2. Test the significance of explanatory variable. (a) There is no evidence that the explanatory variable is significant (p-value > 0.1) (b) There is some evidence that the explanatory variable is significant (0.05 ≤ p-value < 0.1) I (c) There is strong evidence that the explanatory variable is signifi- cant (0.01 ≤ p-value < 0.05) (d) There is very strong evidence that the explanatory variable is significant (0.001 ≤ p-value < 0.01)
2. Test the significance of explanatory variable. (a) There is no evidence that the explanatory variable is significant (p-value > 0.1) (b) There is some evidence that the explanatory variable is significant (0.05 ≤ p-value < 0.1) I (c) There is strong evidence that the explanatory variable is signifi- cant (0.01 ≤ p-value < 0.05) (d) There is very strong evidence that the explanatory variable is significant (0.001 ≤ p-value < 0.01)
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
Please solve both of the attached subparts of the problem. They are all from one question that is above question 1. Refer to that part for data. Thanks!
Transcribed Image Text:**Testing the Significance of an Explanatory Variable**
1. **Understanding p-values for Significance Testing**
- **(a)** There is no evidence that the explanatory variable is significant \
\( \text{(p-value} \geq 0.1) \)
- **(b)** There is some evidence that the explanatory variable is significant \
\( (0.05 \leq \text{p-value} < 0.1) \)
- **(c)** There is strong evidence that the explanatory variable is significant \
\( (0.01 \leq \text{p-value} < 0.05) \)
- **(d)** There is very strong evidence that the explanatory variable is significant \
\( (0.001 \leq \text{p-value} < 0.01) \)
The text outlines different levels of evidence for the significance of an explanatory variable based on the p-value obtained from statistical testing. Each level of p-value corresponds to the strength of evidence against the null hypothesis.
![**Text Transcription for Educational Website**
Since elderly people may have difficulty standing straight, a study aims to predict overall height from height to the knee. Here are data (in centimeters, cm) for five elderly men.
| Knee Height (cm) | 57.7 | 47.4 | 43.5 | 44.8 | 55.2 |
|-------------------|------|------|------|------|------|
| Overall Height (cm)| 192.1 | 153.3 | 146.4 | 162.7 | 169.1 |
Let's assume a regression model:
\[ \text{Overall height} = \beta_0 + \beta_1 \times \text{Knee height} + \epsilon \]
1. **Which R command should you use to fit the model if you have defined variables as:**
```r
knee = c(57.7, 47.4, 43.5, 44.8, 55.2)
overall = c(192.1, 153.3, 146.4, 162.7, 169.1)
```
**Explanation:**
The table provides measurements of knee height and overall height for a sample of five elderly men. The task involves using these measurements to predict overall height based on knee height by applying a regression model. The proposed regression model is a linear equation with coefficients \(\beta_0\) and \(\beta_1\), and \(\epsilon\) representing the error term. The R command needed would typically be used in a statistical analysis to fit this linear regression model using the provided data.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70f1e669-1e2d-4a5b-ad55-41a2890e62c9%2F243b0226-9872-4851-b0f4-d7346f5cd7cc%2Fx6jt7rl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Text Transcription for Educational Website**
Since elderly people may have difficulty standing straight, a study aims to predict overall height from height to the knee. Here are data (in centimeters, cm) for five elderly men.
| Knee Height (cm) | 57.7 | 47.4 | 43.5 | 44.8 | 55.2 |
|-------------------|------|------|------|------|------|
| Overall Height (cm)| 192.1 | 153.3 | 146.4 | 162.7 | 169.1 |
Let's assume a regression model:
\[ \text{Overall height} = \beta_0 + \beta_1 \times \text{Knee height} + \epsilon \]
1. **Which R command should you use to fit the model if you have defined variables as:**
```r
knee = c(57.7, 47.4, 43.5, 44.8, 55.2)
overall = c(192.1, 153.3, 146.4, 162.7, 169.1)
```
**Explanation:**
The table provides measurements of knee height and overall height for a sample of five elderly men. The task involves using these measurements to predict overall height based on knee height by applying a regression model. The proposed regression model is a linear equation with coefficients \(\beta_0\) and \(\beta_1\), and \(\epsilon\) representing the error term. The R command needed would typically be used in a statistical analysis to fit this linear regression model using the provided data.
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