Question

The following observations are given for two variables. x y 2 7 6 10 9 13 11 18 11 21 12 25 7 12 4 7 ​ a. Compute the sample covariance for the above data (rounded to the nearest hundredths). What does the covariance tell us using the context of the problem. (3 points) b. Compute the standard deviation for x (rounded to the nearest hundredths). (2 points) c. Compute the standard deviation for y (rounded to the nearest hundredths). (2 points) d. Compute the sample correlation coefficient (rounded to the nearest hundredths). What does the correlation coefficient tell us that the covariance does not using the context of the problem. (3 points)

Answer 1

The given data is:

x = (2, 7, 6, 10, 9, 13, 11, 18)

y = (11, 21, 12, 25, 7, 12, 4, 7)

a. The formula used for Sample Covariance is:

`S_(XY)=\frac{\sum_(i=1)^(n)(X_(i)-\bar{X})(Y_(i)-\bar{Y})}{n-1}`

where,
`\bar{X}` is mean of X

and
`\bar{Y}` is mean of Y

Firstly calculating
`\bar{X} and \bar{Y}`. We get,

`\bar{X}` = 9.5

`\bar{Y}` = 12.375

Now, Putting all values in above formula. We get,

Sample Covariance
`S_(XY)` = -8.64

b. Standard Deviation is the square root of sum of square of the distance of observation from the mean.

`Standard deviation(\sigma) = \sqrt{(1)/(n)\sum_(i=1)^(n){(x_(i)-\bar{x})^(2)} }`

where,
`\bar{x}` is mean of the distribution.

Using formula we get,

Standard deviation
`(\sigma_(X))` = 4.81

c. Using above formula we get,

Standard deviation
`(\sigma_(Y))` = 7.21

d. Correlation Coefficient is calculate by using formula:

`r_(xy)=(S_(XY))/(S_(X)S_(Y))`

where,
`S_(XY)` = Covariance of X and Y

`S_(X)` = Standard Deviation of X

`S_(Y)` = Standard Deviation of Y

Putting all values in above formula, we get,

Correlation Coefficient
`(r_(xy))` = -0.25

Correlation Coefficient tell us that X and Y has negative week relationship whereas Sample Covariance tell us that X and Y has negative relationship. It does not tell us strength of relationship.