Question

question 4
4 pts

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:

construct a 95% confidence interval for the slope (\\(\beta_1\\)) parameter.

lower limit: (round your answer to 2 decimal places)

upper limit: (round your answer to 2 decimal places)

Explanation:

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

Step 1: Identify the Point Estimate and Standard Error

From the R output, locate the row for the predictor variable temperature to find the estimated slope (\(b_1\)) and its standard error (\(\text{SE}(b_1)\)):

$$ b_1 = -0.22273 $$

$$ \text{SE}(b_1) = 0.04884 $$

Step 2: Determine the Critical Value

The degrees of freedom (\(\text{df}\)) for the simple linear regression model are given in the output as:

$$ \text{df} = 10 $$

For a \(95\%\) confidence interval with \(\text{df} = 10\), the critical value \(t^*\) is obtained from the \(t\)-distribution table:

$$ t^*_{0.025, 10} = 2.228 $$

Step 3: Calculate the Confidence Interval Limits

Calculate the margin of error (\(\text{ME}\)):

$$ \text{ME} = t^* \times \text{SE}(b_1) = 2.228 \times 0.04884 \approx 0.10882 $$

Calculate the lower and upper limits:

$$ \text{Lower Limit} = b_1 - \text{ME} = -0.22273 - 0.10882 = -0.33155 \approx -0.33 $$

$$ \text{Upper Limit} = b_1 + \text{ME} = -0.22273 + 0.10882 = -0.11391 \approx -0.11 $$

Answer:

• Lower Limit: -0.33
• Upper Limit: -0.11