the regression equation, letting overhead width be the predictor (x) variable. Find the best predicted weight of a seal if the overhead width measured from a tograph is 1.7 cm. Can the prediction be correct? What is wrong with predicting the weight in this case? Use a significance level of 0.05. 8.4 175 7.8 180 9.8 258 regression equation is y=+x. und to one decimal place as needed.) 7.2 124 9.3 233 erhead Width (cm) ght (kg) Click the icon to view the critical values of the Pearson correlation coefficient r. 9.1 231 ... e best predicted weight for an overhead width of 1.7 cm is kg. und to one decimal place as needed.) n the prediction be correct? What is wrong with predicting the weight in this case? A. The prediction cannot be correct because there is not sufficient evidence of a linear correlation. The width in this case is beyond the scope of the available sample da B. The prediction cannot be correct because a negative weight does not make sense and because there is not sufficient evidence of a linear correlation. C. The prediction cannot be correct because a negative weight does not make sense. The width in this case is beyond the scope of the available sample data. D. The prediction can be correct. There is nothing wrong with predicting the weight in this case.
the regression equation, letting overhead width be the predictor (x) variable. Find the best predicted weight of a seal if the overhead width measured from a tograph is 1.7 cm. Can the prediction be correct? What is wrong with predicting the weight in this case? Use a significance level of 0.05. 8.4 175 7.8 180 9.8 258 regression equation is y=+x. und to one decimal place as needed.) 7.2 124 9.3 233 erhead Width (cm) ght (kg) Click the icon to view the critical values of the Pearson correlation coefficient r. 9.1 231 ... e best predicted weight for an overhead width of 1.7 cm is kg. und to one decimal place as needed.) n the prediction be correct? What is wrong with predicting the weight in this case? A. The prediction cannot be correct because there is not sufficient evidence of a linear correlation. The width in this case is beyond the scope of the available sample da B. The prediction cannot be correct because a negative weight does not make sense and because there is not sufficient evidence of a linear correlation. C. The prediction cannot be correct because a negative weight does not make sense. The width in this case is beyond the scope of the available sample data. D. The prediction can be correct. There is nothing wrong with predicting the weight in this case.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Question
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![### Predicting Seal Weight Using Overhead Width: A Regression Analysis
#### Problem Statement
Find the regression equation, letting overhead width be the predictor (x) variable. Find the best predicted weight of a seal if the overhead width measured from a photograph is 1.7 cm. Can the prediction be correct? What is wrong with predicting the weight in this case? Use a significance level of 0.05.
#### Data Provided:
- **Overhead Width (cm):** 8.4, 7.8, 9.8, 7.2, 9.3
- **Weight (kg):** 175, 180, 258, 124, 233
#### Task Instructions
1. **Click the icon to view the critical values of the Pearson correlation coefficient r.** (This indicates there is an interactive element on the original website for viewing critical values of the Pearson correlation coefficient.)
2. **Equation of the Regression Line:**
\[
\hat{y} = \_\_\_ + \_\_\_ x
\]
(Round to one decimal place as needed.)
3. **Predicted Weight for Overhead Width of 1.7 cm:**
\[
\hat{y} = \_\_\_ \text{ kg}
\]
(Round to one decimal place as needed.)
4. **Analysis of the Prediction's Validity**
**Question:** Can the prediction be correct? What is wrong with predicting the weight in this case?
**Options:**
- **A.** The prediction cannot be correct because there is not sufficient evidence of a linear correlation. The width in this case is beyond the scope of the available sample data.
- **B.** The prediction cannot be correct because a negative weight does not make sense and because there is not sufficient evidence of a linear correlation.
- **C.** The prediction cannot be correct because a negative weight does not make sense. The width in this case is beyond the scope of the available sample data.
- **D.** The prediction can be correct. There is nothing wrong with predicting the weight in this case.
#### Graphs and Diagrams Explanation
There are no specific graphs or diagrams provided in the text.
### Steps to Solve
1. **Calculation of the Regression Equation**
- Use the formula for the linear regression line:
\[
y =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F39618134-6ceb-48ac-ab5a-e0c16873beed%2Fc9b699ba-2114-4fd4-857b-df3b55391282%2Fu8avy6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Predicting Seal Weight Using Overhead Width: A Regression Analysis
#### Problem Statement
Find the regression equation, letting overhead width be the predictor (x) variable. Find the best predicted weight of a seal if the overhead width measured from a photograph is 1.7 cm. Can the prediction be correct? What is wrong with predicting the weight in this case? Use a significance level of 0.05.
#### Data Provided:
- **Overhead Width (cm):** 8.4, 7.8, 9.8, 7.2, 9.3
- **Weight (kg):** 175, 180, 258, 124, 233
#### Task Instructions
1. **Click the icon to view the critical values of the Pearson correlation coefficient r.** (This indicates there is an interactive element on the original website for viewing critical values of the Pearson correlation coefficient.)
2. **Equation of the Regression Line:**
\[
\hat{y} = \_\_\_ + \_\_\_ x
\]
(Round to one decimal place as needed.)
3. **Predicted Weight for Overhead Width of 1.7 cm:**
\[
\hat{y} = \_\_\_ \text{ kg}
\]
(Round to one decimal place as needed.)
4. **Analysis of the Prediction's Validity**
**Question:** Can the prediction be correct? What is wrong with predicting the weight in this case?
**Options:**
- **A.** The prediction cannot be correct because there is not sufficient evidence of a linear correlation. The width in this case is beyond the scope of the available sample data.
- **B.** The prediction cannot be correct because a negative weight does not make sense and because there is not sufficient evidence of a linear correlation.
- **C.** The prediction cannot be correct because a negative weight does not make sense. The width in this case is beyond the scope of the available sample data.
- **D.** The prediction can be correct. There is nothing wrong with predicting the weight in this case.
#### Graphs and Diagrams Explanation
There are no specific graphs or diagrams provided in the text.
### Steps to Solve
1. **Calculation of the Regression Equation**
- Use the formula for the linear regression line:
\[
y =
Transcribed Image Text:### Critical Values of the Pearson Correlation Coefficient r
This table provides the critical values of the Pearson Correlation Coefficient (r) at two significance levels (α = 0.05 and α = 0.01) for different sample sizes (n). These values are typically used in hypothesis testing to determine whether there is a statistically significant correlation between two variables.
#### Table
| n | α = 0.05 | α = 0.01 |
|-----|----------|----------|
| 4 | 0.950 | 0.990 |
| 5 | 0.878 | 0.959 |
| 6 | 0.811 | 0.917 |
| 7 | 0.754 | 0.875 |
| 8 | 0.707 | 0.834 |
| 9 | 0.666 | 0.798 |
| 10 | 0.632 | 0.765 |
| 11 | 0.602 | 0.735 |
| 12 | 0.576 | 0.708 |
| 13 | 0.553 | 0.684 |
| 14 | 0.532 | 0.661 |
| 15 | 0.514 | 0.641 |
| 16 | 0.497 | 0.623 |
| 17 | 0.482 | 0.606 |
| 18 | 0.468 | 0.590 |
| 19 | 0.456 | 0.575 |
| 20 | 0.444 | 0.561 |
| 25 | 0.396 | 0.505 |
| 30 | 0.361 | 0.463 |
| 35 | 0.335 | 0.430 |
| 40 | 0.312 | 0.402 |
| 45 | 0.294 | 0.378 |
| 50 | 0.279 | 0.361 |
| 60 |
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