Question

The table below lists weights​ (carats) and prices​ (dollars) of randomly selected diamonds. 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​ 95% confidence level with a diamond that weighs 0.8 carats.

Weight	0.3	0.4	0.5	0.5	1.0	0.7
Price	​$516	​$1175	​$1333	​$1416	​$5673	​$2280

Answer 1

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
0.3	516	0.07	2400950.25	413.20
0.4	1175	0.03	792990.25	148.42
0.5	1333	0.00	536556.25	48.83
0.5	1416	0.00	421850.25	43.30
1	5673	0.19	13014056.25	1563.25
0.7	2280	0.02	46010.25	28.60

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	3.4	12393	0.313333333	17212413.5	2245.60
mean	0.57	2065.50	SSxx	SSyy	SSxy

R² =    (Sxy)²/(Sx.Sy) =    0.9350

a)

explained​ variation, = 0.9350 or 93.50%

b)

unexplained​ variation, = 1-0.9350 = 0.0650 or 6.50%

c)

sample size ,   n =   6          
here, x̅ = Σx / n=   0.57   ,     ȳ = Σy/n =   2065.50  
                  
SSxx =    Σ(x-x̅)² =    0.3133          
SSxy=   Σ(x-x̅)(y-ȳ) =   2245.6          
                  
estimated slope , ß1 = SSxy/SSxx =   2245.6   /   0.313   =   7166.80851
                  
intercept,   ß0 = y̅-ß1* x̄ =   -1995.69149          
                  
so, regression line is   Ŷ =   -1995.691   +   7166.8085   *x
                  
SSE=   (SSxx * SSyy - SS²xy)/SSxx =    1118628.3085          
                  
std error ,Se =    √(SSE/(n-2)) =    528.82613          

X Value=   0.8                      
Confidence Level=   95%                      
                          
                          
Sample Size , n=   6                      
Degrees of Freedom,df=n-2 =   4                      
critical t Value=tα/2 =   2.776   [excel function: =t.inv.2t(α/2,df) ]                  
                          
X̅ =    0.57                      
Σ(x-x̅)² =Sxx   0.3                      
Standard Error of the Estimate,Se=   528.826                      
                          
Predicted Y at X=   0.8   is                  
Ŷ =   -1995.691   +   7166.809   *   0.8   =   3737.755

For Individual Response Y                  
standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   612.2577              
margin of error,E=t*std error=t*S(ŷ)=    2.7764   *   612.26   =   1699.8999
                  
Prediction Interval Lower Limit=Ŷ -E =   3737.755   -   1699.90   =   2037.8554
Prediction Interval Upper Limit=Ŷ +E =   3737.755   +   1699.90 =   5437.6552