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 63 inches. is the result close to the actual weight of 522 pounds? use a significance level of 0.05.

chest size (inches) | 58 | 50 | 65 | 59 | 59 | 48
weight (pounds) | 414 | 312 | 490 | 450 | 456 | 260

click the icon to view the critical values of the pearson correlation coefficient r.

\hat{y} = \square + \square x (round to one decimal place as needed.)

what is the best predicted weight of a bear with a chest size of 63 inches?

the best predicted weight for a bear with a chest size of 63 inches is \square pounds.
(round to one decimal place as needed.)

is the result close to the actual weight of 522 pounds?

\bigcirc a. this result is not very close to the actual weight of the bear.
\bigcirc b. this result is close to the actual weight of the bear.
\bigcirc c. this result is exactly the same as the actual weight of the bear.
\bigcirc d. this result is very close to the actual weight of the bear.

Explanation:

Step1: Calculate necessary sums

First, we list the data points:

• \( x \) (chest size): 58, 50, 65, 59, 59, 48
• \( y \) (weight): 414, 312, 490, 450, 456, 280

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

\( \sum x = 58 + 50 + 65 + 59 + 59 + 48 = 339 \)

\( \sum y = 414 + 312 + 490 + 450 + 456 + 280 = 2402 \)

\( \sum xy = (58×414) + (50×312) + (65×490) + (59×450) + (59×456) + (48×280) \)
\( = 23912 + 15600 + 31850 + 26550 + 26904 + 13440 = 138256 \)

\( \sum x^2 = 58^2 + 50^2 + 65^2 + 59^2 + 59^2 + 48^2 \)
\( = 3364 + 2500 + 4225 + 3481 + 3481 + 2304 = 19355 \)

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

Step2: Calculate slope (\( b \)) and intercept (\( a \))

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:
\( b = \frac{6×138256 - 339×2402}{6×19355 - 339^2} \)
First, calculate numerator: \( 6×138256 = 829536 \); \( 339×2402 = 814278 \); numerator \( = 829536 - 814278 = 15258 \)
Denominator: \( 6×19355 = 116130 \); \( 339^2 = 114921 \); denominator \( = 116130 - 114921 = 1209 \)
\( b = \frac{15258}{1209} ≈ 12.6 \) (rounded to one decimal place)

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

\( \bar{x} = \frac{339}{6} = 56.5 \)

\( \bar{y} = \frac{2402}{6} ≈ 400.333 \)

\( a = 400.333 - 12.6×56.5 \)
\( = 400.333 - 711.9 = -311.567 ≈ -311.6 \) (rounded to one decimal place)

So the regression equation is \( \hat{y} = -311.6 + 12.6x \)

Step3: Predict weight for \( x = 63 \)

Substitute \( x = 63 \) into the regression equation:
\( \hat{y} = -311.6 + 12.6×63 \)
\( = -311.6 + 793.8 = 482.2 \)

Step4: Compare with actual weight

Actual weight is 522 pounds. The predicted weight is 482.2 pounds. The difference is \( |522 - 482.2| = 39.8 \) pounds, so it's not very close.

Answer:

The regression equation is \( \hat{y} = -311.6 + 12.6x \).
The best predicted weight for a bear with a chest size of 63 inches is \( 482.2 \) pounds.
The correct option is A. This result is not very close to the actual weight of the bear.