# Thread: ˚F to ˚C. New estimated regression line?

1. ## ˚F to ˚C. New estimated regression line?

GIVEN:
x=˚F, y= deflection adjustment factor (y≥0)
n=15 ∑x(sub i)=1425 ∑y(sub i)=10.68
∑x²(sub i)=139,037.25 ∑x(sub i)y(sub i)=987.645
∑y²(sub i)=7.8518

QUESTION
c. Suppose temperature were measured in ˚C rather than in ˚F. What would be the estimated regression line?

MY WORKING SOLUTION
I calculated the est. regression line for ˚F and got:
y= 1.377-.00376x.

Do I only do this to get the new regression line:
y=[5/9(1.377-32)] - [5/9(.00376-32)]x?

Thanks,
Yvonne

2. Originally Posted by yvonnehr
GIVEN:
x=˚F, y= deflection adjustment factor (y≥0)
n=15 ∑x(sub i)=1425 ∑y(sub i)=10.68
∑x²(sub i)=139,037.25 ∑x(sub i)y(sub i)=987.645
∑y²(sub i)=7.8518

QUESTION
c. Suppose temperature were measured in ˚C rather than in ˚F. What would be the estimated regression line?

MY WORKING SOLUTION
I calculated the est. regression line for ˚F and got:
y= 1.377-.00376x.

Do I only do this to get the new regression line:
y=[5/9(1.377-32)] - [5/9(.00376-32)]x?

Thanks,
Yvonne
No, you have:

$\displaystyle y= 1.377-.00376x_f$

where $\displaystyle x_f$ is the tempratue in Fahrenheit. Now $\displaystyle x_f=(9/5)x_c+32$ where $\displaystyle x_c$ is the corresponding temprature in Centigrade so:

$\displaystyle y= 1.377-.00376((9/5)x_c+32)$

CB