Question

The table below lists measured amounts of redshift and the distances​ (billions of​ light-years) to randomly selected astronomical objects. 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​ 90% confidence level with a redshift of 0.0126.

Redshift   Distance
0.0235   0.33
0.0537   0.74
0.0716   0.98
0.0399   0.54
0.0443   0.63
0.0108   0.15
  

Answer 1

X	Y	XY	X²	Y²
0.0235	0.33	0.007755	0.00055225	0.1089
0.0537	0.74	0.039738	0.00288369	0.5476
0.0716	0.98	0.070168	0.00512656	0.9604
0.0399	0.54	0.021546	0.00159201	0.2916
0.0443	0.63	0.027909	0.00196249	0.3969
0.0108	0.15	0.00162	0.00011664	0.0225

Ʃx =	0.2438
Ʃy =	3.37
Ʃxy =	0.168736
Ʃx² =	0.01223364
Ʃy² =	2.3279
Sample size, n =	6
x̅ = Ʃx/n = 0.2438/6 =	0.040633333
y̅ = Ʃy/n = 3.37/6 =	0.561666667
SSxx = Ʃx² - (Ʃx)²/n = 0.01223 - (0.2438)²/6 =	0.002327233
SSyy = Ʃy² - (Ʃy)²/n = 2.3279 - (3.37)²/6 =	0.435083333
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 0.16874 - (0.2438)(3.37)/6 =	0.031801667

a) Explained variation, SSR = SSxy²/SSxx = (0.0318)²/0.00233 = 0.43457

b) Unexplained variation, SSE = SSyy -SSxy²/SSxx = 0.43508 - (0.0318)²/0.00233 = 0.000513

c)

Standard error, se = √(SSE/(n-2)) = √(0.00051/(6-2)) = 0.0113

Slope, b = SSxy/SSxx = 0.0318/0.00233 = 13.66501

y-intercept, a = y̅ -b* x̅ = 0.56167 - (13.66501)*0.04063 = 0.0064118

Regression equation :

ŷ = 0.0064 + (13.665) x

Predicted value of y at x = 0.0126

ŷ = 0.0064 + (13.665) * 0.0126 = 0.1786

Significance level, α = 0.1

Critical value, t_c = T.INV.2T(0.1, 4) = 2.1318

90% Prediction interval :

Lower limit = ŷ - tc*se*√(1 + (1/n) + ((x-x̅)²/(SSxx)))

= 0.1786 - 2.1318*0.0113*√(1 + (1/6) + ((0.0126 - 0.0406)²/(0.0023))) = 0.1490

Upper limit = ŷ + tc*se*√(1 + (1/n) + ((x-x̅)²/(SSxx)))

= 0.1786 + 2.1318*0.0113*√(1 + (1/6) + ((0.0126 - 0.0406)²/(0.0023))) = 0.2082