We load the data

library(readr)
students <- read_tsv("students2018-2020-tidy.tsv",
    col_types = cols(
        sex = col_factor(levels = c("Male", "Female")),
        handedness = col_factor(levels = c("Left", "Right"))
    )
)
students

# A tibble: 117 x 10
   answer_date id    english_level sex   birthdate  birthplace height_cm
   <date>      <chr> <chr>         <fct> <date>     <chr>          <dbl>
 1 2018-09-17  3e50… I can speak … Male  1993-02-01 -/Turkey         179
 2 2018-09-17  479d… I can unders… Fema… 1998-05-21 Kahramanm…       168
 3 2018-09-17  39df… I can read a… Fema… 1998-01-18 Batman/Tu…        NA
 4 2018-09-17  d2b0… I can read a… Male  1998-08-29 Antalya/T…       170
 5 2018-09-17  f22b… I can read a… Fema… 1998-05-03 Izmir/Tur…       162
 6 2018-09-17  849c… İngilizce bi… Fema… 1995-10-09 Yalova/Tu…       167
 7 2018-09-17  8381… I can speak … Fema… 1997-09-19 Adıyaman/…       174
 8 2018-09-17  b0dd… I can read a… Male  1997-11-27 Bursa/Tur…       180
 9 2018-09-17  2972… I can read a… Fema… 1999-01-02 Istanbul/…       162
10 2018-09-17  72c0… I can read a… Fema… 1998-10-02 Istanbul/…       172
# … with 107 more rows, and 3 more variables: weight_kg <dbl>,
#   handedness <fct>, hand_span <dbl>

Plotting a formula

plot(weight_kg ~ height_cm, data = students)

Reminder of plot() function

• plot(y ~ x) looks like plot(x, y)
• Formulas are nice:
  • plot(y~x, data=dframe)
• In general the defaults are good
  • axis labels are the plotted variable’s names
  • ranges are automatic
• You can choose the horizontal and vertical ranges

Linear models

In science we work by creating models of how nature works

There are several kinds of models

One of the easiest and more commonly used are the linear models

We approximate all our data by a straight line that shows the relationship between some variables, with a formula like \[y=a + b\cdot x\]

A simple model

We want to describe the relationship between weight and height

A basic rule of thumb is that weight in kg is usually equal to height in cm minus 100 \[weight = height - 100\]

That is, the weight is the number of centimeters over one meter

Adding a straight line to a plot

plot(weight_kg ~ height_cm, data = students)
abline(a = -100, b = 1)

A-B-line

This command adds a straight line in a specific position

• abline(h=1) adds a horizontal line in 1

• abline(v=2) adds a vertical line in 2

• abline(a=3, b=4) adds an \(a +b\cdot x\) line

Deciding the model formula

The hard part is to decide

• What is the dependent variable
  • The values that we want to predict
  • The vertical axis
• What are the independent variables
  • The values that we can control
  • The horizontal axis
• What is the formula connecting them

Our model

Here we are using a linear model \[y=a + b\cdot x\] more precisely \[weight = a + b\cdot height\]

Which are the best \(a\) and \(b\) values?

Finding the best line

We draw the figure using a formula

plot(weight_kg ~ height_cm, data=students)

Using the same formula we can get a linear model

lm(weight_kg ~ height_cm, data=students)

Call:
lm(formula = weight_kg ~ height_cm, data = students)

Coefficients:
(Intercept)    height_cm  
   -84.0079       0.8822  

Drawing the tendency line

We draw using abline() with the coefficients of lm()

plot(weight_kg ~ height_cm, data=students)
abline(a = -84.0079, b = 0.8822)

Making the computer work for us

We must avoid copying data manually or using fixed values

model <- lm(weight_kg ~ height_cm, data=students)
model

Call:
lm(formula = weight_kg ~ height_cm, data = students)

Coefficients:
(Intercept)    height_cm  
   -84.0079       0.8822  

The useful part are the coefficients

coef(model)

(Intercept)   height_cm 
 -84.007888    0.882157 

The formula is \(y=a + b\cdot x\)

• \(a\) is coef(model)[1]

• \(b\) is coef(model)[2]

Drawing the tendency line

plot(weight_kg ~ height_cm, data=students)
model <- lm(weight_kg ~ height_cm, data=students)
abline(a=coef(model)[1], b=coef(model)[2])

Easily drawing the tendency line

plot(weight_kg ~ height_cm, data=students)
model <- lm(weight_kg ~ height_cm, data=students)
abline(model)

Predicting with the model

Beyond giving a description of the data, models are often used to get a prediction of what would be the output of the system when we have new data

In this case we need to provide a data.frame with at least one column. The column name must be the same as the one used to create the model. For example

data.frame(height_cm=155:205)

Predicting with the model

model <- lm(weight_kg ~ height_cm, data=students)
guess <- predict(model, newdata=data.frame(height_cm=155:205))

Results

(some data omitted from the table. Get all data at
http://dry-lab.org/static/2017/rubber1.txt)

Graphic Results

plot(length~n_balls, data=rubber)

What are the coefficients?

Remember that straight lines can be represented by the formula \[\text{n_marbles}=A+B\cdot \text{length}\] The coefficient \(A\) is the value where the line intercepts the vertical axis

The coefficient \(B\) is how much length goes up when n_marbles increases. This is called slope

Physical interpretation of the linear model

The formula from Hooke’s Law is \[\text{force}=K\cdot(L-\text{height})\] Since force is the weight of the balls, we can write \[-m g\cdot\text{n_balls}=K\cdot(L-\text{lenght})\] which can be re-written as \[\text{lenght}=\underbrace{L}_{A}+\underbrace{\frac{m g}{K}}_{B}\cdot\text{n_balls}\]

Physical interpretation of coef(model)

When there are no balls, the length of the coil is \(L\), in this case

coef(model)[1]

(Intercept) 
   76.48188 

If the mass of each ball is 20gr, we can find \(K\) as

coef(model)[2]/(0.020 * 9.8)

 n_balls 
25.81566 

Marbles weight \(\approx 21\pm 0.5\) grams