BE01106 Business Statistics Assignment: Data Analysis and Results

This document presents a comprehensive solution to a business statistics assignment, focusing on the analysis of sample data related to property sales. The assignment covers various statistical concepts, including descriptive statistics, percentiles, quartiles, and the interquartile range (IQR). It addresses the identification and handling of outliers, discussing the appropriateness of the mean versus the median as measures of central tendency. The analysis further explores the distribution of the data, examining whether the sold price variable is normally distributed and highlighting reasons for non-normality. The assignment also includes the calculation and interpretation of confidence intervals for the mean sold price and the proportion of brick veneer properties. The document provides detailed calculations, interpretations, and comparisons of confidence intervals, along with relevant references to academic literature.

BE01106 BUSINESS STATISTICS
Assignment
[Pick the date]
Student Name

Assignment Part 1
Sample data
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Assignment Part 2
Task 2
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(a) Number of brick building = 17
(b) Maximum frequency of building type = Brick Vaneer
(c) Proportion of weatherboard building = 22%
Task 3
(a) Sorted data of sold price
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(b) Percentile
(i) 70th percentile
Here,
n=50
P=70
LP = (50+ 1 )∗70
100 =35.7 36 th observation
The 70th percentile would be 36th observation i.e. 645.55 ($’000).
(ii) First and third quartile
 First quartile (25th percentile)
Here,
n=50
P=25
LP = (50+ 1 )∗25
100 =12.75 13 thobservation
The first quartile would be 13th observation i.e. 418 ($’000).
 Third quartile (75th percentile)
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Here,
n=50
P=75
LP = (50+ 1 )∗75
100 =38.25 38 thobservation
The 3rd percentile would be 38th observation i.e. 705 ($’000).
(c) The 70th percentile of the sold price data has come out as $645,550 and implies that 70% of
the sample houses would have a selling price lower than or equal to $645,550.
(d) Inter Quartile range is difference between 3rd and 1st quartile.
Inter Quartile range (IQR) = Q3 – Q1 ¿ 705 – 418=287 ($ ' 000)
Task 4
(a) Descriptive statistics of variable sold prices
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(b) Upper and lower inner fences limits are calculated below:
IFUL=Q3+ ( 1.5∗IQR )
IFUL=705+ ( 1.5∗287 )=1135.5
And
IFLL=Q 1− ( 1.5∗IQR )
IFLL=418− ( 1.5∗287 ) =−12.5
(c) (i) It is apparent that the upper fence limit for the data is 1135.5 whereas the maximum
value in the data tends to exceed this value. Hence, it is apparent that the given data
contains outliers on the positive side. Owing to the presence of these outliers, the mean
can potentially get distorted making it unsuitable as a measure of central tendency.
Hence, the suitable measure would be median which does not get impacted by the
extreme values and hence presents a more accurate measure of central tendency in this
case (Flick, 2015).
(ii) The standard deviation or variation would not be appropriate in this case considering the
presence of outliers and their dependence on mean value. Thus, the appropriate measure of
dispersion would be IQR or inter-quartile range which does not get impacted by the presence of
these outliers and hence presents a more accurate description of dispersion in the sample data
(Hair et. al., 2015).
Task 5
Descriptive statistics
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(a) The three reasons to highlight that the sold price variable is not normally distributed are
highlighted below (Eriksson and Kovalainen, 2015).
 There is presence of skew in the data which in case of normal distribution is zero.
 The measures of central tendency do not coincide which is not acceptable for a normal
distribution.
 The percentage values contained within one standard deviation of the mean is not 68%
which should be the case as per the empirical rule.
Mean +1 standard deviation = 552.13 + 261.59 = 290.54
Mean – 1 standard deviation = 552.13 - 261.59 = 813.73
Total properties within the above selling price range is 36.
Hence, proportion = (36/50) = 72%
(b) Z value for 1.5 comes out to be 0.4322 using the z table.
Hence,
−1. 5 ¿+1.5=0.4322+0.4322=0.8664
It means 86.64% population would fall between -1.5 to +1.5. It indicates that nearly 39
properties from total properties would fall within this range.
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(c) The lower an upper limits are calculated below:
Lower limit = Mean – (1.5 * Standard deviation)
Lower limit =552.13− (1.5∗261.59 ) =159.7466
Upper limit = Mean + (1.5 * Standard deviation)
Upper limit ¿ 552.13+ ( 1.5∗261.59 )=944.5223
There are only 39 properties out of total sample of 50 properties that fall in this range and hence,
this confirms the result of part b.
Task 6
Descriptive statistics
(a) (i) The point estimate of the mean would be $552.13 ($’000).
It means approximately the expected selling price of the population property would be $552,130.
(ii) 90% confidence interval
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Lower limit = Mean – Confidence level ¿ 552.13−65.52=486.61
Upper limit = Mean + Confidence level ¿ 552.13+65.52=617.66
The 90% confidence interval ($’000) = [486.61 617.66]
(iii) We can conclude with 90% confidence that the expected mean of sold price would fall
between the confidence interval [486.61 617.66] ($’000).
(b) If the population mean comes out as $ 650,000, then it implies that the interval computed
above based on the given sample is not accurate. This is because the population mean
should have been included in the above confidence interval but the exclusion of the
population mean from this interval reflects that the sample is not representative of the
population (Hair et. al., 2015).
Task 7
Descriptive statistics for brick veneer properties of the sample.
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