Question

The table shows the number of goals allowed and the total points earned (2 points for a win, and 1 point for an overtime or shoc season. The equation of the regression line is y = -0.600x + 225.826. Use the data to answer the following questions. (a) Find the coefficient of determination, and interpret the result. (b) Find the standard error of the estimate, se, and interpret the result. Goals 212 210 219 223 259 258 273 200 217 199 217 206 256 248 Allowed, x Points, y 106 110 95 91 89 77 47 104 103 103 95 84 68 70 (Round to one decimal place as needed.) A. About 70.3 % of the variation in points earned can be explained by the relationship between number of goals variation is unexplained. OB. About % of the variation in number of goals allowed can be explained by the relationship between numbe remaining variation is unexplained. (b) se = (Round to three decimal places as needed.) Enter your answer in the answer box and then click Check Answer.

Answer 1

1.

The equation of regression line is,

$\widehat{y}=-0.600x+225.826$

	Goals Allowed (x)	Points(y)	$\widehat{y}$	$(y-\widehat{y})^2$
	212	106	98.626	54.37588
	210	110	99.826	103.5103
	219	95	94.426	0.329476
	223	91	92.026	1.052676
	259	89	70.426	344.9935
	258	77	71.026	35.68868
	273	47	62.026	225.7807
	200	104	105.826	3.334276
	217	103	95.626	54.37588
	199	103	106.426	11.73748
	217	95	95.626	0.391876
	206	84	102.226	332.1871
	256	68	72.226	17.85908
	248	70	77.026	49.36468
Total	3197	1242	1243.364	1234.981

Here

n =14

The standard error of the estimate is,

$S_{e}=\sqrt{\frac{\sum (y-\widehat{y})^2}{n-2}}$

1234.981464 14-2

$=\sqrt{\frac{1234.981464 }{12}}$

$\approx 10.145$

The standard error of estimate, S_e indicates approximately how much error you make when you use the predicted value for Y (on the least-squares line) instead of the actual value of Y.

The standard deviation of the estimate is 10.145.

This tells you that the actual points was different from the predicted points (on the least-squares line) by about 10.145

2.

The equation of regression line is,

$\widehat{y}=1.465x-60.269$

	Average Weekly Wages (state),x	Average Weekly Wages(federal) ,y	$\widehat{y}$	$(y-\widehat{y})^2$
	754	1008	1044.341	1320.668281
	769	1054	1066.316	151.683856
	782	1114	1085.361	820.192321
	815	1160	1133.706	691.374436
	843	1192	1174.726	298.391076
	895	1248	1250.906	8.444836
	920	1274	1287.531	183.087961
	938	1309	1313.901	24.019801
Total	6716	9359	9356.788	3497.862568

Here

n = 8

The standard error of the estimate is,

$S_{e}=\sqrt{\frac{\sum (y-\widehat{y})^2}{n-2}}$

$=\sqrt{\frac{3497.862568 }{8-2}}$

$=\sqrt{\frac{3497.862568 }{6}}$

$\approx 24.14$

The standard error of estimate, S_e indicates approximately how much error you make when you use the predicted value for Y (on the least-squares line) instead of the actual value of Y.

The standard deviation of the estimate is 24.14.

This tells you that the actual Average Weekly Wages(federal) was different from the predicted Average Weekly Wages(federal) (on the least-squares line) by about 24.14.