In a study relating the degree of warping, in mm, of a

QUESTION:

In a study relating the degree of warping, in mm, of a copper plate \((y)\) to temperature in \({ }^{\circ}\mathrm C \ (x)\), the following summary statistics were calculated: \(n=40\),

\(\sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}=98,775\), \(\sum_{i=1}^{n}\left(y_{i}-\bar{y}\right)^{2}=19.10\), \(\overline{x}=26.36\), \(\overline{y}=0.5188\), \(\sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)\left(y_{i}-\overline{y}\right)=826.94\).

a. Compute the correlation r between the degree of warping and the temperature.

b. Compute the error sum of squares, the regression sum of squares, and the total sum of squares.

c. Compute the least-squares line for predicting warping from temperature.

d. Predict the warping at a temperature of \(40^\circ \mathrm C\).

e. At what temperature will we predict the warping to be 0.5 mm?

f. Assume it is important that the warping not exceed 0.5 mm. An engineer suggests that if the temperature is kept below the level computed in part (e), we can be sure that the warping will not exceed 0.5 mm. Is this a correct conclusion? Explain.

Equation Transcription:

Text Transcription:

(y)

^oC (x)

n=40

sum_i=1^n(x_i-overline{x})^2=98,775

sum_i=1^n(y_i-overline{y})^2=19.10

overline{x}=26.36

overline{y}=0.5188

sum_ii=1n(x_i-overline{x})(y_i-overline{y})=826.94

40^oC