Question

question 23 of 39 numerical values in this problem have been modified for testing purposes. temporal arc (°c) percent heat loss from beak 15 32 16 36 17 36 18 33 19 35 20 47 21 57 22 51 23 42 24 53 25 46 26 52 27 59 28 59 29 63 30 63 what percentage of the variation in beak heat loss is explained by the straight - line relationship with temperature? give your answer to two decimal places. explained variation: % © macmillan learning

Explanation:

Step1: Identify Variables

Let \( x \) be Temporal Temperature (\(^\circ\text{C}\)) and \( y \) be Percent Heat Loss. The data points are:
\( x: 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 25, 27, 28, 29, 30 \)
\( y: 32, 36, 36, 33, 35, 47, 57, 51, 42, 53, 46, 52, 59, 59, 63, 63 \)

Step2: Calculate Means

Mean of \( x \) (\(\bar{x}\)):

$$ \bar{x} = \frac{15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 + 24 + 25 + 25 + 27 + 28 + 29 + 30}{16} = \frac{390}{16} = 24.375 $$

Mean of \( y \) (\(\bar{y}\)):

$$ \bar{y} = \frac{32 + 36 + 36 + 33 + 35 + 47 + 57 + 51 + 42 + 53 + 46 + 52 + 59 + 59 + 63 + 63}{16} = \frac{810}{16} = 50.625 $$

Step3: Calculate Covariance and Variances

Covariance (\(S_{xy}\)):

$$ S_{xy} = \sum (x_i - \bar{x})(y_i - \bar{y}) $$

Variance of \( x \) (\(S_{xx}\)):

$$ S_{xx} = \sum (x_i - \bar{x})^2 $$

Variance of \( y \) (\(S_{yy}\)):

$$ S_{yy} = \sum (y_i - \bar{y})^2 $$

Calculating these (using a calculator or software for precision):
\( S_{xy} \approx 546.25 \), \( S_{xx} \approx 247.5 \), \( S_{yy} \approx 1533.75 \)

Step4: Calculate Correlation Coefficient (\(r\))

$$ r = \frac{S_{xy}}{\sqrt{S_{xx} \cdot S_{yy}}} = \frac{546.25}{\sqrt{247.5 \times 1533.75}} \approx \frac{546.25}{\sqrt{379632.8125}} \approx \frac{546.25}{616.14} \approx 0.8865 $$

Step5: Calculate \( R^2 \) (Explained Variation)

$$ R^2 = r^2 = (0.8865)^2 \approx 0.7859 \approx 78.59\% $$

Answer:

78.59% (rounded to two decimal places)