(Solved) Business Statistics: Assignment

Running head: BUSINESS STATISTICS ASSIGNMENT_PART II
BUSINESS STATISTICS ASSIGNMENT_PART II
Name of Student
Name of University
Author Note

1BUSINESS STATISTICS ASSIGNMENT_PART II
Table of Contents
Task 2...............................................................................................................................................2
Task 3...............................................................................................................................................3
Task 4...............................................................................................................................................6
Task 5...............................................................................................................................................7
Task 6...............................................................................................................................................8
Task 7...............................................................................................................................................9

2BUSINESS STATISTICS ASSIGNMENT_PART II
Task 2
Table 1 gives the frequency distribution of building type for the selected sample of 50
properties and Table 2 gives the relative frequency for each kind got building. As per Task 1, the
sample selected from the random numbers table as attached, starting from 2nd column of the 87th
row without replacement.
(a) It is seen that 16 of the buildings in the sample are brick buildings.
(b) The most frequently occurring building was found to be brick veneer with a
frequency of 21 in the sample.
Row Labels Frequency of Building Type
Brick 16
Brick Veneer 21
Weatherboard 7
Vacant Land 5
Grand Total 49
Table 1
1 2 3 4
0
5
10
15
20
25
16
21
7
5
Frequency Column Chart of Building Type
Building Type
Frequency
Figure 1

3BUSINESS STATISTICS ASSIGNMENT_PART II
(c) 14.29% of the properties in the sample was found to be weatherboard buildings as
shown in the table and figure below.
Row Labels Relative Frequency of Building Type
Brick 32.65%
Brick Veneer 42.86%
Weatherboar
d
14.29%
Vacant Land 10.20%
Grand Total 100.00%
Table 2
32.65%
42.86%
14.29%
10.20%
Relative Frequency Pie Chart for Building Type
1
2
3
4
Figure 2
Task 3
(a) The following table gives the sorted selling price of the properties sold, in ascending
order, along with the property number.
Property No. Sold Price
198 112
164 152
193 230
344 295
159 300

4BUSINESS STATISTICS ASSIGNMENT_PART II
393 315
347 323
187 331
389 361
277 382
162 385
294 392
354 409
123 415
396 420.5
295 424
272 430
301 432.5
343 439
111 440
370 450
361 459
103 460
184 460
13 475
61 476
275 481.5
254 502
144 552
398 555
235 567.86
139 570
256 582
274 595
138 670
211 725
316 760
271 795
24 881
317 915
332 917
181 936
209 951
210 1036
150 1126
180 2050
158 2650
303

5BUSINESS STATISTICS ASSIGNMENT_PART II
329
Table 3
The percentile formula, 100
P
)1+n(=L P , gives the percentile rank of the sold prices. P is the
percentile for which the corresponding rank of the observed data is to be determined.
Then computing the percentile values for the following, the Lpwas computed and then using the
corresponding prices to the ranks the percentile was computed using the formula:
PIp + Fp × (P(Ip+1) - PIp )
Where, Ip = Rank equal to integer portion of Lp , Fp = Fraction portion of Lp, PIp =Sold Price
corresponding to rank Ip , P(Ip+1) =Sold Price corresponding to rank succeeding Ip
i. For 70th percentile, the Lp = 35.7 and the corresponding price was seen to be 670. The
36th ranked price was found to be 725. Then the 70th percentile was found to be,
670+ 0.7× (725 - 670 ) = 708.5
ii. For the first quartile that is 25th percentile, the Lp was found to be 12.75 and the
corresponding price of 12th rank was seen to be 392. The 13th ranked price was found to
be 409. Then the first quartile was found to be,
392+ 0.75× (409 - 392) = 404.75.
iii. For the third quartile that is 75th percentile, the Lp was found to be 38.25 and the
corresponding price of 38th rank was seen to be 795. The 39th ranked price was found to
be 881. Then the third quartile was found to be,
795+ 0.25× (881 - 795) = 816.5.

6BUSINESS STATISTICS ASSIGNMENT_PART II
(b) The 70th quartile is the value in the data which marks the sold price below which 70% of
the sold prices lie, or in other words 30% of the highest sold prices lie just above the 70th
percentile value. Therefore, 30% of the top most “Sold Price” is greater than 708.5.
(c) The inter-quartile range is given by the difference between the third and first quartile and
was found to be 411.75. It is a measure of the variation present in the data.
Task 4
Sold Price
Mean 608.1991489
Standard Error 64.28995626
Median 460
Mode 460
Standard Deviation 440.7497344
Sample Variance 194260.3284
Kurtosis 11.26274453
Skewness 3.015920302
Range 2538
Minimum 112
Maximum 2650
Sum 28585.36
Count 47
Table 4
(b) The inner fence limits were calculated using the following formulae
IFUL = Q3 + 1.5 x IQR
IFLL = Q1 – 1.5 x IQR
The IFUL was found to be 816.5 + 1.5 x (816.5-404.75) = 1022.375 and the IFLL was found to
be 404.75 - 1.5 x (816.5-404.75) = -212.875

7BUSINESS STATISTICS ASSIGNMENT_PART II
(d) The inner fence limits show that 16 out of 50 of the sample values lie outside the fence and are
hence outliers. This makes up for more than 30% of the sample data. Thus the distribution of the
Sold Price is believed to be non-normal. Therefore it leads one to the conclusions:
i. The median would be a better measure of central tendency than the mean.
ii. The inter-quartile range would serve as the better measure of variance rather than
the standard deviation since it is calculated around the mean, which is not
deemed as an appropriate measure of central tendency in this case.
Task 5
(a) From the table of descriptive statistics in part a of task 4, it can be seen that the
median is much lower than the mean, suggesting an inflated value of the mean, the
kurtosis value is found to be 11.26 which suggests heavy tails and the skewness is 3.01
suggesting the distribution to be far from symmetric, rather being negatively skewed. All
these traits suggest a distribution which does not follow that of a normal distribution.
(b) Assuming that the “Sold Price” population data is normally distributed, the
probability a variable Z following a standard normal distribution is less than 1.5 is 0.9332
and that it is less that -1.5 is 0.0668. Thus probability that Z lies between 1.5 and -1.5 is
the difference of the previously mentioned probabilities, that is, 0.86638. Now, since we
assume that Sold Price follows normal distribution, it can be standardized and then the
chance of the z-scores of price being within the interval (-1.5, 1.5) would be 0.86638,
which is equivalent to chance of the price lying within 1.5 standard deviation from the
mean. Therefore expected number of observations within that range is given by 50
multiplied by 0.86638 which equals 43.319, that is rounded off to 43 observations.

8BUSINESS STATISTICS ASSIGNMENT_PART II
(c) The definition of Z-score is Z = X−μ
σ . Therefore, for Sold Price, X, when Z lies in
(-1.5, 1.5), X lies in (-1.5σ +μ, 1.5σ + μ).
Here, the estimated mean, ¯x
= 608.199, and
estimated standard deviation,
s= √ ∑ ( xi−¯x ) 2
n−1 =440.749.
Estimated interval= ( ¯x
- 1.5 × s , ¯x
+ 1.5 × s ).
Then, the upper bound for price is, 608.199+ 1.5 x 440.749 = 1269.3225,
and the lower bound is 608.199 - 1.5 x 440.749 = -52.9245.
The lower bound can be regarded as 0 since negative values are not
admissible.
45 observations were found to lie within the range of values. This is close enough but not
approximately equal to the expected 43 observations from part b. Hence the conclusion in part
(a) that the distribution is not Normal is supported by the finding.
Task 6
(a)
Sold Price
Mean 608.1991489
Standard Error 64.28995626
Median 460
Mode 460
Standard Deviation 440.7497344
Sample Variance 194260.3284
Kurtosis 11.26274453
Skewness 3.015920302

9BUSINESS STATISTICS ASSIGNMENT_PART II
Range 2538
Minimum 112
Maximum 2650
Sum 28585.36
Count 47
Confidence Level(90.0%) 107.9210046
Table 5
(i) The point estimate of the mean of sold price was found to be
¯x=∑ xi
n =28585. 36
50 = 608.199.
(ii) The 90% confidence interval was found to be ¯x ± margin of error. The margin of
error as from the Excel summary table was found to be 107.921.Thus the interval
was computed as (608.199 – 107.921, 608.199 + 107.921) = (500.278, 716.12).
Here the margin of error 107.921 is obtained from the table above.
(iii) The 90% confidence interval implies that the interval (500.278, 716.12) has a
90% chance of containing the population mean of sold price.
(b) Given the case that the population mean of Sold Price is 650 ($000), the value is found to
lie within the confidence interval specified in part a. This is consistent with the definition
of the confidence interval and hence the interval would be deemed as satisfactory in
being able to aptly specify an interval estimate which contains the population mean.
Task 7
(a) The following table gives a summary of the presence of brick veneer houses in the sample.
Number of Brick Veneer 21
Proportion of Brick Veneer 0.42
Mean Sold Price 477.9

10BUSINESS STATISTICS ASSIGNMENT_PART II
Mean Advertised Price 440.666
7
Total Number of
Properties
50
Confidence interval 1.79
Table 6
(i) The point estimate of the population proportion of veneer brick is given by the
estimate of the mean of a binomial distribution with n =50 and p=0.42, which is
the proportion of brick houses in the sample. The point estimate of proportion is
thus, n.p =50×0.42= 21.
(ii) The 99% confidence interval estimate of the proportion of brick veneer properties
in the population is given by estimated mean ± margin of error, given by critical
z-statistic with alpha (1-0.99) or 0.01 multiplied by the standard error. The margin
of error from the table is 1.79 and confidence interval was thus, (np -1.79, np +
1.79) = (20.82, 21.18).
(b) The formula for calculating calculate confidence interval is as follows:
(sample statistic)  (critical z)  (standard error of the sample statistic)
The 95% confidence interval estimate of the proportion of brick veneer properties
in the population is given by estimated mean ± margin of error, given by critical z-
statistic with alpha (1-0.05) or 0.01 multiplied by the standard error. The standard
error for proportion is the ration of standard deviation with the total sample size.
Standard error is√n × p ×(1− p)/n. The standard error was thus
√50 ×0.42 ×(1−0.42)/50 = 0.0698. The critical value was Z0.05/2 = -1.96. The margin
id error was thus 0.0698× 1.96= 0.179 and confidence interval (np -Z0.05/2 × 0.0698,
np + Z0.05/2 × 0.0698) = (21.136, 20.86).

11BUSINESS STATISTICS ASSIGNMENT_PART II
(c) The confidence interval in part (a) was computed using margin of error 1.79 which was obtained
from the table. This is because Excel corrects the direction of margin of error by taking absolute value.
The margin of error that was computed above is -0.179. This is because critical value is given by referring
to left tail of normal distribution in the normal tables. The negative value of margin of error reverses the
direction of the bounds.