Question

a researcher is examining the impact of rising water temperatures on the availability of dissolved oxygen for aquatic organisms. it is well established in aquatic ecology that warmer water tends to hold less dissolved oxygen. to investigate this relationship, the researcher collects data (n=12) and applies a simple linear regression model in r. the corresponding output from the analysis is presented below.

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call:

lm(formula = dissolved_oxygen ~ temperature)

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residuals:

min 1q median 3q max

-2.25909 -0.58182 0.03636 0.59091 1.95000

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coefficients:

estimate std. error t value pr(>|t|)

(intercept) 14.46364 0.85110

temperature -0.22273 0.04884

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signif. codes: 0 0.001 0.01 0.05 . 0.1 1

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residual standard error: 1.168 on 10 degrees of freedom

multiple r-squared: 0.6753, adjusted r-squared: 0.6428

f-statistic: on 1 and 10 df, p-value:

using the information from the r output above, determine the coefficient of determination (\\(r^2\\)) and the correlation coefficient (r)

coefficient of determination (\\(r^2\\)): (round your answer to 4 decimal places)

correlation coefficient: (round your answer to 4 decimal places)

Explanation:

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

Step 1: Identify the Coefficient of Determination (\(R^2\))

From the R output, locate the value for "Multiple R-squared":

$$ R^2 = 0.6753 $$

Step 2: Calculate the Correlation Coefficient (\(r\))

The correlation coefficient \(r\) is the square root of \(R^2\), with its sign determined by the slope of the regression line (the coefficient of the independent variable temperature):

$$ \text{Slope} = -0.22273 \quad (\text{negative}) $$

$$ r = -\sqrt{R^2} = -\sqrt{0.6753} \approx -0.821766 $$

Rounding to 4 decimal places:

$$ r \approx -0.8218 $$

Answer:

• Coefficient of Determination (\(R^2\)): 0.6753
• Correlation Coefficient (\(r\)): -0.8218