Question

Listed below are amounts of court income and salaries paid to the town justices for a certain town. All amounts are in thousands of dollars. Find the​ (a) explained​ variation, (b) unexplained​ variation, and​ (c) indicated prediction interval. There is sufficient evidence to support a claim of a linear​ correlation, so it is reasonable to use the regression equation when making predictions. For the prediction​ interval, use a 99​% confidence level with a court income of ​$800​,000.

Court Income ​$60 ​$408 ​$1583 ​$1124 ​$268 ​$252 ​$108 ​$169 ​$29

Justice Salary ​$26 ​$40 ​$96 ​$60 ​$46 ​$57 ​$30 ​$27 ​$15

A) Find explained variation. round to 3 decimal places

B) Find unexplained variation. round to 3 decimal places

C) Find indicated prediction interval. round to 3 decimal places

Answer 1

X	Y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
60	26	147882.98	328.01	6964.73
408	40	1336.31	16.90	150.28
1583	96	1296055.75	2692.46	59072.62
1124	60	461644.75	252.46	10795.62
268	46	31171.86	3.57	-333.49
252	57	37077.64	166.12	-2481.83
108	30	113269.64	199.12	4749.17
169	27	75930.86	292.79	4715.06
29	15	172686.42	847.46	12097.28

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	4001	397	2337056.222	4798.9	95729.44
mean	444.56	44.11	SSxx	SSyy	SSxy

a)

explained variation=R² =    (Sxy)²/(Sx.Sy) =    0.817

b)

Unexplained variation= 1-   0.817 =0.183

c)

sample size ,   n =   9          
here, x̅ = Σx / n=   444.56   ,     ȳ = Σy/n =   44.11  
                  
SSxx =    Σ(x-x̅)² =    2337056.2222          
SSxy=   Σ(x-x̅)(y-ȳ) =   95729.4          
                  
estimated slope , ß1 = SSxy/SSxx =   95729.4   /   2337056.222   =   0.0410
                  
intercept,   ß0 = y̅-ß1* x̄ =   25.9014          
                  
so, regression line is   Ŷ =   25.901   +   0.040962   *x
                  
SSE=   (SSxx * SSyy - SS²xy)/SSxx =    877.6625          
                  
std error ,Se =    √(SSE/(n-2)) =    11.19734          

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X Value=   800 (in thousands of dollars)   
Confidence Level=   99%                      
                          
                          
Sample Size , n=   9                      
Degrees of Freedom,df=n-2 =   7                      
critical t Value=tα/2 =   3.499   [excel function: =t.inv.2t(α/2,df) ]                  
                          
X̅ =    444.56                      
Σ(x-x̅)² =Sxx   2337056.2                      
Standard Error of the Estimate,Se=   11.197                      
                          
Predicted Y at X=   800   is                  
Ŷ =   25.901   +   0.041   *   800   =   58.67

For Individual Response Y                  
standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   12.0868              
margin of error,E=t*std error=t*S(ŷ)=    3.4995   *   12.09   =   42.2974
                  
Prediction Interval Lower Limit=Ŷ -E =   58.671   -   42.30   =   16.373 (in thousands of dollars)
Prediction Interval Upper Limit=Ŷ +E =   58.671   +   42.30   =   100.968 (in thousands of dollar)