How to do statistical analysis about curvilinear model in R?

Question

I generated one data

treatment<- c(0,24,36,48)
yield<- c(4.78,9.67,8.02,6.7)
dataA<- data.frame(treatment, yield)

Then I did draw a curvilinear graph.

ggplot(data=dataA, aes(x=treatment, y=yield))+
  stat_smooth(method='lm', linetype=1, se=FALSE, formula=y~poly(x,2), 
  size=0.5, color="Blue") +
  geom_point (col="Black", size=4) +
  scale_y_continuous(breaks = seq(0,12,2), limits = c(0,12)) +
  labs(x="Fertilizer application (kg/ha)", y="Yield (ton/ha)") +
  theme_classic(base_size=18, base_family="serif")+
  theme(axis.line= element_line(size=0.5, colour="black"))+
  windows(width=5.5, height=5)

Also, I analyzed statistical significance.

regression<- lm(yield ~ poly(treatment,2), data=dataA)
summary(regression)

Coefficients:
                    Estimate Std. Error t value Pr(>|t|)  
(Intercept)           7.2925     0.4376  16.663   0.0382 *
poly(treatment, 2)1   1.5441     0.8753   1.764   0.3283  
poly(treatment, 2)2  -3.1137     0.8753  -3.557   0.1745  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.8753 on 1 degrees of freedom
Multiple R-squared:  0.9404,    Adjusted R-squared:  0.8211 
F-statistic: 7.884 on 2 and 1 DF,  p-value: 0.2442

When I checked the equation model in Excel, the model equation was y = -0.006x2 + 0.3234x + 4.8742 but in R, it says y=-3.1137x2 + 1.5441x + 7.2925

Why the model equation is different between Excel and R? Which equation is correct?

To analyze curvilinear, the code, yield ~ poly(treatment,2) would be wrong?

Could you let me know about this?

[Excel]

Thanks,

You may try `lm(yield ~ treatment+ I(treatment^2), data=dataA)` instead. Take a look at https://stackoverflow.com/questions/17457884/using-lmpoly-to-get-formula-coeff . — Park, Nov 01 '22 at 07:37

score 1 · Accepted Answer · answered Nov 01 '22 at 07:41

The Excel and R regressions are not equivalent.

The poly function in R computes orthogonal polynomials by default. In order for the regression to be the same as the Excel equation, you can set the raw argument to TRUE:

regression<- lm(yield ~ poly(treatment, 2, raw = TRUE), data=dataA)
summary(regression)

Output:

Call:
lm(formula = yield ~ poly(treatment, 2, raw = TRUE), data = dataA)

Residuals:
    1        2        3        4
-0.08345  0.50073 -0.66764  0.25036

Coefficients:
                                Estimate Std. Error t value Pr(>|t|)
(Intercept)                      4.863455   0.871291   5.582    0.113
poly(treatment, 2, raw = TRUE)1  0.325773   0.083092   3.921    0.159
poly(treatment, 2, raw = TRUE)2 -0.006098   0.001714  -3.557    0.174

Residual standard error: 0.8753 on 1 degrees of freedom
Multiple R-squared:  0.9404,    Adjusted R-squared:  0.8211
F-statistic: 7.884 on 2 and 1 DF,  p-value: 0.2442

How to do statistical analysis about curvilinear model in R?

1 Answers1