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

• Dec 6th 2008, 03:41 PM
yvonnehr
˚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(Thinking)
• Dec 7th 2008, 01:30 AM
CaptainBlack
Quote:

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(Thinking)

No, you have:

$
y= 1.377-.00376x_f
$

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

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

CB