Question

The wind chill factor depends on wind speed and air temp. This data represents the wind speed (in miles per hour) and wind chill factor at an air temp. of 15 degrees F. Wind speed(x) 5, 10, 15, 20, 25, 30, 35 Wind chill(y) 12, -3, -11, -17, -22, -25, -27 Compute the least squares regression line and correlation coefficient for this data. Predict the wind chill for a wind speed of 50 miles per hour. Determine the wind speed (to the nearest mile per hour) for a wind chill of -45 degrees F. State the percentage of explained variation for the data set. State the value for the sum of the squared residuals.

Answer 1

Ans:

	X	Y	XY	X^2	Y^2
1	5	12	60	25	144
2	10	-3	-30	100	9
3	15	-11	-165	225	121
4	20	-17	-340	400	289
5	25	-22	-550	625	484
6	30	-25	-750	900	625
7	35	-27	-945	1225	729
Total=	140	-93	-2720	3500	2401

slope=(7*(-2720)-140*(-93))/(7*3500-140^2)=-1.2286

intercept=(-93-(-1.2286)*140)/7=11.286

Regression eqn:

y'=-1.2286x+11.286

Correlation cofficient,r=(7*(-2720)-140*(-93))/SQRT((7*3500-140^2)*(7*2401-(-93)^2))=-0.952

	X	Y	y'	(y-y')^2	(y-(-13.2857))^2
1	5	12	5.14315	47.02	639.37
2	10	-3	-0.9997	4.00	105.80
3	15	-11	-7.14255	14.88	5.22
4	20	-17	-13.2854	13.80	13.80
5	25	-22	-19.4283	6.61	75.94
6	30	-25	-25.5711	0.33	137.22
7	35	-27	-31.714	22.22	188.08
Total=	140	-93		108.86	1165.43
	mean(y)=	-13.2857		SSE	SST
			SSR=	1056.57

SSE=sum of squared residual=108.86

SST=1165.43

SSR=1165.43-108.86=1056.57

cofficient of determination,R^2=1056.57/1165.43=0.9066,that means 90.66% of the variation is explained by the above regression model.