The mean height of American women is about 63.4 inches and the standard deviatio

Question

The mean height of American women is about 63.4 inches and the standard deviation is about 1.8 inches. The mean height of American men is about 66.6 inches with a standard deviation of 3.3 inches. The correlation between the heights of husbands and wives is about 0.4.

(A) We want to predict the height of the husband whose wife is 63.4 inches tall. The slope of the appropriate regression line is

(a) 20.1088

(b) 0.2182

(c) 0.7333

(d) 48.8679

(B) The intercept of the appropriate regression line is

a. 0.2182

b. 48.8679

c. 20.1088

d. 0.7333

(C) The predicted value of husbands height when wife’s height is 63.4 inches is

a. 64.1333

b. 66.60002

c. 1275.63122

(D) There are two data points, x1= wife’s height=63 with husband’s height= 70.7067, and another pair with x2=67.4 and y2=78.33322. If you use the regression line computed earlier then
i. The residual at x2 is equal to the residual at x1

ii. The residual at x2 is smaller than the residual at x1

iii. The residual at x2 is larger than the residual at x1

(E) The regression line that you found out in this problem is

i. The regression line of mens’s heights on women’s heights

ii. The regression line of womens’s heights on men’s heights

iii. It does not matter both are the same

please answer A-E if possible!

Explanation / Answer

A) slope b1=r*Sy/Sx =0.4*3.3/1.8=0.7333

option C

B)

intercept bo=Ybar-b1*Xbar =66.6-0.7333*63.4=20.1088

option C

C)

predicted value =20.1088+63.4*0.7333=66.60002

option B

d)

for residual for x1=Yactual -Yprediceted =70.7067-(20.1088+63*0.7333)=4.4

residual for X2 =Yactual -Yprediceted =78.3322-(20.1088+67.4*0.7333)=8.80

iii. The residual at x2 is larger than the residual at x1

E) i. The regression line of mens’s heights on women’s heights

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