Introduction to linear regression

Chapter 7 - Introduction to Linear Regression

* Practice: 7.23, 7.25, 7.29, 7.39
* Graded: 7.24, 7.26, 7.30, 7.40

The relationship between number of calories and amount of calbohydarates in grams that Starbucks food menu items contain seems to have positive correlation.
Explanatory = Calories, Response = Carbs
Fitting a regression line to these data will enable predicting the number of carbs in gram for a given number of calories.
A simple linear model is inadequate for modelling these data as we do not satisfy the criteria for constant variability, despite the relationship is linear and normality in residuals shown.

mean.shoulder <- 107.20
sd.shoulder <- 10.37
mean.height <- 171.14
sd.height <- 9.41

R.hs <- 0.67

# slope
b1 <- R.hs * (sd.height / sd.shoulder)
b1

## [1] 0.6079749

b0 <- mean.height - b1 * mean.shoulder
b0

## [1] 105.9651

\[\hat{height} = 105.965 + 0.608 * shoulder\]
The slope is 0.6079749 – This means that for every 10 cm increase in soulder girth, there will be an aditional 6.08cm to the height. But not all values makes sense as we plug them into linear regression equation. e.g. a person with a shoulder width 0cm (hypothetically) still indicates a height of 105.965 which doesn’t makes sense.

r.squared <- R.hs^2
r.squared

## [1] 0.4489

This indicates that 0.4489 of the variation found in this data is explinaed by the linear model – explained by the soulder girth width.

random.student.s <- 100
student.height <- b0 + (b1 * random.student.s)
student.height

## [1] 166.7626

residual <- 160 - student.height
residual

## [1] -6.762581

This means that this model overestimated the actual value by -6.7625805

Based on scatterplot in 7.15, the original data had a response variable between 85cm~135cm. A measure of 56 is outside the sample, thus it would be inappropriate to use this linear regression for prediction as it is out of the range that was based on.

\[\hat{y} = -0.357 + 4.034 * body.weight\]
Expected heart rate of a cat with 0 kg of body weight is -0.357, This doesn’t make sense. The regression will only return meaningful value when the inputs are in the range.
For each additional kg increase in body weight, we expect an additional 4.034 grams in the heart weight of a cat.
Body weight explains 64.66% of the variability in the heart weight of a cat.
0.8041144

b1 <- (3.9983 - 4.010) / (-0.0883)
b1

## [1] 0.1325028

There maybe a near zero slope level. But the data suggests there is correlation between beauty and teaching evaluation. The p-value in the summary chart suggests that we shoudl reject the H0 (there is no correlation).
In order to use the least linear regression method, we need to check 4 criterias.
- Linearity: There seem to be linearity in the scatter plot.
- Near normal residuals: The residuals appear to be nearly normal.
- Constant variability: Residual plots shows constant variability.
- Independent observations: There is no obvious pattern that suggests these observations were dependent on each other.

In short, We can perform the linear regression.