Question

question 7
1 pts
consider the following r output, which models dissolved oxygen levels based on water temperature.

r
slr_model <- lm(dissolved_oxygen ~ temperature)
slr_model

##

call:

lm(formula = dissolved_oxygen ~ temperature)

##

coefficients:

(intercept) temperature

14.4636 -0.2227

predict the dissolved oxygen when water temperature is 21°c, using the regression model. round your answer to 4 decimal places.

question 8
1 pts
consider the following r output, which models dissolved oxygen levels based on water temperature.

r
slr_model <- lm(dissolved_oxygen ~ temperature)
slr_model

##

call:

lm(formula = dissolved_oxygen ~ temperature)

##

coefficients:

(intercept) temperature

14.4636 -0.2227

the dissolved oxygen level of water at 21°c is found to be 9.0 mg/l. calculate the residual using the regression model. round your answer to 4 decimal places.

Explanation:

⚡ Using what you learned: regression, residuals, and model fit

Step 1: Formulate the Regression Equation

From the R output, the coefficients are:

• Intercept (\(b_0\)): \(14.4636\)
• Slope for temperature (\(b_1\)): \(-0.2227\)

The regression equation is:

$$ \widehat{y} = 14.4636 - 0.2227x $$

Step 2: Predict Dissolved Oxygen for Question 7

Substitute \(x = 21\) into the equation:

$$ \widehat{y} = 14.4636 - 0.2227(21) $$

$$ \widehat{y} = 14.4636 - 4.6767 $$

$$ \widehat{y} = 9.7869 $$

Step 3: Calculate the Residual for Question 8

Using the actual value \(y = 9.0\) and the predicted value \(\widehat{y} = 9.7869\):

$$ \text{Residual} = y - \widehat{y} $$

$$ \text{Residual} = 9.0 - 9.7869 $$

$$ \text{Residual} = -0.7869 $$

Answer:

Question 7:
9.7869

Question 8:
-0.7869