Predicting the predictor in linear regression? I am trying to solve the following Statistics 101 task on linear regression:
Both A and B are metric variables. It is known that the value of A is a good predictor for the value of B and in an experiment the following results are observed:
A       B
1199    33
502     49
617     71
1029    52
487     62
1548    17
1251    49
860     35
891     62
766     45

In the first task I am supposed to predict the corresponding B for A = 900. 
I computed the regression line (slope of -0.033, intercept of 77.720) and therefore got a prediction of approximately 74.747.
Next I am supposed to predict A for B = 40.
My idea was to simply use the same parameters and solve the equation for A like this: B = slope * A + intercept, therefore A = (B - intercept) / slope
Using this I got a prediction A = 1142.085.
My sample solution says the prediction is supposed to be 1030.814. I assume I was supposed to compute another regression line B —> A and use these values to predict A. But I don’t quite understand how this is a better approximation as the task description said that A was known to be a good predictor for B (and not the other way around)?
 A: The slope for A~B minimizes errors between B and the model.
The slope for B~A will be different, and minimize errors between A and the model.
These two approaches will result in different slopes.
For a visual representation, see: https://www.r-bloggers.com/principal-component-analysis-pca-vs-ordinary-least-squares-ols-a-visual-explanation/
Having said that, as someone mentioned, the wording of the question is poor, and doesn't provide context. I don't think you're supposed to go into the real world and just throw whatever model you want at data.
A: I think I can see what your instructor is trying to do, but I'm not sure that I would have used this.
Let's recall first principles. In classical simple linear regression, the independent variable $x$ is measured without error and fixed. For each value of $x$, there is a subpopulation of values of the dependent variable $y$ which are normally distributed and have equal variance.
If we specify that A is your independent variable and B is your dependent variable and the above assumptions hold, then  it is difficult to turn this around in the second part of your problem when you reverse the identities of the two variables. That is to say, regressing B on A, you are making assumptions about B and A. When you regress A on B, the former assumptions you made now do not hold.
Let's suspend the issue about assumptions for a little bit.
Regressing A on B will produce a different slope when regressing B on A. This is easily shown. In the regression equation $B=\alpha_1 + \beta_1A$, 
$\beta_1=\dfrac{\sum (A_i-\bar{A})(B_i-\bar{B})}{\sum (A_i-\bar{A})^2}$.
However, in the regression equation $A=\alpha_2 + \beta_2B$,
$\beta_2=\dfrac{\sum (A_i-\bar{A})(B_i-\bar{B})}{\sum (B_i-\bar{B})^2}$.
You can see that $\beta_1 \neq \beta_2$.
