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Disclaimer: this is directly related to a homework problem. The example I am giving is based off the homework problem (because I am more interested in how to solve it than in what the answer is).

Data given:

  • Number of people surveyed: 50
  • Sum of x values: 1500
  • Sum of y values: 500
  • Point on least squares regression line: (5, 0)

Problem: What is the least squares regression line's predicted y-value at x = 25?

What I've done so far: Based off of the count and sum, I was able to get the mean for x (30) and for y (10). However, it is here that I get stuck. I only know how to calculate the least square regression line (LSRL) when given Sx and Sy (std dev of x and y). But I cannot calculate the std dev of either without having the actual data points!

After some thought, I realized that since I know one point of the LSRL, I could figure out any other point along the line if I knew the slope. But then I realized I do not know how to calculate the slope without first knowing Sx and Sy.

I have reached the conclusion that there are only three possible reasons I do not yet have the answer to this question...

  1. There is a way to calculate the std dev from this data, but I don't know it.
  2. There is another way to calculate the second point without using std dev, but I don't know it.
  3. My professor forgot to give us the data table, and without that I am unable to solve this.
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  • $\begingroup$ It's number two. $\endgroup$
    – Fojtasek
    Commented Oct 25, 2011 at 2:03
  • $\begingroup$ @Fojtasek haha, yeah. I just figured it out and answered my own question. I think the actual act of writing down step by step all the issues I was having is what allowed me to realize the answer XD $\endgroup$
    – Moses
    Commented Oct 25, 2011 at 2:08

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As it turns out, the answer to this question is quite simple, and I had an aha moment shortly after posting this question that revealed the answer to me.

A least squared linear regression model always goes through the point (x-bar,y-bar). Knowing this fact as well as a second point on the line, I was able to solve the problem.

First calculate the slope:

b0 = (y2 - y1) / (x2 - x1)

Then calculate the y-intercept:

b1 = -b0*x + y

Finally, plugin in the desired x-value. Now that you have the slope and y-int, so you are able to solve for any point on the line by plugging in the x or y value into the y=mx+b formula.

~Hope this helps someone else who also had this question~

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