(a) Compute the mean square error using equation s² = MSE = SSE n-2 72.92 (b) Compute the standard error of the estimate using equation s = 8.539 (c) Compute the estimated standard deviation of b₁ using equation ₁ 0.636 (d) Use the t test to test the following hypotheses (a = 0.05): Ho: B₁ = 0 H₂: B₁ = 0 (Round your answer to two decimal places.) MSE = Find the p-value. (Round your answer to four decimal places.) p-value = 0.0145 SSE √n-2 S Σ(x − x) Find the value of the test statistic. (Round your answer to three decimal places.) -5.106 ■ (Round your answer to three decimal places.) (Round your answer to three decimal places.) State your conclusion. O Do not reject H. We cannot conclude that the relationship between x and y is significant. Reject H. We conclude that the relationship between x and y is significant. Reject H. We cannot conclude that the relationship between x and y is significant. O Do not reject Ho. We conclude that the relationship between x and y is significant.

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## Statistical Calculations and Hypothesis Testing

### Consider the following data:

\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x_i & 3 & 12 & 6 & 20 & 14 \\
\hline
y_i & 55 & 40 & 55 & 5 & 15 \\
\hline
\end{array}
\]

### (a) Mean Square Error (MSE)
Compute the mean square error using the equation:
\[ s^2 = \text{MSE} = \frac{\text{SSE}}{n - 2} \]
- Result: **72.92**

### (b) Standard Error of the Estimate
Compute the standard error of the estimate using the equation:
\[ s = \sqrt{\text{MSE}} = \sqrt{\frac{\text{SSE}}{n - 2}} \]
- Result: **8.539**

### (c) Estimated Standard Deviation of \( b_1 \)
Compute the estimated standard deviation of \( b_1 \) using the equation:
\[ s_{b_1} = \frac{s}{\sqrt{\sum(x_i - \bar{x})^2}} \]
- Result: **0.636**

### (d) Hypothesis Testing
Use the t-test to evaluate the following hypotheses (\(\alpha = 0.05\)):
- \( H_0: \beta_1 = 0 \) 
- \( H_a: \beta_1 \neq 0 \)

#### Test Statistic
- Value: **-5.106**

#### p-value
- Result: **0.0145**

#### Conclusion
- **Reject \( H_0 \). We conclude that the relationship between x and y is significant.**
Transcribed Image Text:## Statistical Calculations and Hypothesis Testing ### Consider the following data: \[ \begin{array}{|c|c|c|c|c|c|} \hline x_i & 3 & 12 & 6 & 20 & 14 \\ \hline y_i & 55 & 40 & 55 & 5 & 15 \\ \hline \end{array} \] ### (a) Mean Square Error (MSE) Compute the mean square error using the equation: \[ s^2 = \text{MSE} = \frac{\text{SSE}}{n - 2} \] - Result: **72.92** ### (b) Standard Error of the Estimate Compute the standard error of the estimate using the equation: \[ s = \sqrt{\text{MSE}} = \sqrt{\frac{\text{SSE}}{n - 2}} \] - Result: **8.539** ### (c) Estimated Standard Deviation of \( b_1 \) Compute the estimated standard deviation of \( b_1 \) using the equation: \[ s_{b_1} = \frac{s}{\sqrt{\sum(x_i - \bar{x})^2}} \] - Result: **0.636** ### (d) Hypothesis Testing Use the t-test to evaluate the following hypotheses (\(\alpha = 0.05\)): - \( H_0: \beta_1 = 0 \) - \( H_a: \beta_1 \neq 0 \) #### Test Statistic - Value: **-5.106** #### p-value - Result: **0.0145** #### Conclusion - **Reject \( H_0 \). We conclude that the relationship between x and y is significant.**
**ANOVA Table for Hypothesis Testing**

**Objective**: Use the F test to assess the hypotheses at a 0.05 level of significance. This involves an analysis of variance (ANOVA) table.

### ANOVA Table

| Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F    | p-value |
|---------------------|----------------|--------------------|-------------|------|---------|
| Regression          | 1901.25        | 1                  | 1901.25     | 26.07| 0.015   |
| Error               | 218.75         | 3                  | 72.92       |      |         |
| Total               | 2120           | 4                  |             |      |         |

### Calculations

- **Test Statistic**: \( F = 26.07 \)
- **p-value**: \( p = 0.015 \)

### Conclusion Options

1. Do not reject \( H_0 \). We cannot conclude that the relationship between \( x \) and \( y \) is significant.
2. Reject \( H_0 \). We cannot conclude that the relationship between \( x \) and \( y \) is significant.
3. Do not reject \( H_0 \). We conclude that the relationship between \( x \) and \( y \) is significant.
4. **Reject \( H_0 \). We conclude that the relationship between \( x \) and \( y \) is significant.**

**Selected Conclusion**: Option 4 is chosen, indicating that there is a significant relationship between \( x \) and \( y \) since the p-value (0.015) is less than the significance level (0.05).
Transcribed Image Text:**ANOVA Table for Hypothesis Testing** **Objective**: Use the F test to assess the hypotheses at a 0.05 level of significance. This involves an analysis of variance (ANOVA) table. ### ANOVA Table | Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F | p-value | |---------------------|----------------|--------------------|-------------|------|---------| | Regression | 1901.25 | 1 | 1901.25 | 26.07| 0.015 | | Error | 218.75 | 3 | 72.92 | | | | Total | 2120 | 4 | | | | ### Calculations - **Test Statistic**: \( F = 26.07 \) - **p-value**: \( p = 0.015 \) ### Conclusion Options 1. Do not reject \( H_0 \). We cannot conclude that the relationship between \( x \) and \( y \) is significant. 2. Reject \( H_0 \). We cannot conclude that the relationship between \( x \) and \( y \) is significant. 3. Do not reject \( H_0 \). We conclude that the relationship between \( x \) and \( y \) is significant. 4. **Reject \( H_0 \). We conclude that the relationship between \( x \) and \( y \) is significant.** **Selected Conclusion**: Option 4 is chosen, indicating that there is a significant relationship between \( x \) and \( y \) since the p-value (0.015) is less than the significance level (0.05).
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