用R建立一个回归模型看一看明年学硕、教育学的国家线
建立数据框
gjx <- data.frame(
year = c(2017,2018,2019,2020,2021,2022),
grade = c(310,320,325,331,337,351)
)
gjx
拟合模型
fit1 <- lm(grade~year,data = gjx)
anova(fit1)
summary(fit1)
F=133.55, p=0.0003203, adjusted R-square=0.97
it seems our model works just fine
confint(fit1)
*the table below seems to be the 95% interval of the coefficient *
| 2.5 % | 97.5 % | |
|---|---|---|
| (Intercept) | -18420.406517 | -11156.393483 |
| year | 5.687247 | 9.284182 |
回归诊断
高斯-马尔科夫假设的诊断
par(mfrow=c(2,2))
plot(fit1)
i don't really know this G-M hypothesis shit
其他乱七八糟的诊断
just here for fun, stop doing such boring、 fun-killing、awkward、annoying、painful especially math-ish things plz
-----wise man
回归方程可视化
library(ggplot2)
ggplot(gjx,
aes(x=year, y=grade,
color="#5e616d"))+
geom_point()+ #绘画散点图
stat_smooth(method = lm,color="black")+ #在散点图加回归拟合线
annotate( "text",
label = "R^2=0.97",
parse=T,x=2019,y=300)+ #在图上添加R方
annotate("text",
label = "y=-14790 + 7.486x",
x=2019,y=305) #在图上添加方程
ggplot2 is amazing
reference：铭记yu心, R语言|回归分析（一） ———R语言数据分析系列(一), CSDN
2017~2022年的预测值
predict(fit1,
data.frame(year=2017:2022),
interval = 'prediction',
level = 0.95)
new <- data.frame(year=c(2023))
### 用于预测的数据名必须与回归中自变量名称相同
well... what could i say since the model nearly precisely predicted the grades of the limit
2023年预测值
predict(fit1,new,interval = 'prediction',level = 0.95)
| fit | lwr | upr | |
|---|---|---|---|
| 2023 | 355.2 | 344.9209 | 365.4791 |
*it seems that the limit of first try of PEE would be at least 344, which is not going to happen. and the top of the limit would be, well, 365, then 365 it is *
画个图吧
x <- c(gjx$year,2023)
y <- c(gjx$grade,365)
plot(gjx$year,gjx$grade,
cex=1, # 图形大小
pch=18, # 点类型
xlim=c(2017,2023), # x轴范围
ylim=c(310,370) # y轴范围
xlab='年份' # x、y的坐标名称, lz太懒了没有加上
ylab='分数')
lines(x,y,lwd=1.5,col='gray') #画折线
points(2023,365,cex=2,pch=17,col='red') #添加2023年的新点
pic
looks like a exp regression is more suitable? someome who know how to realise it contact me