Question.4 In this Question, Value1K$ refers the value of randomly selected prop
ID: 3241041 • Letter: Q
Question
Question.4
In this Question, Value1K$ refers the value of randomly selected properties listed in year 2015 and Value2K$ refers the values of the same properties listed in year 2016. All the values are in thousands of dollars.
Data for Question.4
Property#
Value1$K
Value2$K
1
1584
1656
2
1656
1764
3
1710
1620
4
1710
1890
5
1674
1818
6
1746
1710
7
1746
1980
8
1764
1890
9
1656
1728
10
1782
1980
a. Test the hypothesis that the mean of the difference in the listed values of the properties in the two years is non-existent. What is the name of this test? Explain why this test is the appropriate test.
b. What is the Confidence Interval for the mean of the difference in the values of the properties in the two years? Is it consistent with the conclusion reached in part ‘a’?
c. Based on this Confidence Interval in 'b' above, state how much more or less, is the mean of the difference in values for years 2016 and 2015. Calculate the percentage changes in the 'minimum', 'maximum' and 'mean' values of the properties in the in the year 2016 compared to year 2015.
d. Calculate the p-Value for this test. What conclusion would you draw?
e. Now for the same data, Test the hypothesis that the median value of the difference in values in the two years is non-existent. What is the name of this test?
f. Specify with diagrams and justify which methodology would be more appropriate? Methodology used in part ‘a’ or in part ‘e’. What are the names of these respective methodologies?
Property#
Value1$K
Value2$K
1
1584
1656
2
1656
1764
3
1710
1620
4
1710
1890
5
1674
1818
6
1746
1710
7
1746
1980
8
1764
1890
9
1656
1728
10
1782
1980
General Instructions When you perform a test of hypothesis, you must always use the 4-step approach i. Sl the "Nill'' and Alternate" hypotheses, ii. S2: manually calculate value of the test statistic, iii. S3: specify the level of significance and the critical value of the statistic, iv S5. use appropriate decision rule and then reach a conclusion about not rejecting or rejecting the mull hypothesis. S5: If asked to calculate p-value, do so and relate the p-value to the level of significance in reaching your conclusion. If you use MiniTab to perform the hypothesis test, you must paste the relevant output into your assignment. This output simply verifies and occasionally replaces the manual computation of the test statistic, p-1alue or the confidence internal. You must supply all the required steps, mentioned above, to make vour testing procedure standard and complete. If Confidence Coefficient (CC) and Level of Significance LS) are not specified, assume the default values of 95% and 5% respectively. Use precision level of only 4 Decimal Digits (DD) and no more or no less, when calculations are dome with a calculator.Explanation / Answer
a.
i) State the hypothesis -
Null Hypothesis H0: Mean of the difference value of randomly selected properties listed in year 2015 and 2016 is 0 (non-existent).
Alternative Hypothesis H1: Mean of the difference value of randomly selected properties listed in year 2015 and 2016 is not 0 (existent).
ii) Calculation of test statistic
Value1$K is (1584,1656,1710,1710,1674,1746,1746,1764,1656,1782)
Value2$K is (1656,1764,1620,1890,1818,1710,1980,1890,1728,1980)
Value1$K-Value2$K is (-72, -108, 90, -180, -144, 36, -234, -126, -72, -198)
Mean of Value1$K-Value2$K is -100.8
Standard deviation of Value1$K is 60.7157
Standard deviation of Value2$K is 128.6297
Standard error of the mean difference, SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(60.71572/10) + (128.62972/10)] = 44.98
Degree of freedom of the test, DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (60.71572/10 + 128.62972/10)2 / { [ (60.71572 / 10)2 / (10 - 1) ] + [ (128.62972 / 10)2 / (10 - 1) ] }
= 13 (rounding to nearest integer)
test statistic, t = Mean Difference/SE = -100.8/44.98 = -2.2410
iii) Level of significance and critical value -
Level of significance = 5% = 0.05
For a two-tail test, the Level of significance = 0.05/2 = 0.025
Critical value of test statistic, t at Level of significance = 0.025 and df = 13 is -2.1604
iv) Decision rule -
As the t stat (-2.2410) is less than critical value of t (-2.1604), we reject the null hypothesis and conclude that Mean of the difference value of randomly selected properties listed in year 2015 and 2016 is not 0 (existent).
v) p-value -
p-value for t = -2.2410 and df = 13 is 0.0216. As p-value is less than the level of significance 0.025, we reject the null hypothesis and conclude that Mean of the difference value of randomly selected properties listed in year 2015 and 2016 is not 0 (existent).
This test is called two-sample t-test. This test is apprpriate because
b.
Standard error of the mean difference is 44.98
Z-value at 95% confidence interval is 1.96
95% confidence interval is (-100.8-1.96*44.98, -100.8+1.96*44.98)
= (-188.9608, -12.6392)
As, the mean difference of 0 does not lie in the 95% confidence interval, we are 95% confident that mean difference is not 0 and it is consistent with the conclusion reached in part ‘a’.
c.
mean of the difference in values for years 2016 and 2015 = -100.8
mean of the difference is (-100.8 + 188.9608 = 88.1608) more than the lower limit of the 95% confidence interval.
mean of the difference is (100.8 - 12.6392 = 88.1608) less than the upper limit of the 95% confidence interval.
'minimum' values of the properties in the in the year 2015 = 1584
'minimum' values of the properties in the in the year 2016 = 1620
% change in 'minimum' values of the properties iin the year 2016 compared to year 2015 = (1620-1584)*100/1584 = 2.2727 %
'maximum' values of the properties in the in the year 2015 = 1782
'maximum' values of the properties in the in the year 2016 = 1980
% change in 'minimum' values of the properties iin the year 2016 compared to year 2015 = (1980-1782)*100/1782 = 11.1111 %
'mean' values of the properties in the in the year 2015 = 1702.8
'mean' values of the properties in the in the year 2016 = 1803.6
% change in 'minimum' values of the properties iin the year 2016 compared to year 2015 = (1803.6-1702.8)*100/1702.8 = 5.9197 %
d.
p-value for t = -2.2410 and df = 13 is 0.0216. As p-value is less than the level of significance 0.025, we reject the null hypothesis and conclude that Mean of the difference value of randomly selected properties listed in year 2015 and 2016 is not 0 (existent).
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