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. 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. 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 means

Let $x$ be temperature and $y$ be percent heat - loss from beak.
$n = 16$
$\bar{x}=\frac{15 + 16+\cdots+30}{16}=\frac{360}{16}=22.5$
$\bar{y}=\frac{32 + 36+\cdots+63}{16}=\frac{738}{16}=46.125$

Step2: Calculate slope ($b_1$)

$b_1=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2}$
$\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})=(15 - 22.5)(32 - 46.125)+(16 - 22.5)(36 - 46.125)+\cdots+(30 - 22.5)(63 - 46.125)=420$
$\sum_{i = 1}^{n}(x_i-\bar{x})^2=(15 - 22.5)^2+(16 - 22.5)^2+\cdots+(30 - 22.5)^2 = 260$
$b_1=\frac{420}{260}\approx1.615$

Step3: Calculate intercept ($b_0$)

$b_0=\bar{y}-b_1\bar{x}$
$b_0 = 46.125-1.615\times22.5$
$b_0=46.125 - 36.338=9.787$

Answer:

$\hat{y}=9.787 + 1.615x$
For the multiple - choice part:
For every degree Celsius, the toucan will lose about 1.6% more heat through its beak on average.