To calculate the covariance, we will use the formula for covariance between two variables $X$ and $Y$:

$\text{Cov}(X, Y) = \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})(Y_i - \bar{Y})$

In this case:

• The sample mean of $X$, denoted $\bar{X}$, is $0.035$.
• The sample mean of $Y$, denoted $\bar{Y}$, is $0.12$.

We also have six pairs of $(X_i, Y_i)$ values. Let's plug each pair into the formula and compute each term individually.

Given:

1. $(X_1, Y_1) = (-0.1, 0.21)$
2. $(X_2, Y_2) = (0.2, 0.07)$
3. $(X_3, Y_3) = (0.05, 0.3)$
4. $(X_4, Y_4) = (-0.05, -0.03)$
5. $(X_5, Y_5) = (0.02, -0.08)$
6. $(X_6, Y_6) = (0.09, 0.25)$

Step-by-Step Calculation

1. For $(X_1, Y_1)$:
   $(X_1 - \bar{X})(Y_1 - \bar{Y}) = (-0.1 - 0.035)(0.21 - 0.12) = (-0.135)(0.09) = -0.01215$

2. For $(X_2, Y_2)$:
   $(X_2 - \bar{X})(Y_2 - \bar{Y}) = (0.2 - 0.035)(0.07 - 0.12) = (0.165)(-0.05) = -0.00825$

3. For $(X_3, Y_3)$:
   $(X_3 - \bar{X})(Y_3 - \bar{Y}) = (0.05 - 0.035)(0.3 - 0.12) = (0.015)(0.18) = 0.0027$

4. For $(X_4, Y_4)$:
   $(X_4 - \bar{X})(Y_4 - \bar{Y}) = (-0.05 - 0.035)(-0.03 - 0.12) = (-0.085)(-0.15) = 0.01275$

5. For $(X_5, Y_5)$:
   $(X_5 - \bar{X})(Y_5 - \bar{Y}) = (0.02 - 0.035)(-0.08 - 0.12) = (-0.015)(-0.2) = 0.003$

6. For $(X_6, Y_6)$:
   $(X_6 - \bar{X})(Y_6 - \bar{Y}) = (0.09 - 0.035)(0.25 - 0.12) = (0.055)(0.13) = 0.00715$

Summing All Terms

Now, summing these six products: $-0.01215 + (-0.00825) + 0.0027 + 0.01275 + 0.003 + 0.00715 = 0.0052$

Final Covariance

Since we have six data points, we divide the sum by $n = 6$: $\text{Cov}(X, Y) = \frac{0.0052}{6} = 0.0008667$

So, the covariance is approximately: $\text{Cov}(X, Y) \approx 0.00087$

Would you like more details, or do you have any questions?

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Here are 5 additional questions to explore:

1. How does changing the mean of one variable affect covariance?
2. How would variance differ from covariance in a data set?
3. How can covariance be standardized to calculate correlation?
4. What does a negative covariance indicate about two variables?
5. How would you interpret a covariance of 0?

Tip: Covariance gives insights into the direction of the linear relationship between variables but not the strength—correlation is typically used to understand both.