Question

question 22 of 39
numerical values in this problem have been modified for testing purposes.

temperature (°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

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.
\\(\hat{y}=\\) %
adapted from a graph in glenn j. tattersall et al., heat exchange from

Explanation:

Step 1: Calculate necessary sums

First, we need to find the sum of \( x \) (temperature), sum of \( y \) (percent heat loss), sum of \( xy \), and sum of \( x^2 \). Let's denote \( x_i \) as the temperature and \( y_i \) as the percent heat loss.

The data points are:
\( (15, 32), (16, 36), (17, 36), (18, 33), (19, 35), (20, 47), (21, 57), (22, 51), (23, 42), (24, 53), (25, 46), (26, 52), (27, 59), (28, 59), (29, 63), (30, 63) \)

Number of data points \( n = 16 \)

Sum of \( x \): \( \sum x = 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30 \)
\[ \begin{align*} \sum x&=\frac{(15 + 30)\times16}{2}\\ &=\frac{45\times16}{2}\\ &= 45\times8\\ &= 360 \end{align*} \]

Sum of \( y \): \( \sum y = 32 + 36 + 36 + 33 + 35 + 47 + 57 + 51 + 42 + 53 + 46 + 52 + 59 + 59 + 63 + 63 \)
Let's calculate step by step:
\( 32+36 = 68 \); \( 68 + 36 = 104 \); \( 104+33 = 137 \); \( 137+35 = 172 \); \( 172+47 = 219 \); \( 219+57 = 276 \); \( 276+51 = 327 \); \( 327+42 = 369 \); \( 369+53 = 422 \); \( 422+46 = 468 \); \( 468+52 = 520 \); \( 520+59 = 579 \); \( 579+59 = 638 \); \( 638+63 = 701 \); \( 701+63 = 764 \)
So \( \sum y = 764 \)

Sum of \( xy \): \( \sum xy = 15\times32 + 16\times36 + 17\times36 + 18\times33 + 19\times35 + 20\times47 + 21\times57 + 22\times51 + 23\times42 + 24\times53 + 25\times46 + 26\times52 + 27\times59 + 28\times59 + 29\times63 + 30\times63 \)
Calculate each term:
\( 15\times32 = 480 \); \( 16\times36 = 576 \); \( 17\times36 = 612 \); \( 18\times33 = 594 \); \( 19\times35 = 665 \); \( 20\times47 = 940 \); \( 21\times57 = 1197 \); \( 22\times51 = 1122 \); \( 23\times42 = 966 \); \( 24\times53 = 1272 \); \( 25\times46 = 1150 \); \( 26\times52 = 1352 \); \( 27\times59 = 1593 \); \( 28\times59 = 1652 \); \( 29\times63 = 1827 \); \( 30\times63 = 1890 \)

Now sum these terms:
\( 480+576 = 1056 \); \( 1056+612 = 1668 \); \( 1668+594 = 2262 \); \( 2262+665 = 2927 \); \( 2927+940 = 3867 \); \( 3867+1197 = 5064 \); \( 5064+1122 = 6186 \); \( 6186+966 = 7152 \); \( 7152+1272 = 8424 \); \( 8424+1150 = 9574 \); \( 9574+1352 = 10926 \); \( 10926+1593 = 12519 \); \( 12519+1652 = 14171 \); \( 14171+1827 = 15998 \); \( 15998+1890 = 17888 \)
So \( \sum xy = 17888 \)

Sum of \( x^2 \): \( \sum x^2 = 15^2 + 16^2 + 17^2 + 18^2 + 19^2 + 20^2 + 21^2 + 22^2 + 23^2 + 24^2 + 25^2 + 26^2 + 27^2 + 28^2 + 29^2 + 30^2 \)
We know that \( \sum_{k = 1}^{n}k^2=\frac{n(n + 1)(2n + 1)}{6} \), here we can calculate from \( k = 15 \) to \( k = 30 \) as \( \sum_{k = 15}^{30}k^2=\sum_{k = 1}^{30}k^2-\sum_{k = 1}^{14}k^2 \)

\( \sum_{k = 1}^{30}k^2=\frac{30\times31\times61}{6}= 9455 \)

\( \sum_{k = 1}^{14}k^2=\frac{14\times15\times29}{6}= 1015 \)

So \( \sum x^2=9455 - 1015 = 8440 \)

Step 2: Calculate slope \( b \) and intercept \( a \)

The formula for the slope \( b \) of the least - squares regression line \( \hat{y}=a + bx \) is:
\( b=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2} \)

The formula for the intercept \( a \) is:
\( a=\bar{y}-b\bar{x} \), where \( \bar{x}=\frac{\sum x}{n} \) and \( \bar{y}=\frac{\sum y}{n} \)

First, calculate \( \bar{x}=\frac{360}{16}=22.5 \)

\( \bar{y}=\frac{764}{16}=47.75 \)

Now calculate \( b \):
\[ \begin{align*} b&=\frac{16\times17888-360\times764}{16\times8440 - 360^2}\\ &=\frac{286208-275040}{135040 - 129600}\\ &=\frac{11168}{5440}\\ &= 2.053 \end{align*} \]

Calculate \( a \):
\( a = 47.75-2.053\times22.5 \)
\[ \begin{align*} a&=47.75 - 46.1925\\ &= 1.5575 \end{align*} \]

So the equation of the least - squares regression line is \( \hat{y}=1.5575 + 2.053x \)

Step 3: Predict \( y \) at \( x = 25 \)

Substitute \( x = 25 \) into the regression equation:
\( \hat{y}=1.5575+2.053\times25 \)
\[ \begin{align*} \hat{y}&=1.5575 + 51.325\\ &= 52.8825 \end{align*} \]

Rounding to one decimal place, \( \hat{y}\approx52.9 \)

Answer:

\( 52.9 \)