correlation and Regression using R programming
Tags: tutorial, datascience, newbie, devops
Published: 2021-06-26 00:32:25

## How do I calculate correlation using R programming??
To calculate correlation between two data sets is quite easy all you need is to make use of cor() function and insert the parameters `x`and `y`. 
### Example1
What is the correlation between X and Y if y = (100,94,150,160,180), and x = (23,21,32,40,45).
### Solution
#### Codes>>
```R
y=c(100,94,150,160,180)
x =c(23,21,32,40,45)
cor(x,y)
```
#### Result>>
```
> y=c(100,94,150,160,180)
> x =c(23,21,32,40,45)
> cor(x,y)
[1] 0.978522
```
### Example2
What is the correlation between X and Y if y =(100,94,150,160,180,80), and x = (23,21,32,40,45, 10).
### Solution
####Codes>>
```R
y=c(100,94,150,160,180,80)
x =c(23,21,32,40,45, 10)
cor(x,y)
```
#### Result>>
```
> y=c(100,94,150,160,180,80)
> x =c(23,21,32,40,45, 10)
> cor(x,y)
[1] 0.972529
```

### How to calculate the regression
To deal with regression there are two values very important they are : the beta and the alpha. they are used to form the equation that connect the two items. 
i.e y=bx + a. where a and b represent the alpha and beta respectively.
note: alpha is the intercept why b is the slope. 
  
### How do I calculate the beta and alpha of regression using R programming.
To calculate your alpha and beta using code we make use of `lm()` function.
where the parameters are place inside the lm() functions. let's take a look at some examples
#### Example1
If x = [23,21,32,40,45, 10] and y = [100,94,150,160,180,80]. Obtain the linear model of the relationship between y and x.
Solution
#### Codes>>
```R
y=c(100,94,150,160,180,80)
x =c(23,21,32,40,45, 10)
lm(y ~ x)
```
#### Result>>
```
> y=c(100,94,150,160,180,80)
> x =c(23,21,32,40,45, 10)
> lm(y ~ x)

Call:
lm(formula = y ~ x)

Coefficients:
(Intercept)            x  
     39.693        3.075  
```
*Conclusion*: from the result above alpha =39.693 while Beta=3.075.
and the equation is Y=3.075x+39.693

#### Example2
If x = [23,21,32,40,45] and y = [100,94,150,160,180]. Obtain the linear model of the relationship between y and x.
#### Solution
##### Codes>>
```R
y=c(100,94,150,160,180)
x =c(23,21,32,40,45)
lm(y ~ x)
```
##### Result>>
```
>  y=c(100,94,150,160,180)
>  x =c(23,21,32,40,45)
>  lm(y ~ x)

Call:
lm(formula = y ~ x)

Coefficients:
(Intercept)            x  
     22.071        3.563  
```
Conclusion: from the result above alpha =22.071 while Betaa=3.563.
and the equation is Y=3.563x+22.071.

#### Example3
If x = [23,21,32,40,45, 10] and y = [100,94,150,160,180,80]. What would be the value of y if x = 20.
#### Solution.
Note the we already write codes for this item in Example1. and our equation is Y=3.075x+39.693. so let's find the value of y when x=20.
##### codes>>
```R
x=20
Y=3.075*x+39.693 
print(Y)
```
#### Result>>
```
> x=20
> Y=3.075*x+39.693 
> print(Y)
[1] 101.193
```
### Example4
If x = [23,21,32,40,45, 10] and y = [100,94,150,160,180,80]. What would be the value of x if y = 200
#### Solution.
Note the we already write codes for this item in Example1. and our equation is Y=3.075x+39.693. since we need to find x when y=200 therefore we have to make x the subject of the formula. 
Working that out we get x=(Y-39.693)/3.075. Now let's write the codes.
#### Codes>>
```R
Y=200
x=(Y-39.693)/3.075
print(x)
```
#### Result>>>
```
> Y=200
> x=(Y-39.693)/3.075
> print(x)
[1] 52.13236
```

Yeah. that is how easy it is to use codes to solve most of the problem relating with correletion and regressions. I hope you find this article helpful? Alright you can chat me up if you have any doubt or observation on `07045225718` or `09153036869` it is still your guy Maxwizard. Enjoy coding!