Regression/Deviation Homework - Direction, Strength, Slope & Intercept

[Calculator14]

Homework Statement

Data were obtained on watermelon plants; variables of interest were the depth of the roots (inches) and the weight of the watermelon (ounces). A total of 100 plants were measured, with the following summary results: depth mean = 16, weight mean = 75, depth standard deviation = 8, weight standard deviation = 20, and correlation = 0.68. Assume that the relationship is linear. [10 points total]

a. Describe the relationship’s direction and strength. [2 points]
b. Determine the value for the slope of the least squares regression line for predicting weight.
c. Determine the value for the intercept of this least squares regression line.
d. State this regression equation.
e. When the roots are 20 inches deep, predict the watermelon weight.
f. One of the data points was depth = 20, weight = 86. Find the residual for this point.
g. Give the coordinates for two points that are on the regression line. [2 points]
h. If we formed the regression equation to predict root depth, would the resulting regression line have the same slope? (Yes or No)

Homework Equations

I THINK I know a variety of the equations I have to use, for I usually am given a data table to work with. Please help, the wording is very confusing and I have been on this question for two hours.

The Attempt at a Solution

the direction is positive for a. and I am not sure about it's strength but I do know it is 68%
b = a (sy/sx) for b.?
For number c. I understand I have to find b but I am not sure how due to the wording

I think h. is yes but I am not sure :/

Please help me!:(

 

[KataKoniK]

a. The direction of the relationship is positive, meaning that as the depth of the roots increases, the weight of the watermelon also increases. The strength of the relationship is moderate, with a correlation coefficient of 0.68.

b. To determine the slope of the least squares regression line, you can use the formula b = r (Sy/Sx), where b is the slope, r is the correlation coefficient, Sy is the standard deviation of the dependent variable (weight), and Sx is the standard deviation of the independent variable (depth). Plugging in the values, we get b = 0.68 * (20/8) = 1.7.

c. To determine the intercept of the least squares regression line, you can use the formula a = y - bx, where a is the intercept, y is the mean of the dependent variable (weight), b is the slope, and x is the mean of the independent variable (depth). Plugging in the values, we get a = 75 - (1.7 * 16) = 48.8.

d. The regression equation is y = 48.8 + 1.7x, where y is the predicted weight and x is the depth of the roots.

e. When the roots are 20 inches deep, the predicted weight would be y = 48.8 + (1.7 * 20) = 81.8 ounces.

f. To find the residual, you need to first calculate the predicted weight using the regression equation. In this case, the predicted weight would be y = 48.8 + (1.7 * 20) = 81.8 ounces. The residual is then the difference between the actual weight (86 ounces) and the predicted weight (81.8 ounces), which is 4.2 ounces.

g. Two points on the regression line would be (16, 75) and (20, 81.8). These coordinates represent the mean values for depth and weight, and the predicted weight for a depth of 20 inches, respectively.

h. No, the slope of the regression line to predict root depth would be different. This is because the regression line is based on the relationship between depth and weight, and switching the variables would result in a different slope.