Question

The table shows the total 4-year cost (in thousands of dollars) and the Graduation rate (nearest percentage) from 4 randomly selected colleges/universities in the U.S. 129 Cost, x Graduation Rate, y 104 62 88 30 69 35 45 The mean of x is 97.5 The standard deviation of x is 25.410 The mean of y is 43 The standard deviation of y is 14.119 The deviations and the Z-scores for each x and y value are given below: Note: I tried my best with the tables below. BlackBoard has some issues with some of the font and symbols that were used. Zx Zy (x - x) 31.5 6.5 -9.5 -28.5 (y - y) 2 19 1.240 0.256 -0.374 -1.122 0.142 1.346 -0.921 -0.567 a) Show the equations along with plugging the values into the corresponding equation on how the mean of the y values and the standard deviation of the y values were obtained. You may use a table above to help you with the standard deviation. b) Find the correlation coefficient. You may use any of the tables above for help or any other equation method (Just stating the answer from a graphing calculator/technology will receive only 0 or 1 point for this question). Round the answer to 3 decimal places.

Answer 1

(a) Mean of X =

ΣΧ, 129 + 104 + 88 + 69 -= 97.5

Mean of Y = $\frac{\sum Y_{i}}{n}=\frac{45+62+30+35}{4}=43$

Standard Deviation of X = $\sqrt{\frac{\sum (X_{i}-\bar X)^{2}}{n-1}}=\sqrt{\frac{(129-97.5)^{2}+(104-97.5)^{2}+(88-97.5)^{2}+(69-97.5)^{2}}{3}}=25.41$

Standard Deviation of Y = $\sqrt{\frac{\sum (Y_{i}-\bar Y)^{2}}{n-1}}=\sqrt{\frac{(45-43)^{2}+(62-43)^{2}+(30-43)^{2}+(35-43)^{2}}{3}}=14.119$

(b) The corrlation Coefficient =$\frac{S_{xy}}{\sqrt{S_{xx}*S_{yy}}}$

$S_{xy} = n*\sum XY- \sum X * \sum Y$

$S_{xx} = n*\sum X^{2}- (\sum X)^{2}$

$S_{yy} = n*\sum Y^{2}- (\sum Y)^{2}$

We have the followign table:

	X	Y	XY	X2	Y2
	129	45	5805	16641	2025
	104	62	6448	10816	3844
	88	30	2640	7744	900
	69	35	2415	4761	1225
SUM	390	172	17308	39962	7994

correlation coefficient = $r = \frac{2152}{\sqrt{7748*2392}}=0.50$