## 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>>

```
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>>

```
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>>

```
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>>

```
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>>

```
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>>

```
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!

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