25 |
2014 |
2.9 |
5423.9 |
8.41 |
2015 |
1.5 |
5713.4 |
2.25 |
Total |
56.5 |
83637.4 |
|
Mean |
2.7 |
3982.7 |
|
Median |
2.5 |
4127.6 |
|
Mode |
2.4 |
#VALUE! |
|
variance |
1.42 |
1530783.22 |
|
Std. derivation |
1.19 |
1237.24 |
|
Central tendency calculation
Mean= = total of observations/actual number of observations
= 56.5 / 20 = 2.85
From Excel (Mean is 2.7)
Median: Middle value in case of even series
= (N / 2)
= 20 / 2
= 10th Observation So the median is 2.4 and the Excel value is 2.5
Mode: The value which is repeated the most number of times in the given series so the model value is 2.4.
Variance = âx2 / n - Mean2
= 180.49 / 20 - (2.7) 2
= 9.02 - 7.29
= 1.73 and by using Excel the value is 1.42.
Standard Deviation:
âvariance = 1.73
= 1.31 and by using Excel the std dev is 1.19.
Quartiles,
Q1 = ¼th of total observation
= ¼th * 20 is the 5th term so the value from the list is 2963.0
Q3 = ¾th * 20 which is 15th term and the value is 4632.8
Calculation for secondary variables (Ordinary index)
Mean |
3982.7 |
Median |
4127.6 |
Mode |
0 |
Variance |
1530783.22 |
Standard deviation |
1237.24 |
Quartile |
1.9 is 5th term and 2.3 |
D) Correlation Coefficient
|
Rate of inflation (%) |
All-Ordinaries index |
Rate of inflation (%) |
1 |
|
All-Ordinaries index |
0.038875116 |
1 |
Correlations |
|
|
|||
|
Rate of inflation (%) |
All-Ordinaries index |
|||
Rate of inflation (%) |
Pearson Correlation |
1 |
-.011 |
||
Sig. (2-tailed) |
|
.963 |
|||
The sum of Squares and cross-products |
32.952 |
-342.143 |
|||
Covariance |
1.648 |
-17.107 |
|||
N |
21 |
21 |
|||
All-Ordinaries index |
Pearson Correlation |
-.011 |
1 |
||
Sig. (2-tailed) |
.963 |
|
|||
The sum of Squares and cross-products |
-342.143 |
30615662.571 |
|||
Covariance |
-17.107 |
1530783.129 |
|||
N |
21 |
21 |
|||
E) Estimating Regression Line
F) Estimating Coefficient of Determination
The coefficient of association indicates the interdependence of two factors, whereas negative correlation implies both being inversely proportional (Priebe and Spink, 2012). Though constructive, one illustrates a direct relation. Throughout this context, the significance of the correlation is determined as
R2 = 0.9892
This states that there has been appropriate variation among the purchasing power of customers with dollar currency and the rate of inflation for that specific year.
G) Testing the Significance of the Relationship
The significance could be checked on two faces, i.e. at 5 percent and 95 percent, respectively. In this scenario, the coefficient association was measured as 0.0388 at a significance point of 5 percent. This indicates that somewhere in the ordinary measure and inflation rate there is still no distinction.
Hypothesis= H0 : β1 = 0
= Ha : β1 â 0
Level of significance α = .05
Test statistic: t = b1 / Sb1
Rejection rule = Reject H0 in case if p-value ⤠.05 or |t| > 3.182 (With 3 degrees of freedom)
Value of test statistic t = b1 / Sb1 = 5 / 1.08 = 4.63
Rejecting H0 in case t = 4.63. 3.182, which can reject H0
H) Value of the Standard Error
The standard error could be described as a statistical term that is being used to measure a sample's accuracy and is also expressed by an assignment of surveys (Sekaran and Bougie, 2016). In the context of the above data, specific formulas can be implemented such as:
Thus, from the inflation figures, ê¹est.. = â0.9892 = 0.9946
CONCLUSION
In conclusion, it has been outlined according to the above analysis that analysis and statistical techniques are important aspects of decision-making within a company. Some techniques also help to evaluate the situations of the company and the sector, which benefit in reducing the chances of problems, and if any arise, better solutions are provided. In or out of certain adverse business and sector conditions, various internal and external factors are also evaluated. The entire study analyses multiple possible scenarios, utilizing different methods for improved understanding and review.