Business Statistics Assignment: Frequency, Percentiles, and Confidence

This business statistics assignment solution provides a comprehensive analysis of a dataset. It begins with frequency and relative frequency analysis, including the creation of a frequency column chart and a relative frequency pie chart. The solution then calculates percentiles, quartiles, and the interquartile range (IQR) for sold price data, interpreting their meaning and significance. Descriptive statistics are computed, and the appropriate measures of central tendency and variation are identified, along with an analysis of the data's distribution, including skewness and kurtosis. Further, the assignment includes calculations and interpretations of confidence intervals for the mean sold price and the proportion of brick veneer properties. The solution demonstrates the application of statistical concepts to real-world data and provides insightful interpretations of the results.

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BUSINESS STATISTICS
Assignment Part II
Task 2
Frequency Column Chart and Relative Frequency Pie-Chart are shown below:
Frequency Column Chart
V4 Building Type Frequency Relative frequency
1 Brick 12 0.24
2 Brick Veneer 20 0.4
3 Weatherboard 15 0.3
4 Vacant land 3 0.06
TOTAL 50 1.00
Brick
24%
Brick Veneer
40%
Weatherboard
30%
Vacant land
6%
Relative Frequency Pie-Chart
(a)A total of 12 properties in my sample consist of brick buildings.
(b) The most frequency building type in my sample is brick veneer which accounts for 40% of
the total sample or in total 20 buildings.

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(c)It is apparent from the relative frequency and also the corresponding pie chart that 30% of the
properties in my sample consist of weatherboard buildings.
Task 3
(a) Sold price data (in $ 000’s) is shown below:

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(b) Percentile location formula
Lp= ( n+ 1 ) P
100

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(i) The 70th Percentile
P=70
n=47
Lp= ( n+ 1 ) P
100 = ( 47 +1 ) 70
100 =33.6
Hence, 70th percentile = 34th value = $ 715,000
(ii) The first and third quartiles
 First quartile
First quartile Q1= ( 47+1 )∗25
100 =12
Hence, first quartile = 12th value = $ 460,000
 Third quartile
Third quartileQ3= ( 47+1 )∗75
100 =36
Hence, third quartile = 36th value = $ 736,000
c) The 70 percentile tends to represent that 70% of the houses sold would have a price lower than
or equal to the corresponding value of 70th percentile. Thus, only 30% of the houses have a
selling price in excess of $ 715,000 based on the given sample.
d) The value of Inter- Quartile Range for the variable sold price would be given below:
Inter−Quartile Range=Third quartile−First quartile
IQR=736000−460000= $ 276,000

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The IQR is a measure of dispersion as it highlights the range of the middle 50% of the values. It
is often considered to be a superior metric in comparison to range as the range may be impacted
by outliers which is not true for IQR.
Task 4
(a) Descriptive Statistics for sold price is shown below:
(b) The upper and lower inner fence limits based on the IQR and quartile values is computed
below:
IFUL=Q3 +1.5∗IQR=736,000+ ( 1.5∗276,000 )=$ 1,150,000
IFL L=Q1−1.5∗IQR=736,000− ( 1.5∗276,000 ) =$ 322,000
c) i) Based on the above computed limits coupled with the descriptive statistics, it is apparent
that the appropriate measure of central tendency would be the median of the sold price. This is
because there are some outliers on the positive side which would tend to distort the mean and
make it unsuitable as a central tendency measure.

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ii) The appropriate variation measure to be used would be IQR since it focuses on the middle
50% of the data and tends to ignore the extreme values which potentially would distort other
measures of variation such as standard deviation.
Task 5
a) Based on the descriptive statistics, it is apparent that the population is not normally
distribution. Three piece of evidence which tend to support the above conclusion are
represented below.
 There is presence of positive skew or right tail while in case of normal distribution the
skew is supposed to be zero.
 The kurtosis value is not equal to +3 which is the standard value expected for normal
distribution.
 There is no coincidence of the mean, median and mode which is essential for a normal
distribution.
b) P(Z<1.5) = 0.9332 (based on Z table)
P(Z<-1.5) = 0.0668 (based on Z table)
P(-1.5<Z<1.5) = 0.9332-0.0668 = 0.8664 or 86.64%
Values that would be expected to lie between the Z=-1.5 and Z=1,5 are 86.64% of 47 = 41
c) Mean – 1,5*standard deviation = 666378-1.5*423435 = $ 31,225
Mean + 1.5*standard deviation = 666378 +1.5*423435 = $ 1,301,530
Based on the given sample of sold price, it is apparent that out of the 47 sold price data available,
44 properties tend to lie in the above range. The answer in the given case is greater than the
predicted answer in b) which lends support to the conclusion drawn in (a).
Task 6
a) The requisite descriptive statistics for sold price is indicated below.

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i) The point estimate of the mean sold price would be the mean value itself which is equal to
$666,378
ii) Lower limit of the 90% confidence interval = 666378-103681 = $ 562,696
Higher limit of the 90% confidence interval = 666378-103681 = $ 770,059
iii) There is a 90% chance that the mean sold price of the population would lie between $
562,696 and $ 770,059.
b) The value $ 650,000 tends to lie in the 90% confidence interval that is derived above and
hence it would be considered a satisfactory value.
Task 7
The descriptive statistics for brick veneer properties is indicated below.