Use the following to answer the questions below:
Students in a small statistics course wanted to investigate if forearm length (in cm) was useful for predicting foot length (in cm). The data they collected are displayed in the provided scatterplot (with regression), and the computer output from the analysis is provided.
Use three decimal places when reporting the results from any calculations, unless otherwise specified.
The regression equation is Foot (cm) = 9.22 + 0.574 Forearm (cm)
| Predictor | Coef | SE Coef | T | P |
| Constant | 9.216 | 4.521 | 2.04 | 0.066 |
| Forearm (cm) | 0.5735 | 0.1578 | 3.63 | 0.004 |
| Source | DF | SS | MS | F | P |
| Regression | 1 | 44.315 | 44.315 | 13.20 | 0.004 |
| Residual Error | 11 | 36.916 | 3.356 | | |
| Total | 12 | 81.231 | | | |
Predicted Values for New Observations
| Forearm (cm) | Fit | SE Fit | 95% CI | 95% PI |
| 28 | 25.274 | 0.513 | (24.144, 26.403) | (21.086, 29.461) |
Use the ANOVA table to compute and interpret
R2.
▸ 0.298
About 30% of the variability in foot lengths in this sample is explained by the person's forearm length.
▸ 0.206
About 21% of the variability in foot lengths in this sample is explained by the person's forearm length.
▸ 0.546
About 55% of the variability in foot lengths in this sample is explained by the person's forearm length.
▸ 0.454
About 45% of the variability in foot lengths in this sample is explained by the person's forearm length.
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