BE01106 Business Statistics

This document presents a solved assignment for the course BE01106 Business Statistics. It covers various statistical concepts and techniques, including descriptive statistics (mean, median, mode, standard deviation, quartiles, interquartile range, skewness, kurtosis), inferential statistics (confidence intervals for mean and proportion), and outlier detection. The assignment uses a dataset of house sold prices to illustrate these concepts. The solutions demonstrate the calculation and interpretation of these statistical measures, along with discussions on the distribution of the data and the selection of appropriate measures of central tendency and dispersion. The assignment also includes the construction of confidence intervals for both the mean and proportion, providing a range of plausible values for the population parameters. The detailed solutions provide a comprehensive guide for students studying business statistics.

Running Head: BE01106 - BUSINESS STATISTICS
BE01106 – Business Statistics
Name of the Student
Name of the University
Author Note

1BE01106 - BUSINESS STATISTICS
Table of Contents
Task 1...............................................................................................................................................2
Task 2...............................................................................................................................................2
Task 3...............................................................................................................................................3
Task 4...............................................................................................................................................5
Task 5...............................................................................................................................................6
Task 6...............................................................................................................................................7
Task 7...............................................................................................................................................9

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Task 1
Task 2
Brick
Brick veneer
Weatherboard
Vacant land
More
0
5
10
15
20
25
Histogram
Frequency
Type of Building
Frequency
Figure 2.1: Column Chart for Type of Building

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26%
44%
20%
10%
Relative Frequency Pie Chart
Brick
Brick veneer
Weatherboard
Vacant land
More
Figure 2.1: Pie Chart for Type of Building
(a) Thirteen properties in the selected sample consist of brick buildings.
(b) Brick Veneer type of building occurs most frequently in the selected sample.
(c) 20 percent of the properties in the selected sample consist of weatherboard buildings.
Task 3
(a) The sorted data for sold price is given in the following table 3.1.
Table3.1: Sold Price
112 342 432.5 525 1175
114 350 446 560 1200
125 354 453 563 1215
152 357 455 610 1225
239.5 358.5 460 615 1606
265 385 461.25 618 1885
295 400 475 670 2020
305 400 481 882.5
323 410 490 951
340 420.5 510 1060

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(b) i) In the dataset, there are three missing values. Thus, the total number of values (n)
decreases to 47. Thus, here, the 70th percentile is the (n+1) P
100 , where P is the percentile
number. Thus, the 70th percentile becomes,
( 48+1 ) 70
100 =34.3 ≃34.
The 34th value in the sorted data is the 70th percentile. Thus, the 70th percentile is
610.
ii) The first quartile (Q1) is the ((48+1) 1
4 )th value = 12.25 ≃ 12. Thus, the first
quartile is 350.
The third quartile (Q3) is the ¿)th value = 36.75 ≃ 37. Thus, the third quartile is
670.
(c) The 70th percentile that has been determined here is $ 610,000. This means that $
610,000 is higher than 70 percent of the sold price of the houses.
(d) The inter-quartile range is given by (Q3 – Q1) = (670 – 350) = 420. Thus, the
inter-quartile range is 420,000. This means that half of the sold price of the houses lie
between $ 670,000 and $ 350,000.

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Task 4
(a) The table of descriptive statistics for the sold price of the houses is given below in
table 4.1.
Table 4.1: Sold Price ($000)
Mean 597.697
Standard Error 63.507
Median 455
Mode 400
Std. Deviation 435.381
Sample Variance
189556.
7
Kurtosis 2.84
Skewness 1.75
Range 1908
Minimum 112
Maximum 2020
Sum
28091.7
5
Count 47
(b) IFUL = Q3 + 1.5 x IQR
= 670 + 1.5 x 420
= 1300.
IFLL = Q1 – 1.5 x IQR
= 350 – 1.5 x 420
= -280.
(c) i) The inner fence limits are evaluated in the previous part. Any values beyond these
limits are considered as outliers. It can be seen that that the inner fence upper limit has
been obtained as 1300 and from table 4.1 it can be seen that the maximum value of the
data is 2020. Thus, there is the presence of outliers to the selected data. Hence, median
will be considered as a better measure if central tendency

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ii) Following the above discussion, the most appropriate measure of dispersion for
the sample “Sold Price” data is the inter-quartile range and the quartile deviation.
Task 5
(a) From table 4.1 it can be seen that the mean of the data is greater than the median
of the data, which is again greater than the mode of the data. Thus, it can be said that the
distribution of the sold prices are not normally distributed as in a normal distribution, the
mean, median and mode of the data are equal.
It can be seen from the measure of skewness (1.75), that the data is positively
skewed. There are more houses with less sold price. Thus, the data is not normally
distributed as a normal distribution is always symmetric.
It can also be seen that the kurtosis of the distribution is 2.84 which is positive.
Thus, the distribution is leptokurtic but a normal distribution is always mesokurtic
(Kurtosis = 0). Thus, from here also, it can be concluded that the data is not normally
distributed.
(b) The proportion of values that lie under z = +1.5 as estimated from the normal
probability table is 0.9332 and the proportion of values that lie under z = -1.5 is (1 –
0.9332) = 0.0668. Thus, the proportion of values that lie between z = -1.5 and z = +1.5 is
(0.9332 – 0.0668) = 0.8664. Thus, the number of values of the variable “sold price” in the
sample selected that can be expected to lie between this range is (0.8664 * 50) = 43.32 ≃
44.
(c) Using the mean and standard deviation from the descriptive statistics table 4.1, the
bound for the 1.5 standard deviation spread from the mean can be calculated and is given
by;

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(Mean ± 1.5 * Standard Deviation)
= 597.697 ± 1.5 * 435.381
= (-56.07, 1250.07)
Thus, the number of observations of the variable “sold Price” in the sample
selected that fall within this range is 44. This count matches exactly with the answer
calculated for part (b). In part (a), it was established that the sample was not drawn from
a normal population but by assuming normality, all the estimates has matched with the
original values. Thus, it contradicts the conclusion drawn in part (a).
Task 6
(a) The descriptive statistics measures for the variable “Sold Price” assuming a 90
percent confidence interval is given in table 6.1.
Table 6.1: Sold Price ($000)
Mean 597.697
Standard Error 63.507
Median 455
Mode 400
Standard Deviation 435.381
Sample Variance
189556.
7
Kurtosis 2.84
Skewness 1.75
Range 1908
Minimum 112
Maximum 2020
Sum
28091.7
5
Count 47
Confidence
Level(90.0%) 106.606
(i) A point estimate of the mean sold price of the population properties is
given by the sample mean as the sample mean is an unbiased estimator of the

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population mean. Thus, the point estimate for the population mean of the variable
sold price is given by $ 597,697.
(ii) A 90 percent confidence interval can be obtained as;
(Sample Mean ± 90 percent confidence Level) = (597.697 ± 106.606) = (491.090,
704.090).
(iii) The confidence interval obtained in part (ii) states that there is a 90
percent possibility that the sold price of the houses will lie within this range. That
is, there is a 90 percent possibility that the sold prices of the houses will lie
between $ 491,090 and $ 704,090.
(b) If the population mean is actually 650 ($000), then the interval estimate obtained
in part (a) will be satisfactory as the population mean lies within the interval estimate.

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Task 7
(a) The descriptive statistics table for the brick veneer properties in the sample
selected is given in table 7.1.
Table 7.1: Brick Veneer
Mean 0.44
Standard Error
0.07091
2
Median 0
Mode 0
Standard Deviation
0.50142
7
Sample Variance
0.25142
9
Kurtosis -2.02037
Skewness
0.24928
9
Range 1
Minimum 0
Maximum 1
Sum 22
Count 50
Confidence
Level(99.0%)
0.19004
2
(i) A point estimate of the proportion of brick veneer properties in the population can
be estimated by the sample proportion as the sample proportion is an unbiased estimator
of the population proportion. Thus, the estimate of the population proportion of brick
veneer properties is given by 0.44.
(ii) A 99 percent confidence interval estimate of the proportion of brick veneer
properties in the population is given by (Sample Mean ± 90 percent confidence Level) =
0.44 ± 0.190) = (0.250, 0.630).
(b) The 95 percent confidence interval estimate of the proportion of brick veneer properties
in the population is given by the formula:

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(sample statistic)  (critical z or t)  (standard error of the sample statistic)
= 0.44  1.96  0.071
= (0.301, 0.579)
(c) The confidence interval obtained in part (a) was a 99 percent confidence interval
and the confidence interval obtained in part (b) is a 95 percent confidence interval. Thus,
it can be expected that the precision will be high for the confidence interval obtained in
part (a) than that obtained manually in part (b).