Question

the chart below tracks starting salaries for health care workers since the year 2000. using linear regression, which equation is the line of best fit?

a

b

c

Explanation:

Step1: Recall the linear regression formula

The general form of a linear regression equation is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y - intercept.

Step2: Analyze the data trend

As the year (\(x\)) increases, the starting salary (\(y\)) also increases. So the slope \(m\) should be positive.
In option B: \(y = 27541.81x+839.72\), the slope \(m = 27541.81>0\)
In option C: \(y = 27541.81x - 839.72\), the slope \(m = 27541.81>0\)
In option A: \(y=839.72x - 27541.81\), the slope \(m = 839.72\) (much smaller than the other two options, and when \(x = 2000\), \(y=839.72\times2000-27541.81=1679440 - 27541.81=1651898.19\) which is not in line with the data values in the table)
Let's assume \(x\) represents the year. When \(x = 2000\)
For option B: \(y=27541.81\times2000 + 839.72=55083620+839.72 = 55084459.72\) (not in line with the data \(y = 28000\))
For option C: Let \(x\) be the number of years after 2000. If \(x = 0\) (year 2000), \(y=27541.81\times0-839.72=- 839.72\) (not correct). If we consider \(x\) as a mis - labeled variable (maybe a calculation error in the problem setup), but if we use the formula \(y=mx + b\) and assume a positive relationship between \(x\) (year) and \(y\) (salary) with a reasonable slope - intercept combination.
If we use the formula \(y=839.72x+27541.81\) (assuming a typo in option A's sign). When \(x = 0\) (year 2000), \(y = 27541.81\approx28000\) (considering rounding), when \(x = 1\) (year 2001), \(y=839.72\times1+27541.81=28381.53\approx29750\) (not a perfect fit but better than other options with wrong sign in intercept or slope - intercept combination)

Answer:

A. \(y = 839.72x+27541.81\)