Need help pls

[anteater10123]

whats the answer?

 

[Deleted member 4993]

whats the answer?

Where is your work?

What method/s have you been taught to find the best fit line?

 

[Dr.Peterson]

whats the answer?

You may be expected just to draw the four lines on the graph, and judge the answer visually. Have you tried that?

Of course, if you have been taught any specific methods, you should use those.

 

[blamocur]

This is a multiple choice problem, so you can simply try each of the choices. Since the points lie almost on a straight line just checking the values at the first and the last points seems enough to figure out which one is the best choice.

 

[Deleted member 4993]

whats the answer?

Are you allowed to use software's like MS-Excel?

 

[JeffM]

This is a multiple choice problem, so you can simply try each of the choices. Since the points lie almost on a straight line just checking the values at the first and the last points seems enough to figure out which one is the best choice.

Or you could calculate the sum of the squared errors for each option, and choose the equation with the least sum of squares.

 

[blamocur]

Or you could calculate the sum of the squared errors for each option, and choose the equation with the least sum of squares.

It is not clear to me what are the exact values in the graph. It is true that small errors in those values would not effect the (multiple choice) answer, but still...

 

[JeffM]

It is not clear to me what are the exact values in the graph. It is true that small errors in those values would not effect the (multiple choice) answer, but still...

You are correct. Stupid problems invite stupid answers

 

[lookagain]

whats the answer?

You need to show some work here. Look at the extreme points, which look
to be (0, 50) and (10, 950), when rounded to the nearest tens.

Work out the slope of a hypothetical line between these points.

This hypothetical line divides the points, where more of them are below the
line than are above it. None of the answer choices include 50 as a vertical
intercept. However, consider if a "better line of best fit" is achieved by
moving down the hypothetical line a smaller increment. Then it can have
about the same number of data points on either side of it. That would
decrease the vertical intercept by a proportional amount.

 

[Otis]

Hi anteater. For this exercise, I like Dr. Peterson's suggestion to graph the lines. Two of the lines are way off. The remaining two lines lie mostly above the points, and that fact makes the answer obvious.