GO Classes DPP | STATISTICS |COVARIANCE

Answer 1

By applying formula of covariance :

Cov(X,Y) = E(XY) - E(X)E(Y)

E(XY) = (80+200+360+560+800)/5 = 400

E(X) = 6

E(Y) = 60

applying in formula we got,

Cov(X,Y) = E(XY) - E(X)E(Y)

Cov(X,Y) = 400-360 = 40

Cov(X,Y) = 40

Comments

correction--

Cov {sample​} = n / n-1 ​× Cov {population​}

so, Cov {sample​} = (5 / 4 ) *  40 = 50

 

• Population covariance: 40

• Sample covariance: 50

---

There is no claim about the given data is a 'sample'.

---

yes i agree , i just came to know by  default is population cov (what i think)

Answer 2

Given :
\[
\begin{array}{ccc}
\text{Student} & \text{Study Hours (X)} & \text{Test Score (Y)} \\
1 & 2 & 40 \\
2 & 4 & 50 \\
3 & 6 & 60 \\
4 & 8 & 70 \\
5 & 10 & 80 \\
\end{array}
\]

There are two formulas for covariance depending on the context:

1. Population covariance:

$$
\operatorname{Cov}(X, Y)=\frac{\sum\left(X_i-\tilde{X}\right)\left(Y_i-\tilde{Y}\right)}{n}
$$

2. Sample covariance:

$$
\operatorname{Cov}(X, Y)=\frac{\sum\left(X_i-\bar{X}\right)\left(Y_i-\bar{Y}\right)}{n-1}
$$

Since we are dealing with a sample of 5 students (not the entire population), we should divide by $n-1=4$.
 

Step 1: Calculate means

$$
\begin{aligned}
& \bar{X}=(2+4+6+8+10) / 5=30 / 5=6 \\
& \bar{Y}=(40+50+60+70+80) / 5=300 / 5=60
\end{aligned}
$$

Step 2: Compute $(\mathrm{Xi}-\overline{\mathrm{X}})$ and $(\mathrm{Yi}-\overline{\mathrm{Y}})$

\[
\begin{array}{ccccc}
X & Y & X_i - \overline{X} & Y_i - \overline{Y} & (X_i - \overline{X})(Y_i - \overline{Y}) \\
2 & 40 & -4 & -20 & 80 \\
4 & 50 & -2 & -10 & 20 \\
6 & 60 & 0 & 0 & 0 \\
8 & 70 & 2 & 10 & 20 \\
10 & 80 & 4 & 20 & 80 \\
\end{array}
\]

Step 3: Sum of products

\begin{aligned}
&\Sigma[(X i-\bar{X})(Y i-\bar{Y})]=80+20+0+20+80=200\\
&\operatorname{Cov}(X, Y)=\frac{200}{4}=50
\end{aligned}

Comments

200 will divide by 5 instead of 4 !

---

No we have to Divide 4 not  by  5

If you have access to the textbook, you should look in Chapter 5, Section 5.7 the sample covar covariance formula 

ref Wackerly, Mendenhall, Scheaffer (2014)

https://mqala.co.za/nisa/Mathematical%20Statistics%20-%207th%20Edition%20-%20Wackerly.pdf

---

I completely agree with the sample , population covariance formulas but the point is - 

There is no claim that the given data is a 'sample' in the question it self!. Why we are assuming it to be a sample?

Data analyst is studying the habits "OF" 5 students not a 'sample of' 5 student from the class . for his experiment population itself is 5 here.

I hope we can arrive to the same page.

---

ok i got it

thanks for correcting me👍