Business Statistics Assignment

This document is an assignment on Business Statistics. It includes various tasks such as frequency distribution, percentile calculation, descriptive statistics, confidence interval estimation, and more. The assignment provides insights into the analysis of sale price data and the interpretation of results. College/University: Not mentioned

BUSINESS STATISTICS
[DATE]

Assignment Part 1
Student Id = 4610559
Last three digits = 559
Column = 5
Row = 59
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Assignment Part II
Task 2
Frequency distribution column chart for building type (V4 variable)
Row Labels Count of V4
1 (Brick) 17
2 (Brick veneer) 19
3 (Weatherboard) 10
4 (Vacant land) 4
Grand Total 50
1 (Brick) 2 (Brick veneer) 3 (Weatherboard) 4 (Vacant land)
0
2
4
6
8
10
12
14
16
18
20
17
19
10
4
Building Type
Building Type
Frequency
Relative frequency pie chart for building type (V4 variable)
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1 (Brick); 17; 34%
2 (Brick veneer); 19;
38%
3 (Weatherboard);
10; 20%
4 (Vacant land); 4;
8%
Building Type
1 (Brick)
2 (Brick veneer)
3 (Weatherboard)
4 (Vacant land)
(a) Number of brick building in sample = 17
(b) Most frequently occurred building type in sample = Brick veneer
(c) Proportion of total properties that consists weatherboard building in sample =0.20
Task 3
(a) Sorted ‘Sold Price’ data is indicated below.
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(b) Percentile location formula along with the associated rules would be taken into
consideration to determine the following.
(i) 70th percentile
Percentile location formula
Here,
P=70
N=50 ¿
Now,
70 th percentile= ( 50+1 ) × 70
100
¿ 35.7 term 36 term=936($ ' 000)
(ii) The first and third quartile
The first quartile would be 25th percentile.
Percentile location formula
Here,
P=25
N=50 ¿
Now,
25 th percentile= ( 50+1 ) × 25
100
¿ 12.75 term 13th term=501($ ' 000)
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The third quartile would be 75th percentile.
Percentile location formula
Here,
P=75
N=50 ¿
Now,
75 th percentile= ( 50+1 ) × 75
100
¿ 38.25 term 38 term=1125 ($ ' 000)
(c) The 70th percentile indicates that 70% of the sample property would have sold price same
as or lower than $936,000 (Hillier, 2016).
(d) Inter quartile range or IQR of the sample for sold price data
Inter Quartile Range = Third quartile – First Quartile
Inter Quartile Range=1125−501=624( $' 000)
Inter quartile range is indicative of the aspect that middle 50% of sample properties sold have
prices which would fall within $530,000 range (Flick, 2015).
Task 4
(a) Descriptive statistics for sale price is shown below.
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(b) The lower inner fence and upper inner fence limit for the sale price are computed below.
The upper inner fence ( IFUL )=Third Quartile + { ( 1.5 )∗ ( IQR ) }=1125+ ( 1.5∗624 )=2061
The lower inner fence ( IFLL ) =First Quartile – { ( 1.5 )∗( IQR ) }=501− ( 1.5∗624 ) =−435
Therefore, the lower inner fence and upper inner fence limit for the Sold Price is -$435,000
and $2,061,000 respectively.
(c) (i) From the descriptive statistics, it is apparent that there is significant amount of positive
skew that is present in the sample sale price data owing to which mean would not be a
suitable measure of central tendency as it would be impacted by outliers. Median is the
preferred measure of central tendency in this case since it is not impacted by outliers
(Hair et. al., 2015).
(ii) With regards to measure of dispersion, standard deviation is not a suitable choice as it is
impacted by outliers, which is the case here owing to presence of high skew value. IQR or
Interquartile percentile would be the preferred measure of dispersion since it is not impacted
by outliers (Eriksson and Kovalainen, 2015).
Task 5
(a) Descriptive statistics for sale price is indicated below.
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(a) The sale price data is not normally distributed. The requisite evidences in this regards are
listed below (Flick, 2015).
 Normal distribution does not have skew and in present case, the sale price is highly
skewed as evident from the significantly positive value (+2.13) of skew.
 For normal distribution, the measures of central tendency (mean, median and mode) must
be coincident. However, in present scenario, the mean, median and mode are different
and hence, the sold price data is not from normal distribution.
 On the basis of empirical rule, 68% of the values should lie within one standard deviation
of the mean which is also not satisfied with regards to sale price data.
(b) 1.5 standard deviation of the mean
Between z = - 1.5 and z = + 1.5 value
The z value for 1.5 would be 0.4322 and hence,
-1.5 to +1.5 = 0.4322+0.4322 =0.8664
It indicates that 86.64% of sample data values would fall between z = - 1.5 and z = + 1.5.
The respective number of sample data points = 86.64 % * number of observations
¿ 86.64 %∗43=37.25 37 values apporximately
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(c) The lower and upper limit is computed as shown below.
Lower limit = Mean + {(1.5) * Standard deviation} = 39.88
Upper limit = Mean + {(1.5) * Standard deviation} =1478.29
It can be said based on the above shown lower and upper limit values that 40 values of the
sale price data lies within this limit. Hence, the value does not match from the value
computed in part (b).
Task 6
Descriptive statistics for sale price is indicated below.
(a)
(i) Point estimate for the mean “Sold Price” = $759.09 ($’000) or $759,088.37
(ii) 90% confidence interval is computed as shown below.
90 % confidence interval=Mean ±Confidence level
Lower limit ($ 000’s) ¿ 759.09−122.98=636.11
Upper limit ($ 000’s) ¿ 759.09+122.98=882.07
Therefore, the 90% confidence interval would be [636.11 882.07] ($ 000’s).
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(ii) It can be said with 90% confidence that the average sale price of all the houses
included in the population of interest would fall between $636,110 and $882,070
(Hair et. al., 2015).
(b) Yes, it can be concluded from the confidence interval estimation that $650,000 falls
within the 90% confidence interval and thus, the interval estimation obtained in part a is
satisfactory (Eriksson and Kovalainen, 2015).
Task 7
Descriptive statistic for the brick veneer properties
(a) The point estimate and 99% confidence interval is computed below.
(i) Point estimate for the mean = 0.38
(ii) 99% confidence interval
99 % confidence interval=Mean ± Confidence level
Lower limit ¿ 0.38−0.19=0.19
Upper limit ¿ 0.38+0.19=0.57
Therefore, the 99% confidence interval would be [019 0.57].
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(b) 95% confidence interval
Formula for 95% confidence interval
= (Sample Statistic) + {Critical z or t * Standard Error}
Population distribution is normally distributed and therefore, the z value will be used.
For 95% confidence interval, the z value = 1.96
Lower limit = 0.38 – (1.96*0.07) = 0.0516
Upper limit = 0.38 + (1.96*0.07) = 0.5159
The 95% confidence interval [0.0516 0.5159]
Based on empirical rule
μ ± σ 68% Particular Symbol Amount
μ ±2 σ 95% Mean μ 0.38
μ ±3 σ 99.70% Standard deviation σ 0.49
95% confidence interval
Lower limit = 0.38 – (2*0.49) = - 0.6006
Upper limit = 0.38 + (2*0.49) = 1.3606
(c) Comparing the two computations in part (b) and part (a), it is apparent that the confidence
interval in part a is wider than the one computed in part b. This is on expected lines since
the confidence interval in part a has a higher precision and owing to the higher precision
tends to include a wider interval in comparison to confidence interval computed in part b
(Hillier, 2016).
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References
Eriksson, P. & Kovalainen, A. (2015). Quantitative methods in business research (3rd ed.).
London: Sage Publications.
Flick, U. (2015). Introducing research methodology: A beginner's guide to doing a research
project (4th ed.). New York: Sage Publications.
Hair, J. F., Wolfinbarger, M., Money, A. H., Samouel, P., & Page, M. J. (2015). Essentials of
business research methods (2nd ed.). New York: Routledge.
Hillier, F. (2016). Introduction to Operations Research. (6th ed.). New York: McGraw Hill
Publications
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