In a study relating the degree of warping, in mm, of a copper plate (y) to temperature in °c (x), the following summary statistics were calculated:

n=40,  x¯=26.36, y¯=0.5188,∑ni=1(xi−x¯)2=98,775, ∑ni=1(yi−y¯)2=19.75,∑ni=1(xi−x¯)(yi −y¯)=806.94.

a. compute the least-squares line for predicting warping from temperature. round the answers to four decimal places.

y = ? ? + ? ? x

b. predict the warping at a temperature of 40°c. round the answer to three decimal places.

? ? mm

c. at what temperature will we predict the warping to be 0.5 mm? round the answer to two decimal places.

? ? °c

a)

y = 0.3035 + 0.0082x

b)

0.6315 mm

c)

x = 23.9634 °C

Step-by-step explanation:

a. Compute the least-squares line for predicting warping from temperature. Round the answers to four decimal places.

We need to find an equation of the form

y = b + mx

where m is the slope and b the Y-intercept.

The slope m can be computed with the formula

Replacing the values in our formula (we will round at the end of the calculations)

the Y-intercept b is computed with the formula

therefore we have

and the least-squares line rounded to 4 decimals would be

y = 0.3035 + 0.0082x

b. Predict the warping at a temperature of 40°C. Round the answer to three decimal places.

We simply replace x with 40 to get

y = 0.3035 + 0.0082*40 = 0.6315 mm

c. At what temperature will we predict the warping to be 0.5 mm? Round the answer to two decimal places

Here, we replace y with 0.5 and solve for x

0.5 = 0.3035 + 0.0082x ===> x = (0.5-0.3035)/0.0082 ===>

x = 23.9634 °C

i think it will be zero as slope has a formula of difference between initial and final distance divided by itz time interval

x = -5

step-by-step explanation: