Call: lm(formula = qualified ~ weight + relate + weight:relate, data = data) Res

Question

Call:
lm(formula = qualified ~ weight + relate + weight:relate, data = data)

Residuals:
   Min     1Q Median     3Q    Max
-3.65 -1.15 -0.15   0.85   2.85

Coefficients:
              Estimate Std. Error t value Pr(>|t|)   
(Intercept)    4.51164    1.01676   4.437 1.62e-05 ***
weight         0.63836    0.63416   1.007    0.316   
relate         0.59788    0.63888   0.936    0.351   
weight:relate -0.09788    0.39435 -0.248    0.804   
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.298 on 172 degrees of freedom
Multiple R-squared: 0.06599,   Adjusted R-squared: 0.0497
F-statistic: 4.051 on 3 and 172 DF, p-value: 0.008192

write down the equation that R gives and interpret all the coefficients and the p-value associated with the coefficcients.

Explanation / Answer

The regression equation is formed using the coefficients values as shown

Qualified = 4.51164 +0.63836*weight + 0.59788*relate -0.09788*weightRelate

The values of the coeffiecients can be interpreted as

For every unit increase in the value of weight , the value of qualified increases by 0.63836

For every unit increase in the value of relate , the value of qualified increases by 0.59788

All the p values that are less than signficance level of 0.05 are considered signficant for the equation , here none of the p values are less than 0.05 , hence the variables do not contribute significantly in explaining the variation of the model

This is also shown by the low rsquare value which is just 0.06599 , this means that the model is able to explain only 6% variation in the data

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.