Question

question 22 of 39 > that beak heat loss is higher at higher temperatures and that the relationship is roughly linear. note: the numerical values in this problem have been modified for testing purposes. use the equation of the least - squares regression line to predict beak heat loss, as a percentage of total body heat loss from all sources, at a temperature of 25°c. give your answer to one decimal place.

temperat ure (°c)	percent heat loss from beak
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

$hat{y}=$

Explanation:

Step1: Calculate means

Let \(x\) be temperature and \(y\) be percent heat loss from beak.
\(\bar{x}=\frac{15 + 16+\cdots+30}{16}=\frac{360}{16}=22.5\)
\(\bar{y}=\frac{32 + 36+\cdots+63}{16}=\frac{739}{16}=46.1875\)

Step2: Calculate slope \(b_1\)

\[

$$\begin{align*} \sum_{i = 1}^{16}(x_i-\bar{x})(y_i - \bar{y})&=(15 - 22.5)(32-46.1875)+(16 - 22.5)(36 - 46.1875)+\cdots+(30 - 22.5)(63 - 46.1875)\\ &= 592.5\\ \sum_{i=1}^{16}(x_i-\bar{x})^2&=(15 - 22.5)^2+(16 - 22.5)^2+\cdots+(30 - 22.5)^2\\ &=170 \end{align*}$$

\]
\(b_1=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i-\bar{x})^2}=\frac{592.5}{170}\approx3.4853\)

Step3: Calculate intercept \(b_0\)

\(b_0=\bar{y}-b_1\bar{x}=46.1875-3.4853\times22.5=46.1875 - 78.41925=- 32.23175\)

Step4: Form regression equation

The regression - line equation is \(\hat{y}=b_0 + b_1x=-32.23175+3.4853x\)

Step5: Predict for \(x = 25\)

\(\hat{y}=-32.23175+3.4853\times25=-32.23175 + 87.1325 = 54.90075\approx54.9\)

Answer:

54.9