Descriptive Statistics and Interval Estimation for Sold Price and Brick Veneer Properties

This report provides a detailed analysis of the sold price and brick veneer properties using descriptive statistics and interval estimation techniques. It includes point estimates, confidence intervals, and measures of central tendency and dispersion. The report also discusses the normality of the data and provides graphical summaries.

Statistics
Student Name:
Instructor Name:
Course Number:
14 September 2018

Task 1
Task 1 has been done in excel. The excel file is attached together with this word document.
Task 2 (10 marks)
Use Excel to produce a Frequency Column Chart (4 marks) and a Relative Frequency Pie-
Chart (3 marks) for your sample to show the number and proportion, respectively, of each
building type.
Use these graphical summaries to answer the following questions:

(a) How many properties in your sample consist of brick buildings? (1 mark)
Answer
32 properties consist of brick buildings
(b) Which building type occurs most frequently in your sample? (1 mark)
Answer
Brick veneer occurs most frequently
(c) What proportion of properties in your sample consists of weatherboard buildings? (1
mark)
Answer
38% of the properties consists of weatherboard buildings
Task 3 (10 marks)
(a) Use Excel to sort your sample “Sold Price” data and paste into your MS Word
assignment document. (1 mark)
Answer
Sorted out sold price
Sold Price
112 387.5 590 805
132 400 615 831
245 410 621 845
265 424 627.5 865
300 440 670 880
305 441 675 882
319 445 711 900
348 501 718 1424
350 505 736 1800
351 552 745 2650
361 555 755
363 570 800
(b) Use the percentile location formula;
100
P
)1+n(=LP , and the three associated rules (Slide 11 of Week 2 Seminar,
Session 1) (1 mark) to determine:
(i) The 70th percentile. (1 mark)
Answer Remember to show all working!

LP = ( n+1 ) P
100 → ( 46+1 ) 70
100 = 47∗70
100 =32.9 ≈ 33
Therefore the 70th percentile is located at rank 33 which is 736.00
(ii) The first and third quartiles. (2 mark)
Answer
First quartile ( 25th percentile )= ( 46+1 ) 25
100 = 47∗25
100 =11.7 5 ≈ 12
Therefore the 25th percentile is located at rank 12 which is 363.00
Third quartile ( 75 th percentile )= ( 46 +1 ) 75
100 = 47∗75
100 =35.25 ≈ 35
Therefore the 75th percentile is located at rank 35 which is 755.00
(c) Briefly explain what the 70th percentile that you have determined informs you about
your sample “Sold Price” data. (2 mark)
Answer
70th percentile means that 70% of the properties sold made were sold out at $736,000
or below and that 30% of the properties were sold scored above$736,000.
(d) Determine the Inter-Quartile Range of your sample “Sold Price” data and provide a brief
explanation of what information this statistic provides about your sample data. (3
marks)
Answer
InterQuartile Range ( IQR ) =Q 3 – Q 1→ 755−363=392
Thus the inter-quartile range is $392,000. This value implies that the difference in the
75th percentile and 25th percentile is $392,000.
Task 4 (10 marks)
(a) Use Excel to produce a Descriptive Statistics table for your sample “Sold Price” data
and paste into your MS Word assignment document. (4 marks)
Answer
Sold Price
Mean 635.3696
Standard Error 63.63855

Median 562.5
Mode #N/A
Standard Deviation 431.6176
Sample Variance 186293.8
Kurtosis 10.67923
Skewness 2.792241
Range 2538
Minimum 112
Maximum 2650
Sum 29227
Count 46
(b) Use results from Task 3 to determine manually for this data, the upper and lower
inner fence limits;
IFUL = Q3 + 1.5 x IQR (1.5 marks)
Answer
IFUL=Q3+ 1.5 x IQR→ 755+1.5∗431.6176=755+647.4264=1402.4264
and IFLL = Q1 – 1.5 x IQR (1.5 marks)
Answer
IFUL=Q1−1.5 x IQR→ 363−1.5∗431.6176=363−647.4264=−284.4264
(c) Based on the limits calculated in (b), choose from the numerical summary measures
provided in the Descriptive Statistics table, and/or measures calculated
previously in Task 3;
(i) an appropriate measure of central tendency, and, (1 mark)
Answer
Median is the appropriate measure of central tendency
(ii) an appropriate measure of dispersion for your sample “Sold Price” data. (1
mark)
Answer
Inter-Quartile range is the most appropriate measure of dispersion
Provide a brief explanation of the reasoning behind your choice in both cases. (1 mark)
Answer
The two measures have been chosen based on the fact that the dataset had
some outliers and it is the mentioned measures that are least affected by the
outliers.
Task 5 (10 marks)
Remember to show all working! Failure to do so will result in the loss of marks.

(a) From the Descriptive Statistics table obtained in Task 5, identity three pieces of evidence
that indicate whether your sample “Sold Price” data has been obtained from a normally
distributed population or not. What is your conclusion? Note: Make sure only one
piece of evidence relates to the shape of the sample data. ( 2 marks)
Answer
The three pieces of evidence that indicate whether the dataset is normally distributed or
not
 Skewness
 Kurtosis
 Standard deviation
The skewness value is given as 2.792241. This value is greater than 1 implying that the
distribution is highly skewed. The kurtosis value is greater than 3 implying that it is a leptokurtic
distribution.
(b) Regardless of your conclusion in above, assume the “Sold Price” population data is
normally distributed. Applying the Standard Normal tables, calculate how many “Sold
Price” observations in your sample would expect to lie within 1.5 standard deviations of
the mean (i.e. between z = –1.5 and z = +1.5). (4 marks)
Answer
Probability (z > -1.5) = 0.0668
Probability (z < 1.5) = 0.9332
So the proportion of observations that lie within 1.5 SD are;
0.9332-0.0668 = 0.8664 = 86.64%
(c) Use the mean and standard deviation from the Descriptive Statistics table of Task 5 to
calculate the bound for 1.5 standard deviation spread from the mean. Using the “Sold
Price” sample data, manually count the number of observations fall within the bound.
State whether this count matches, approximately, your answer to (b) and hence whether
this result confirms (or not) your conclusion in (a). (4 marks)
Answer
Mean=635.3696
Standard deviation ( SD)=431.6176

Upper bound =Mean+1.5∗SD →635.3696+ 1.5∗431.6176=635.3696+647.4264=1282.796
Lower bound= Mean−1.5∗SD → 635.3696−1.5∗431.6176=635.3696−647.4264=−12.0568
We need to count the number of observations between -12.0568 and 1282.796. We
have 43 observations in the range representing 93.48%. This though does not match the
one in a) above it is close to.
Task 6 (10 marks)
Remember to show all working! Failure to do so will result in the loss of marks.
(a) Use Excel to produce a Descriptive Statistics table for the “Sold Price” variable in your
sample suitable for constructing an interval estimate of the population mean “Sold
Price”. (2 marks) Hence determine:
Answer
Sold Price
Mean 635.3696
Standard Error 63.63855
Median 562.5
Mode #N/A
Standard Deviation 431.6176
Sample Variance 186293.8
Kurtosis 10.67923
Skewness 2.792241
Range 2538
Minimum 112
Maximum 2650
Sum 29227
Count 46
(i) A point estimate of the mean “Sold Price” of the population of properties. (1
mark)
Answer
The point estimate of the mean is 635.3696
(ii) A 90% confidence interval estimate of the mean “Sold Price” of the population of
properties. (2 marks)
Answer
C . I : Mean ± Zα /2
σ
√n →635.3696 ± 1.645∗431.6176
√50 =635.3696± 1.645∗63.63855=635.3696 ± 10
Upper limit: 635.3696+104.685414=740.055014

Lower limit: 635.3696−104.685414=530.684186
Thus the 90% confidence interval is [530.6842, 740.0550]
(iii) Make a brief verbal statement explaining the meaning of the confidence interval
estimate obtained in (ii) in the context of the variable in this task. (3 marks)
Answer
We are 90% confident that the true population mean for the properties sold price
is between $530,684.2 and $740,055.0.
(b) If the population mean “Sold Price” is actually 650 ($000s), would you consider the interval
estimate obtained in (a), to be satisfactory? Explain why or why not. (2 marks)
Answer
I will have consider the the interval estimate obtained in (a), to be satisfactory since 650
($000s) is within the interval. It is actually inside the 90% confidence interval.
Task 7 (10 marks)
Remember to show all working! Failure to do so will result in the loss of marks.
(a) Use Excel to produce a Descriptive Statistics table for the brick veneer properties in your
sample suitable for constructing an interval estimate of the population proportion of
brick veneer properties. Hence determine: (2 marks)
Answer
Brick veneer
Mean 0.38
Standard Error 0.069341
Median 0
Mode 0
Standard Deviation 0.490314
Sample Variance 0.240408
Kurtosis -1.81429
Skewness 0.509877
Range 1
Minimum 0
Maximum 1
Sum 19
Count 50
Confidence Level(99.0%) 0.18583
(i) A point estimate of the proportion of brick veneer properties in the population. (1
mark)
Answer
The point estimate of the proportion of brick veneer properties in the population is
38% (0.38)
(ii) A 99% confidence interval estimate of the proportion of brick veneer properties in
the population. (1 mark)

Answer
The 99% confidence interval estimate of the proportion of brick veneer properties
in the population is 0.858.
(b) Using the following formula:
(sample statistic)  (critical z or t)  (standard error of the sample statistic)
Use the rule of thumb for good normal approximation (Slide 3 of Week 7 Session 2)
for proportion, then the Empirical Rule (Slide 8 of Week 5 Session 1) for a Normal
distribution to determine a 95% confidence interval estimate of the proportion of brick
veneer properties in the population. (4 marks)
Answer
Standard Error of Sample Proportion(SE)= √ 0.38 (1−0.38 )
50 = √0.004712=0.0686
C . I : ^P ± Zα/ 2 SE
Zα/2 =1.96
Zα /2 SE=Margin Error → 1.96∗0.0686=0.1345
C . I : ^P ± Zα/ 2 SE → 0.38± 0.1345
Upper limit : 0.38+ 0.1345=0.5145
Low e r limit :0.38−0.1345=0.2455
Thus the 95% confidence is [0.2455, 0.5145]
(b) Compare, in terms of the precision, the interval manually calculated in (b) with the
interval obtained from the Descriptive Statistics table in (a). Explain why the direction of
the change in precision is expected. (2 marks)
Answer
As can be seen, the 99% confidence has a wider interval than the 95% confidence
interval calculated in (b). This is expected since the 99% has a higher degree of
precision.