QuestionPredict heart weight (g) from body weight (kg) using regression.
(i) (a) Model: $\hat{y}=-0.339+4.028 x$ (b) Intercept: Expected heart weight at 0 kg is $-0.339$ g. (c) Slope: Each kg increases heart weight by 4.028 g. (d) $R^{2}$ : Body weight explains $64.65\%$ of heart weight variability. (e) Correlation coefficient: $0.8049$

Studdy Solution

STEP 1

Assumptions1. The regression model is linear. . The dependent variable (heart weight) is represented by $y$ .
3. The independent variable (body weight) is represented by $x$ .
4. The intercept and slope are estimated from the dataset and are represented by the coefficients in the regression output.
5. The standard error, $t$ value, and $p$ value are also provided in the regression output.
6. The residual standard error (), coefficient of determination (R^), and adjusted coefficient of determination (R_{adj}^) are provided.

STEP 2

The linear model can be written using the coefficients from the regression output. The intercept is -0.339 and the slope (coefficient for body weight) is4.028.
$\hat{y} = -0.339 +4.028x$

STEP 3

The intercept can be interpreted as the expected heart weight (in grams) of a cat with a body weight of0 kilograms. In this case, the intercept is -0.339, which means the expected heart weight of a cat with a body weight of0 kilograms is -0.339 grams.

STEP 4

The slope can be interpreted as the change in the expected heart weight (in grams) for each additional kilogram in a cat's body weight. In this case, the slope is4.028, which means for each additional kilogram in a cat's weight, we expect their hearts to be heavier by4.028 grams on average.

STEP 5

The coefficient of determination, $R^2$ , can be interpreted as the proportion of the variability in the dependent variable (heart weight) that is explained by the independent variable (body weight). In this case, $R^2$ is64.65%, which means body weight explains64.65% of the variability in the weights of cats' hearts.

STEP 6

The correlation coefficient, $r$ , can be calculated from the square root of $R^2$ . However, we need to determine the sign of $r$ . Since the slope of the regression line is positive, the correlation coefficient should also be positive.
$r = \sqrt{R^2}$

STEP 7

Plug in the value for $R^2$ to calculate the correlation coefficient.
$r = \sqrt{64.65\%}$

STEP 8

Convert the percentage to a decimal value.
$64.65\% =0.6465$ $r = \sqrt{0.6465}$

STEP 9

Calculate the correlation coefficient.
$r = \sqrt{.6465} =.804$ The correlation coefficient, rounded to four decimal places, is.804.

Was this helpful?

Studdy Solution

STEP 1

STEP 2

The linear model can be written using the coefficients from the regression output. The intercept is -0.339 and the slope (coefficient for body weight) is4.028.
$\hat{y} = -0.339 +4.028x$

STEP 3

The intercept can be interpreted as the expected heart weight (in grams) of a cat with a body weight of0 kilograms. In this case, the intercept is -0.339, which means the expected heart weight of a cat with a body weight of0 kilograms is -0.339 grams.

STEP 4

The slope can be interpreted as the change in the expected heart weight (in grams) for each additional kilogram in a cat's body weight. In this case, the slope is4.028, which means for each additional kilogram in a cat's weight, we expect their hearts to be heavier by4.028 grams on average.

STEP 5

STEP 6

The correlation coefficient, $r$ , can be calculated from the square root of $R^2$ . However, we need to determine the sign of $r$ . Since the slope of the regression line is positive, the correlation coefficient should also be positive.
$r = \sqrt{R^2}$

STEP 7

Plug in the value for $R^2$ to calculate the correlation coefficient.
$r = \sqrt{64.65\%}$

STEP 8

Convert the percentage to a decimal value.
$64.65\% =0.6465$ $r = \sqrt{0.6465}$

STEP 9

Calculate the correlation coefficient.
$r = \sqrt{.6465} =.804$ The correlation coefficient, rounded to four decimal places, is.804.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

The linear model can be written using the coefficients from the regression output. The intercept is -0.339 and the slope (coefficient for body weight) is4.028.y^=−0.339+4.028x\hat{y} = -0.339 +4.028xy^​=−0.339+4.028x

STEP 3

The intercept can be interpreted as the expected heart weight (in grams) of a cat with a body weight of0 kilograms. In this case, the intercept is -0.339, which means the expected heart weight of a cat with a body weight of0 kilograms is -0.339 grams.

STEP 4

The slope can be interpreted as the change in the expected heart weight (in grams) for each additional kilogram in a cat's body weight. In this case, the slope is4.028, which means for each additional kilogram in a cat's weight, we expect their hearts to be heavier by4.028 grams on average.

STEP 5

STEP 6

The correlation coefficient, rrr, can be calculated from the square root of R2R^2R2. However, we need to determine the sign of rrr. Since the slope of the regression line is positive, the correlation coefficient should also be positive.r=R2r = \sqrt{R^2}r=R2​

STEP 7

Plug in the value for R2R^2R2 to calculate the correlation coefficient.r=64.65%r = \sqrt{64.65\%}r=64.65%​

STEP 8

Convert the percentage to a decimal value.64.65%=0.646564.65\% =0.646564.65%=0.6465r=0.6465r = \sqrt{0.6465}r=0.6465​

STEP 9

Calculate the correlation coefficient.r=.6465=.804r = \sqrt{.6465} =.804r=.6465​=.804The correlation coefficient, rounded to four decimal places, is.804.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

The linear model can be written using the coefficients from the regression output. The intercept is -0.339 and the slope (coefficient for body weight) is4.028.
$\hat{y} = -0.339 +4.028x$

The correlation coefficient, $r$ , can be calculated from the square root of $R^2$ . However, we need to determine the sign of $r$ . Since the slope of the regression line is positive, the correlation coefficient should also be positive.
$r = \sqrt{R^2}$

Plug in the value for $R^2$ to calculate the correlation coefficient.
$r = \sqrt{64.65\%}$

Convert the percentage to a decimal value.
$64.65\% =0.6465$ $r = \sqrt{0.6465}$

Calculate the correlation coefficient.
$r = \sqrt{.6465} =.804$ The correlation coefficient, rounded to four decimal places, is.804.