Will standard units be affected by multiplying a negative constant?

Question

So I'm reasonably sure that for part (a) the standard unit won't change because we basically only multiplied the numbers in list 1 by 2 and added a constant 1.

For part (B) though, I just want to make sure my rationale is correct.

the operation that we would have to do to get the numbers in (ii) is we multiply -2 then add 1 to every number in (i).

My intuition tells me that when we multiply a negative number this changes the SU compare to the numbers in list (i) right?

Thank you guys!

Answer 1

At $a)$ we have two lists 1. and 2. Each of the data can be converted to standard units:

$$x_{ij}^*=\frac{x_{ij}-\overline x_j}{\sigma_j}$$

$i$ is the index for the i-th value. $j$ is the index for list 1 or list 2.

Let $\overline x_1$ the average from list 1. From the text below you can evaluate that at $a)$ the average from list 2 is

$\overline x_2=2 \overline x_1+1 \quad (1)$

And the standard deviation is $\sigma_2=2\sigma_1 \quad (2)$

The converted values (standard units) has to be equal

$$\frac{x_{i1}-\overline x_1}{\sigma_1}=\frac{x_{i2}-\overline x_2}{\sigma_2}$$

Now we can insert the expression for $\overline x_2$ and $\sigma_2$ from (1) and (2) and see if the equation is equal. For this purpose we take the first values from list 1 and list 2: $x_{11}=1, x_{12}=3$

$$\frac{1-\overline x_1}{\sigma_1}=\frac{3-(2\overline x_1+1)}{2\sigma_1}$$

Are the two expressions equal? If yes, then the two lists 1 and 2 are equal after conversion. The same can be done at $b)$.

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At $b)$ the average of the converted data is

$\overline x^c=-2\overline x+1$

It is related to the expected value for random variables:

$\mathbb E(aX+b)=a\cdot \mathbb E(X)+b$

In general we have the empirical variance $\hat s^2(x)$. Then $\hat s^2(ax+b)=a^2\cdot \hat s^2(x)$

It is related to the formula for random variables: $Var(aX+b)=a^2\cdot Var(X)$

Thus in b) the variance of the converted data is four-times ($a^2= (-2)^2$) of the variance of the original data since $a=-2$. Therefore the standard deviation double: $\sqrt{4}=2$ . We see the negative sign of the factore doesn´t change sign of the standard deviation.

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