Simple linear regression with prediction

> # Input data

## Example input data
> head(my_df)
               Y            X
[1,] -0.39643897  0.014797410
[2,] -1.11042940 -1.634525956
[3,] -0.02086076 -0.340575102
[4,]  1.90382401  0.117504381
[5,]  0.62557591  1.042435648
[6,] -1.40269719 -0.006455178

## Data has 50 rows, two columns
> dim(my_df)
[1] 50  2

######

## Model formula here is Y depends on X
> lm(data=my_df, " Y ~ X ")

Call:
lm(formula = " Y ~ X ", data = my_df)

Coefficients:
(Intercept)            X
 -1.076e-17    5.600e-01

> model = lm(data=my_df, " Y ~ X ")

## You can get more information using the `summary()` function
> summary(model)

Call:
lm(formula = " Y ~ X ", data = my_df)

Residuals:
     Min       1Q   Median       3Q      Max
-1.65443 -0.57053 -0.06958  0.47720  1.83802

Coefficients:
              Estimate Std. Error t value Pr(>|t|)
(Intercept) -1.076e-17  1.184e-01   0.000        1
X            5.600e-01  1.196e-01   4.683 2.35e-05 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.8371 on 48 degrees of freedom
Multiple R-squared:  0.3136,    Adjusted R-squared:  0.2993
F-statistic: 21.93 on 1 and 48 DF,  p-value: 2.352e-05

### The adjusted R-squared and p-values of X indicate that overal the model seems moderately correlated and the association between X and Y is significant.


### Predicting new observations

## Get some new data
> new_data = data.frame(X = seq(-3, 3, 0.5) )
> new_data
      X
1  -3.0
2  -2.5
3  -2.0
4  -1.5
5  -1.0
6  -0.5
7   0.0
8   0.5
9   1.0
10  1.5
11  2.0
12  2.5
13  3.0

## Predict using existing model

## $fit shows you the values of your prediction for your new input Xs
> predict(model, new_data, se.fit=TRUE)
$fit
            1             2             3             4             5
-1.680000e+00 -1.400000e+00 -1.120000e+00 -8.400000e-01 -5.600000e-01
            6             7             8             9            10
-2.800000e-01 -1.075513e-17  2.800000e-01  5.600000e-01  8.400000e-01
           11            12            13
 1.120000e+00  1.400000e+00  1.680000e+00

$se.fit
        1         2         3         4         5         6         7         8
0.3777751 0.3215416 0.2668595 0.2149163 0.1682676 0.1326235 0.1183807 0.1326235
        9        10        11        12        13
0.1682676 0.2149163 0.2668595 0.3215416 0.3777751

$df
[1] 48

$residual.scale
[1] 0.8370783