Question

Research was conducted on the amount of training for 5K and the time a contestant took to run the race. The researcher recorded the number of miles during training ( a 1 month period) and the time to complete the 5K. The results are below. Miles Trained 63 72 62 66 75 21 Time (Minutes) 17.5 24.6 20.5 31.5 28.8 27.1 a) Give the correlation coefficient. b) Give the equation of the regression line. Round the values to 4 deimal places. This is in the form y = ax + b, so the slope goes in the first box and the y-inercept in the 2nd box. y = X + c) Predict the time in the 5K if someone trained 62 miles. Use the rounded values entered for the slope and intercept of the regression line to compute the predicted y-value. y = d) Give the residual for 62 miles trained. This is the actual y value for the runner who trained for miles minus the predicted value. Residual

Answer 1

Answer a:

Let x represent the Miles Trained and y represent the Time

The following table shows the calculations –

	Miles Trained, (x)	Time (minutes), (y)	x^2	y^2	xy
	63	17.5	3969	306.25	1102.5
	72	24.6	5184	605.16	1771.2
	62	20.5	3844	420.25	1271
	66	31.5	4356	992.25	2079
	75	28.8	5625	829.44	2160
	21	27.1	441	734.41	569.1
Total	359	150	23419	3887.76	8952.8

Total number of observations, n = 6

Mean of x, $\bar{x}$ = 359/6 = 59.8333

Mean of y, $\bar{y}$ = 150/6 = 25

Standard Deviation of x, Sx = {($\sum$(x^2)/n) - ($\bar{x}$^2)}^0.5 = {(23419/6) – (59.8333^2)}^0.5 = 17.9762

Standard Deviation of y, Sy = {($\sum$(y^2)/n) - ($\bar{y}$^2)}^0.5 = {(3887.76/6) – (25^2)}^0.5 = 4.7916

Covariance between x and y, Cov.(x, y) = ($\sum$xy/n) - ($\bar{x}$$\bar{y}$) = (8952.8/6) – (59.8333 x 25) = -3.6992

Correlation coefficient, r = Cov.(x, y) / (Sx . Sy) = -3.6992 / (17.9762 x 4.7916) = -0.0429

(rounded up to 4 decimal places)

Answer b:
The general way of obtaining a regression equation (least squares) is given below -

let the the equation of be line regression y = atba yi at bai tet s² = {(Wi-a-bxi) 2 estemate Therefore, we and bob the by method a least squares. Performing the derirations get, 852 Sb 85² Sa >> I-a - b*) -O sadaqaletogldo >> ? (y; - Q -bx;) -0 DOLOG BIRORROR => ny na tn bt a) Ixig; = a&x; & bl{xi) atbñ = I di Yi = (y-ba) & di&o & 2 y – ba =nył + b ( 1 2 - n*? & the value b sh 7 Zi – não 9 a.] si sa b sh yai 3 * [substituting a. Sy SA de correlation wlfficient sy- Standard deviation py sari 17 11 I a 199, b- cov. (ay) vor. (a)

where b = a, is the slope of the regression equation and a = b, is the Y - Intercept of the equation

Therefore, Slope a = Cov.(x, y) / Var.(x) = -3.6992 / (17.9762^2) = -0.0114

Y - Intercept, b = $\bar{y}$ - b$\bar{x}$ = 25 - ((-0.0114) x 59.8333) = 25.6821

The regression equation is -

$\hat{y}$ (predicted value) = -0.0114x + 25.6821

Answer c:
when x = 62 miles

$\hat{y}$ = ((-0.0114) x 62) + 25.6821 = 24.9753

If someone trained 62 miles the time (predicted) taken to cover 5K is 24.9753 minutes

Answer d:
Actual time taken to cover 5K when someone trained 62 miles is 20.5 minutes

Residual = Actual y value - Predicted y value = 20.5 - 24.9753 = -4.4753 minutes

(All values are rounded up to 4 decimal places)