Question

year, x tuition, y The data in the chart show the cost of tuition at a state university for the years 1997-2001. Fit a regression line to the data, where x is the number of years after 1997. 1. (1998) 10,287 Then find the coefficient of correlation. Finally, use the 2. (1999) 10,787 0. (1997) 9651 Choose the correct regression line. O A. y-610.7x+9621 O C. y 610.7x-9621 O B. y 9621x+610.7 O D. y 9621 The coefficient of correlation is (Round to the nearest thousandth as needed) Click to select your answer(s)

Answer 1

a) The correct regression line is : y = 610.7x+9621

Since this line gives values of y for x = 0, 1, 2, 3, 4 which are closest than the other lines.

b) COMPUTATION TABLE :

x	y	x2	xy	y2
0	9651	0	0	93141801
1	10287	1	10287	105822369
2	10787	4	21574	116359369
3	11278	9	33834	127193284
4	12209	16	48836	149059681
10	54212	30	114531	591576504

Now, the coefficient of correlation is :

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2-(\sum x)^2][n\sum y^2-(\sum y)^2]}}$

i.e., $r=\frac{5*114531-10*54212}{\sqrt{[5*30-10^2][5*591576504-54212^2]}}$

i.e., $r=\frac{30535}{\sqrt{50*18941576]}}$

i.e., $r=\frac{30535}{\sqrt{947078800}}$

i.e., $r\approx0.992$

Therefore, the coefficient of correlation is = 0.992.

c) Since the coefficient of correlation is very close of 1, therefore the regression line is a good fit.

d) Putting x = 7 in the above equation we get,

y(7) = 610.7*7+9621

i.e., y(7) $\approx$ 13896

Therefore, the predicted cost of tuition in 2004 is $13896.

e) Putting x = 10 in the above equation we get,

y(7) = 610.7*10+9621

i.e., y(7) = 15728

Therefore, the predicted cost of tuition in 2004 is $15728.

f) Putting x = 13 in the above equation we get,

y(7) = 610.7*13+9621

i.e., y(7) $\approx$ 17560

Therefore, the predicted cost of tuition in 2004 is $17560.