A data analyst performed a linear regression analysis to model the relationship between sepal width (Sepal.Width) and sepal length (Sepal.Length) using the built-in iris dataset with the following R code: model |t|) (Intercept) 3.41895 0.25356 13.48

Question

A data analyst performed a linear regression analysis to model the relationship between sepal width (Sepal.Width) and sepal length (Sepal.Length) using the built-in iris dataset with the following R code:

model <- lm(Sepal.Width ~ Sepal.Length, data = iris)

summary(model)
Call:
lm(formula = Sepal.Width ~ Sepal.Length, data = iris)

Residuals:
    Min      1Q  Median      3Q     Max 
-1.1095 -0.2454 -0.0167  0.2763  1.3338

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept)   3.41895    0.25356   13.48   <2e-16 ***
Sepal.Length -0.06188    0.04297   -1.44    0.152    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.4343 on 148 degrees of freedom
Multiple R-squared:  0.01382,    Adjusted R-squared:  0.007159 
F-statistic: 2.074 on 1 and 148 DF,  p-value: 0.1519

What is the scientific conclusion assuming all assumptions are met?

BlackTom AI · Accepted Answer

The p-value for Sepal.Length is 0.152 (greater than the usual α = 0.05), so we fail to reject the null hypothesis that the slope equals zero; there is no statistically significant linear relationship.

1) Relevant concepts
- Hypothesis test for regression slope: H0: β1 = 0 (no effect) vs HA: β1 ≠ 0 (effect).
- t-statistic = Estimate / Std. Error; corresponding p-value determines statistical significance.
- 95% confidence interval for slope: Estimate ± t* × SE; if it contains 0, the effect is not significant.
- R-squared measures proportion of variance explained.

2) Step-by-step reasoning and calculations
- Coefficient estimate for Sepal.Length: β̂1 = −0.06188.
- Standard error: SE = 0.04297.
- Compute t-statistic: t = β̂1 / SE = −0.06188 / 0.04297 ≈ −1.44 (matches output).
- p-value reported = 0.152. Since 0.152 > 0.05, do not reject H0.
- 95% CI: use t* ≈ 1.976 for df = 148. Margin = 1.976 × 0.04297 ≈ 0.0849. CI = −0.06188 ± 0.0849 → (−0.1468, 0.0230). This interval includes 0, reinforcing non-significance.
- R-squared = 0.01382, so Sepal.Length explains only ~1.38% of the variance in Sepal.Width — an extremely small effect.

3) Why the correct answer is correct
- The slope estimate is small and its confidence interval includes zero; the p-value is not below conventional thresholds. Under the model assumptions, there is no statistically reliable evidence that sepal length affects sepal width.

4) Why other conclusions would be incorrect
- Claiming a significant negative or positive effect is incorrect because p > 0.05 and the CI contains 0.
- Claiming a strong relationship is incorrect because R-squared is ≈1.4%, indicating negligible explained variance.
- Claiming practical importance is not supported because the estimated change (≈ −0.06 cm per 1 cm) is both statistically non-significant and tiny.

Conclusion: The appropriate scientific conclusion is that there is no evidence to suggest that sepal length affects sepal width, which matches the provided answer.

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