A researcher is interested if the number of pets people own (the x-variable) is related to their stress on a scale of 0-80 (the y-variable). Given these sample data points:

A data table for X and Y variables
	xn	yn
n1	2	60
n2	1	50
n3	3	40
n4	4	70
n5	4	50

Calculate and show all your work:

1. cov_xy(the covariance ofxand y)
2. s_x(the standard deviation ofx)
3. s_y(the standard deviation ofy)
4. r_xy(the correlation ofxand y)

For the same data-set:

1. Calculate the regression prediction equation ( = bx + a).
2. Provide an interpretation for the slope (b) and the y-intercept (a).
3. If someone has three pets (x = 3), what would you predict their stress to be (the y-variable)?
4. If someone has eight pets (x = 8), what would you predict their stress to be (the y-variable)?Why might calculating this be a problem?

Please show work using the method provided in the following examples so I can better understand. Thank you!

A researcher is interested if the number of exercise in hours (x) is related to energy (y). Given these sample data points...

xn yn n1 6 53 n2 3 42 n3 4 47 n4 7 38

A. Covariance - cov= (x - )( y - )/(N-1)

Step 1: Calculate deviation scores for the X variable: 6 - 5 = +1 3 - 5 = -2 4 - 5 = -1 7 - 5 = +2 Step 2: Calculate deviation scores for the X variable: 53 - 45 = +8 42 - 45 = -3 47 - 45 = +2 38 - 45 = -7 Step 3: Cross-multiply the deviations scores of x and y 1 x 8 = 8 -2 x -3 = 6 -1 x 2 = -2 2 x -7 = -14 Step 4: Add the cross-products 8 + 6 + -2 + -14 = -2 Step 5: Dvide by N-1 cov= -2/3 cov= -0.67

B. sx (the standard deviation of xn)

Step 1: Find the mean o (6 + 3 + 4 + 7)/4 = 20/4 = 5

Step 2: Deviation Scores o 6 - 5 = +1 o 3 - 5 = -2 o 4 - 5 = -1 o 7 - 5 = +2

Step 3: Square the deviation scores o (1) = 1 o (-2) = 4 o (-1) = 1 o (2) = 4

Step 4: Add all the squared deviation scores together o 1 + 4 + 1 + 4 = 10 Step 5: divide by (n-1) o 10/(4-1) = 10/3 = 3.33 Variance= 3.33 Step 6: Take square root of sample variance o (3.33)= 1.3 Standard Deviation = 1.83

C. sy (the standard deviation of yn) Step 1: Find the mean o (53 + 42 + 47 + 38)/4 = 180/4 = 45 Step 2: Deviation Scores o 53 - 45 = +8 o 42 - 45 = -3 o 47 - 45 = +2 o 38 - 45 = -7 Step 3: Square the deviation scores o (8) = 64 o (-3) = 9 o (2) = 4 o (-7) = 49 Step 4: Add all the squared deviation scores together o 64 + 9 + 4 + 49 = 126 Step 5: divide by (N-1) o 126/(4-1) = 126/3 = 42 Variance= 42 Step 6: Take square root of sample variance o (42)= 6.48 Standard Deviation = 6.48 d. rxy (the correlation of xn and yn) Step 1: rxy = cov/(SxSy) o rxy = -0.67/(1.83)(6.48) o rxy = -0.67/11.86 = -.06 rxy = -0.06

2. For the same data-set,

A. Calculate the regression prediction equation ( = bx + a). Step 1: Find b b = cov/Sx2 b = -0.67 /3.33 b= -0.20

Step 2: Find a o a = - b() o a = 45 - [(-0.20)(5)] o a = 45 - (-1) o a = 46 = -0.20xi + 46

B. Provide an interpretation for the slope (b) and the y-intercept (a). For every one more hour people exercise, I would predict their energy goes down .20 points (b = -.20). If someone does not exercise at all (x = 0), I would predict their energy is 60 points (a = 60) C. If someone works out 6 hours a day (x = 6) what would you predict her energy to be (y)? = -.20(6) + 46 = -1.2 + 46 = 44.8 D. If someone works out 9 hours a day (x = 9) what would you predict her energy to be (y)? = -.20(9) + 46 = -1.8 + 46 = 44.2 Why might calculating this be a problem? Provide YOUR explanation here!