Question

the data show the chest size and weight of several bears. find the regression equation, letting chest size be the independent (x) variable. then find the best predicted weight of a bear with a chest size of 51 inches. is the result close to the actual weight of 442 pounds? use a significance level of 0.05.
chest size (inches) | 45 | 50 | 43 | 43 | 52 | 52
weight (pounds) | 352 | 374 | 275 | 314 | 440 | 367
click the icon to view the critical values of the pearson correlation coefficient r.
(round to one decimal place as needed.)
is the result close to the actual weight of 442 pounds?
a. this result is close to the actual weight of the bear.
b. this result is very close to the actual weight of the bear.
c. this result is exactly the same as the actual weight of the bear.
d. this result is not very close to the actual weight of the bear.

Explanation:

Step1: Calculate necessary sums

First, we list the data points:

• \( x \) (chest size): 45, 50, 43, 43, 52, 52
• \( y \) (weight): 352, 374, 275, 314, 440, 367

Calculate \( \sum x \), \( \sum y \), \( \sum xy \), \( \sum x^2 \):

\( \sum x = 45 + 50 + 43 + 43 + 52 + 52 = 285 \)

\( \sum y = 352 + 374 + 275 + 314 + 440 + 367 = 2122 \)

\( \sum xy = (45×352) + (50×374) + (43×275) + (43×314) + (52×440) + (52×367) \)
\( = 15840 + 18700 + 11825 + 13502 + 22880 + 19084 = 101831 \)

\( \sum x^2 = 45^2 + 50^2 + 43^2 + 43^2 + 52^2 + 52^2 \)
\( = 2025 + 2500 + 1849 + 1849 + 2704 + 2704 = 13631 \)

\( n = 6 \) (number of data points)

Step2: Calculate slope (\( b \)) and intercept (\( a \)) of regression line

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

Substitute the values:
\( b = \frac{6×101831 - 285×2122}{6×13631 - 285^2} \)
\( = \frac{610986 - 604770}{81786 - 81225} \)
\( = \frac{6216}{561} \approx 11.08 \)

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

\( \bar{x} = \frac{285}{6} = 47.5 \)

\( \bar{y} = \frac{2122}{6} \approx 353.67 \)

\( a = 353.67 - 11.08×47.5 \)
\( = 353.67 - 526.3 \approx -172.63 \)

So the regression equation is \( \hat{y} = -172.63 + 11.08x \)

Step3: Predict weight for \( x = 51 \)

Substitute \( x = 51 \) into the regression equation:
\( \hat{y} = -172.63 + 11.08×51 \)
\( = -172.63 + 565.08 \approx 392.45 \)

Step4: Compare with actual weight (442 pounds)

The predicted weight is approximately 392.45 pounds, and the actual weight is 442 pounds. The difference is \( |442 - 392.45| = 49.55 \) pounds, which is not very small. So the result is not very close to the actual weight.

Answer:

D. This result is not very close to the actual weight of the bear.