Question

numerical values in this problem have been modified for testing purposes.

temperal ure (°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:

%

Explanation:

Step1: Identify Variables

Let \( x \) be Temperature (°C) and \( y \) be Percent Heat Loss. The data points are:
\( (15, 32), (16, 36), (17, 36), (18, 33), (19, 35), (20, 47), (21, 57), (22, 51), (23, 42), (24, 53), (25, 46), (26, 52), (27, 59), (28, 59), (29, 63), (30, 63) \)

Step2: Calculate Mean of \( x \) and \( y \)

Mean of \( x \) (\( \bar{x} \)):
\( \bar{x} = \frac{15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30}{16} = \frac{368}{16} = 23 \)

Mean of \( y \) (\( \bar{y} \)):
\( \bar{y} = \frac{32 + 36 + 36 + 33 + 35 + 47 + 57 + 51 + 42 + 53 + 46 + 52 + 59 + 59 + 63 + 63}{16} = \frac{800}{16} = 50 \)

Step3: Calculate \( SS_{xx} \), \( SS_{yy} \), \( SS_{xy} \)

\( SS_{xx} = \sum (x_i - \bar{x})^2 \)
\( (15-23)^2 + (16-23)^2 + \dots + (30-23)^2 = 240 \)

\( SS_{yy} = \sum (y_i - \bar{y})^2 \)
\( (32-50)^2 + (36-50)^2 + \dots + (63-50)^2 = 1768 \)

\( SS_{xy} = \sum (x_i - \bar{x})(y_i - \bar{y}) \)
\( (15-23)(32-50) + (16-23)(36-50) + \dots + (30-23)(63-50) = 600 \)

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

\( r = \frac{SS_{xy}}{\sqrt{SS_{xx} \cdot SS_{yy}}} = \frac{600}{\sqrt{240 \cdot 1768}} \approx \frac{600}{\sqrt{424320}} \approx \frac{600}{651.4} \approx 0.921 \)

Step5: Calculate \( R^2 \) (Coefficient of Determination)

\( R^2 = r^2 = (0.921)^2 \approx 0.848 \) (or 84.80% when multiplied by 100)

Answer:

84.80%