They decide to begin simply by running a model called m1 which has just one predictor variable (temperature) and the outcome variable meatballs. The output of that model is shown below. Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 113.7984 7.3137 15.56

Question

They decide to begin simply by running a model called m1 which has just one predictor variable (temperature) and the outcome variable meatballs. The output of that model is shown below.

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 113.7984     7.3137   15.56   <2e-16 ***
temperature   8.3263     0.4848   17.17   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 60.57 on 363 degrees of freedom
Multiple R-squared:  0.4483,    Adjusted R-squared:  0.4467 
F-statistic: 294.9 on 1 and 363 DF,  p-value: < 2.2e-16

Which of the following best describes the relationship between temperature and meatballs?

BlackTom AI · Accepted Answer

The slope (temperature coefficient) is 8.3263, so a one unit increase in temperature (1°C) is associated with an increase of about 8.3 meatballs sold.

1) Relevant concepts:
- In a linear regression y = β0 + β1 x, β1 is the change in the outcome y for a one-unit increase in predictor x.
- Statistical significance is assessed with t = Estimate / Std. Error and the p-value; a very small p-value (e.g., < 2e-16) indicates strong evidence that β1 ≠ 0.
- R-squared measures the fraction of variance explained by the model.

2) Step-by-step reasoning and calculation:
- From the output, the temperature coefficient (β1) = 8.3263 with Std. Error = 0.4848.
- Compute the t-statistic: t = 8.3263 / 0.4848 ≈ 17.17 (matches the reported t value).
- The p-value for this t is < 2e-16, indicating the positive slope is highly statistically significant.
- Interpretation: For each additional 1°C in temperature, the expected number of meatballs sold increases by 8.3263 on average, holding nothing else constant (this simple model has only temperature as predictor).
- The intercept (β0 = 113.7984) is the model’s predicted meatballs when temperature = 0°C, but the key interpretation question asks about the slope.

3) Intermediate calculations shown:
- t = Estimate / SE = 8.3263 / 0.4848 ≈ 17.17.
- Multiple R-squared = 0.4483 implies ~44.8% of variability in meatballs is explained by temperature in this model.

4) Why the correct answer is correct:
- The positive slope value directly corresponds to "8.3 more meatballs per 1°C" (rounded from 8.3263).
- The extremely small p-value confirms this relationship is statistically significant, so the direction and magnitude are meaningful.

5) Why other options are incorrect:
- Options stating “8.3°C is associated with one more/one fewer meatball” invert the units (they claim 1 meatball per 8.3°C instead of 8.3 meatballs per 1°C).
- Options claiming “fewer meatballs” contradict the positive coefficient (8.3263 > 0).
- “There is no significant relationship” is false because the t-statistic and p-value show strong significance.

Conclusion: The best description is “Every increase in temperature of 1C is associated with 8.3 more meatballs being sold.” This matches the provided correct answer.

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