Linear Regression Application, Interpolation and Extrapolation

Question

Use the data and story to answer the following questions

The information below shows the number of state-registered automatic weapons and the murder rate for several Northwestern states.

x = 11.3, 8.1, 6.7, 3.4, 2.9, 2.4, 2.5, 0.6

y = 13.5, 10.9, 9.7, 6.6, 6.5, 5.8, 6.5, 4.7

x = thousands of automatic weapons

y = murders per 100,000 residents

A) How many murders per 100,000 residents can be expected in a state with 1.5 thousand automatic weapons?

Answer = _____ (Round to 3 decimal places)

B) How many murders per 100,000 residents can be expected in a state with 2.5 thousand automatic weapons?

Answer = _____ (Round to 3 decimal places)

Answer 1

Hi Anchal,

If we are assuming a linear relationship, which I suspect we are given the question title, then we can use the slope-intercept equation:

y=mx + b

m=slope

b=y-intercept

To get slope, recall formula:

m=(y₂-y₁)/(x₂-x₁)

We can use any two ordered pairs. I'll go with first and last:

(x₁,y₁)= (11.3,13.5)

(x₂,y₂)= (0.6, 4.7)

m=(4.7-13.5)/(0.6-11.3)

m=0.822

Now, we can plug this into our slope-intercept equation and use a point of our choosing for x and y. I'll use the third pair (6.7,9.7):

y=mx + b

9.7= 0.822(6.7) + b

9.7= 5.510 + b

Subtract 5.510 on both sides:

b= 4.190

Now, you have the regression equation:

y= 0.822x + 4.190

We will use this to solve parts (A) and (B). Murders were the response variable. Automatic weapons were the explanatory. Recall that explanatory implies "x," so plug the values you were given in for x:

(A) y^=0.822(1.5) + 4.190

y^= 5.423

(B) y^=0.822(2.5) + 4.190

y^=6.245

I hope this helps.