MHB Statistics - Finding a relationship?

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The discussion focuses on establishing relationships between the means and standard deviations of two variables, y and x, defined by the equation y_j = ax_j + b. It is clarified that the mean of y, denoted as \(\overline{y}\), can be expressed as \(\overline{y} = a\overline{x} + b\). Additionally, the standard deviation of y, \(\sigma_y\), is found to be \(\sigma_y = a\sigma_x\). This indicates that the mean of y is a linear transformation of the mean of x, while the standard deviation of y is scaled by the constant a. Overall, the relationships derived align with statistical intuition regarding linear transformations.
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let a and b be constants and let $$y_j = ax_j+b$$ for $$j = 1,2...n$$. 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?
 
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Let's look at the means first...we know:

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

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:

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

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
 
would it be that they are both squared?
 
I don't really understand. This is my first statistic class. is ybar supposed to be the symbol for the mean?
 
MarkFL said:
Let's look at the means first...we know:

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

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:

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

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 $y_i - \overline{y}$
 
Yes, the bar over a variable represents the mean.

This is what I was suggesting you do:

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

Now for the deviation:

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

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