Question

1

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 0.0716 0.0391 0.0105 Redshift 0.0235 0.0539 0.0443 Distance 0.32 0.75 1.02 0.57 0.61 0.15 a. Find the explained variation. (Round to six decimal places as needed.) b. Find the unexplained variation. (Round to six decimal places as needed.) c. Find the indicated prediction interval billion light-years< y< | | billion light-years (Round to three decimal places as needed.)

Answer 1

X	Y	XY	X2	Y2
0.0235	0.32	0.00752	0.000552	0.1024
0.0539	0.75	0.040425	0.002905	0.5625
0.0716	1.02	0.073032	0.005127	1.0404
0.0391	0.57	0.022287	0.001529	0.3249
0.0443	0.61	0.027023	0.001962	0.3721
0.0105	0.15	0.001575	0.00011	0.0225

Sample size, n =	6
$\small \Sigma$ x =	0.2429
$\small \Sigma$ y =	3.42
$\small \Sigma$ xy =	0.171862
$\small \Sigma$ x2 =	0.01218557
$\small \Sigma$ y2 =	2.4248
$\small \bar x$ = $\small \Sigma$x/n =	0.040483333
$\small \bar y$= $\small \Sigma$y/n =	0.57
SSxx = $\small \Sigma$x2 - ($\small \Sigma$x)2/n =	0.002352168
SSyy = $\small \Sigma$y2 - ($\small \Sigma$y)2/n =	0.4754
SSxy = $\small \Sigma$xy - ($\small \Sigma$x)($\small \Sigma$y)/n =	0.033409

a) Explained variation, SSR = SSxy²/SSxx  = 0.474524

b) Unexplained variation, SSE = SSyy - SSxy²/SSxx = 0.000876

c) Slope, b = SSxy/SSxx = 14.20349025

y-intercept, a = $\small \bar y$ -b*$\small \bar x$ = -0.00500463

Regression equation :

$\small \hat y$ =  -0.005005 + 14.203490 x

Predicted value of y at X =0.0126

$\small \hat y$ =  -0.005005 + 14.203490 * 0.0126 = 0.173959

Standard error, se = $\small \sqrt$((SSyy -SSxy²/SSxx )/(n-2)) = 0.01480

At $\small \alpha$ = 0.10 and df = n-2 = 4, critical value, t_c = T.INV.2T(0.10, 4 ) =2.1318

90% Prediction interval for the mean

SSrr 1 (0.0126 0.04048)2 + 0.173959 ± 2.1318 * 0.0148 /1 + 0.002352 0.135, 0.213)