Question

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=19.50,∑ni=1(xi−x¯)(yi−y¯)=846.94. Compute the least-squares line for predicting warping from temperature. (Round the final answers to three decimal places.) yˆ = + x

Answer 1

The least-squares line for predicting warping from temperature is given by the formula \( \hat{y} = b_0 + b_1x \), where:

\[ b_1 = \frac{\sum_{i=1}^{n}(x_i - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i - \bar{x})^2} \]

\[ b_0 = \bar{y} - b_1\bar{x} \]

Using the provided summary statistics:

\[ b_1 = \frac{846.94}{98,775} \approx 0.0086 \]

\[ b_0 = 0.5188 - 0.0086 \times 26.36 \approx 0.303 \]

Therefore, the least-squares line is \( \hat{y} = 0.303 + 0.0086x \).