Question

question 21 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. 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 is the equation of the least - squares regression line for predicting beak heat loss, as a percentage of total body heat loss from all sources, from temperature? use decimal notation. give the values of the intercept and slope to three decimal places. use the letter x to represent the value of the temperature. $hat{y}=$ explain in specific language what the slope of this line says about the relationship between beak heat loss and temperature. for every degree celsius, the toucan will lose about 2.1% more heat through its beak on average. for every degree celsius, the toucan will lose about 2.1% less heat through its beak on average for every degree celsius, the toucan will lose about 1.6% less heat through its beak on average for every degree celsius, the toucan will lose about 1.6% more heat through its beak on average.

Explanation:

Step1: Calculate the means

Let \(x\) be temperature and \(y\) be percent heat loss from beak.
\(\bar{x}=\frac{15 + 16+\cdots+30}{16}=\frac{336}{16}=21\)
\(\bar{y}=\frac{32 + 36+\cdots+63}{16}=\frac{733}{16}=45.8125\)

Step2: Calculate the slope \(b_1\)

\[

$$\begin{align*} b_1&=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i-\bar{x})^2}\\ \sum_{i = 1}^{16}(x_i-\bar{x})(y_i - \bar{y})&=(15 - 21)(32-45.8125)+(16 - 21)(36 - 45.8125)+\cdots+(30 - 21)(63-45.8125)\\ &=(- 6)\times(-13.8125)+(-5)\times(-9.8125)+\cdots+9\times17.1875\\ &=82.875 + 49.0625+\cdots+154.6875\\ &=336.5\\ \sum_{i=1}^{16}(x_i-\bar{x})^2&=(15 - 21)^2+(16 - 21)^2+\cdots+(30 - 21)^2\\ &=(-6)^2+(-5)^2+\cdots+9^2\\ &=36 + 25+\cdots+81\\ &=208 \end{align*}$$

\]
\(b_1=\frac{336.5}{208}\approx1.618\)

Step3: Calculate the intercept \(b_0\)

\(b_0=\bar{y}-b_1\bar{x}=45.8125-1.618\times21\)
\(b_0=45.8125 - 33.978=11.835\)
The equation of the least - squares regression line is \(\hat{y}=11.835 + 1.618x\)

Step4: Interpret the slope

The slope \(b_1 = 1.618\approx1.6\). This means that for every degree Celsius increase in temperature, the toucan will lose about \(1.6\%\) more heat through its beak on average.

Answer:

\(\hat{y}=11.835+1.618x\)
For the multiple - choice part: D. For every degree Celsius, the toucan will lose about \(1.6\%\) more heat through its beak on average.