Statistics - Finding a relationship?

[shamieh]

let a and b be constants and let \(\displaystyle y_j = ax_j+b\) for \(\displaystyle j = 1,2...n\). What are the relationships between the means of y and x, and the standard deviations of y and x?

I'm not sure what they are wanting here?

 

[MarkFL]

Let's look at the means first...we know:

\(\displaystyle \overline{y}=\frac{1}{n}\sum_{j=1}^n\left(y_j\right)\)

Can you use the definition of $y_j$ and the properties of sums to obtain a relationship between $\overline{y}$ and $\overline{y}$?

Once you have the above, then look at:

\(\displaystyle \sigma_y=\sqrt{\frac{\sum\limits_{j=1}^n\left(y_j-\overline{y}\right)^2}{n}}\)

Replace $y_j$ and $\overline{y}$ with the expressions in terms of $x_j$ and $\overline{x}$, and you should be able to establish a relationship between $\sigma_y$ and $\sigma_x$. :D

 

[shamieh]

would it be that they are both squared?

 

[shamieh]

I don't really understand. This is my first statistic class. is ybar supposed to be the symbol for the mean?

 

[shamieh]

Let's look at the means first...we know:

\(\displaystyle \overline{y}=\frac{1}{n}\sum_{j=1}^n\left(y_j\right)\)

Can you use the definition of $y_j$ and the properties of sums to obtain a relationship between $\overline{y}$ and $\overline{y}$?

Once you have the above, then look at:

\(\displaystyle \sigma_y=\sqrt{\frac{\sum\limits_{j=1}^n\left(y_j-\overline{y}\right)^2}{n}}\)

Replace $y_j$ and $\overline{y}$ with the expressions in terms of $x_j$ and $\overline{x}$, and you should be able to establish a relationship between $\sigma_y$ and $\sigma_x$. :D

Sorry for multiple posts. I think I see it now. the deviation of the $i^{th}$ observation, $y_i$, from the sample mean $\overline{y}$ is the difference between them $y_i - \overline{y}$

 

[MarkFL]

Yes, the bar over a variable represents the mean.

This is what I was suggesting you do:

\(\displaystyle \overline{y}=\frac{1}{n}\sum_{j=1}^n\left(y_j\right)=\frac{1}{n}\sum_{j=1}^n\left(ax_j+b\right)=a\frac{1}{n}\sum_{j=1}^n\left(x_j\right)+\frac{1}{n}bn=a\overline{x}+b\)

Now for the deviation:

\(\displaystyle \sigma_y=\sqrt{\frac{\sum\limits_{j=1}^n\left(y_j-\overline{y}\right)^2}{n}}=\sqrt{\frac{\sum\limits_{j=1}^n\left(ax_j+b-a\overline{x}-b\right)^2}{n}}=a\sqrt{\frac{\sum\limits_{j=1}^n\left(x_j-\overline{x}\right)^2}{n}}=a\sigma_x\)

Both of these results should agree nicely with intuition too. :D