Question

In: Economics

Appraised Value Property Size Age 467.8 0.2249 44 363.9 0.2169 56 429.9 0.1631 27 540.5 0.4672...

Appraised Value	Property Size	Age
467.8	0.2249	44
363.9	0.2169	56
429.9	0.1631	27
540.5	0.4672	18
405.8	0.2523	47
375.2	0.2214	86
318.2	0.1864	46
748.4	0.5062	7
212.8	0.2274	59
632.4	0.1385	13
349.4	0.1749	58
356.5	0.4216	49
359.6	0.2518	43
273.9	0.1158	17
303.8	0.1685	62
280.9	0.1746	58
390.4	0.3873	46
617.5	0.6578	47
313.9	0.1718	58
361.8	0.1425	75

1. Construct a 95% confidence interval estimate for the the mean appraised value for houses that have a land area of 0.15 acres and is 35 years old.

2. Construct a 95% prediction interval estimate for the the mean appraised value for houses that have a land area of 0.15 acres and is 35 years old.

Need process.

Expert Solution

It shall be noted that regression of Appraisal Value on Property Size and Age is given as follows, as estimated using Excel - Data Analysis Tool Pack based Regression

1)

Using the data analysis tool pack - Descriptive statistics, compute the descriptive summary of the variable - Appraisal value

Appraised Value

Mean	405.13
Standard Error	30.15225147
Median	362.85
Mode	#N/A
Standard Deviation	134.8449679
Sample Variance	18183.16537
Kurtosis	1.071007329
Skewness	1.197199183
Range	535.6
Minimum	212.8
Maximum	748.4
Sum	8102.6
Count	20
Confidence Level(95.0%)	63.10938761

The mean value is 405.13 and the confidence level (95%) is 63.10938761

Hence, the 95% confidence interval estimate for the mean of the appraised value for house is given as:

Lower Limit = 405.13 - 63.10938761 = 342.0206124

Upper Limit = 405.13 + 63.10938761 = 468.2393876

Thus, 95% confidence interval estimate is [342.0206124 , 468.2393876]

2)

Estimate the regression of Appraised value on Property Size and Age of the House.

The result is:

SUMMARY OUTPUT

Regression Statistics
Multiple R	0.761188758
R Square	0.579408326
Adjusted R Square	0.529926952
Standard Error	92.45223613
Observations	20

ANOVA
	df	SS	MS	F	Significance F
Regression	2	200174.0706	100087.0353	11.70962496	0.000635061
Residual	17	145306.0714	8547.415965
Total	19	345480.142

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	428.6181491	73.52043378	5.829918664	2.00871E-05	273.5035926	583.7327055	273.5035926	583.7327055
Property Size	449.8693026	150.5123253	2.988920021	0.008247175	132.3160539	767.4225512	132.3160539	767.4225512
Age	-3.101505555	1.06908272	-2.901090342	0.009939578	-5.357072932	-0.845938177	-5.357072932	-0.845938177

The Predicted value of Appraised value at 0.15 land size and 35 years as age of house is:

428.6181491 + 449.8693026*0.15 + (-3.101505555*35) = 492.9970389

Using the 95% coefficient of Intercept, Property Size and Age , compute the Lower Prediction Limit as given below:

273.5035926 + 132.3160539*0.15 + (-5.357072932*35) = 105.8534481

Using the 95% coefficient of Intercept, Property Size and Age , compute the Upper Prediction Limit as given below:

583.7327055 + 767.4225512*0.15 + (-0.845938177*35) = 669.238252

Thus, 95% prediction interval estimate is [105.8534481 , 669.238252]

Rahul Sunny answered 1 year ago

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