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The data show the chest size and weight of several bears. Find the regression equation, letting chest size be the independent [tex]$(x)$[/tex] variable. Then find the best predicted weight of a bear with a chest size of 53 inches. Is the result close to the actual weight of 485 pounds? Use a significance level of 0.05.

\begin{tabular}{l|cccccc}
\hline
Chest size (inches) & 44 & 59 & 55 & 59 & 57 & 44 \\
\hline
Weight (pounds) & 403 & 645 & 563 & 587 & 548 & 406 \\
\hline
\end{tabular}

Click the icon to view the critical values of the Pearson correlation coefficient [tex]$r$[/tex].

1. What is the regression equation?
\begin{equation}
\hat{y} = \square + \square \times x
\end{equation}
(Round to one decimal place as needed.)

2. What is the best predicted weight of a bear with a chest size of 53 inches?
The best predicted weight for a bear with a chest size of 53 inches is [tex]$\square$[/tex] pounds.
(Round to one decimal place as needed.)

3. Is the result close to the actual weight of 485 pounds?
A. This result is close to the actual weight of the bear.
B. This result is not very close to the actual weight of the bear.
C. This result is very close to the actual weight of the bear.
D. This result is exactly the same as the actual weight of the bear.

Sagot :

GinnyAnsweridn
### Step-by-Step Solution

#### Given Data:

| Chest size (inches) | 44 | 59 | 55 | 59 | 57 | 44 |
|---------------------|-----|-----|-----|-----|-----|-----|
| Weight (pounds) | 403 | 645 | 563 | 587 | 548 | 406 |

We need to find the regression equation letting the chest size be the independent variable [tex]\( x \)[/tex] and the weight be the dependent variable [tex]\( y \)[/tex]. Then, using this equation, we have to predict the weight of a bear with a chest size of 53 inches and see how close this prediction is to the actual weight of 485 pounds.

#### Step 1: Find the Regression Equation

The regression equation is generally of the form:
[tex]\[ \hat{y} = b_0 + b_1 x \][/tex]

Where:
- [tex]\( b_0 \)[/tex] is the y-intercept
- [tex]\( b_1 \)[/tex] is the slope

From the given data and calculations (not shown here), we have:
- [tex]\( b_0 = -190.2 \)[/tex] (rounded to one decimal place)
- [tex]\( b_1 = 13.5 \)[/tex] (rounded to one decimal place)

Thus, the regression equation is:
[tex]\[ \hat{y} = -190.2 + 13.5x \][/tex]

#### Step 2: Predict the Weight for a Chest Size of 53 inches

Let [tex]\( x = 53 \)[/tex]:
[tex]\[ \hat{y} = -190.2 + 13.5 \times 53 \][/tex]

Calculating this:
[tex]\[ \hat{y} = -190.2 + 715.5 = 525.3 \][/tex]

Thus, the predicted weight for a bear with a chest size of 53 inches is:
[tex]\[ 525.3 \text{ pounds} \][/tex]

#### Step 3: Determine if the Predicted Weight is Close to the Actual Weight of 485 Pounds

Calculate the difference between the predicted weight and the actual weight:
[tex]\[ \text{Difference} = |525.3 - 485| = 40.3 \][/tex]

Given that the difference is 40.3 pounds, we can assess the closeness.

#### Conclusion:

Given the multiple-choice options:
A. This result is close to the actual weight of the bear.
B. This result is not very close to the actual weight of the bear.
C. This result is very close to the actual weight of the bear.
D. This result is exactly the same as the actual weight of the bear.

Option B is appropriate:
B. This result is not very close to the actual weight of the bear.

#### Final Answers:
1. The regression equation is: [tex]\( \hat{y} = -190.2 + 13.5x \)[/tex]
2. The best predicted weight for a bear with a chest size of 53 inches is [tex]\( 525.3 \)[/tex] pounds.
3. Is the result close to the actual weight of 485 pounds? Answer: B. This result is not very close to the actual weight of the bear.
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