Answer :
### Problem 11.3.5: Probability that the Estimated Slope is within 1.5 Units of the True Slope
Given:
- Total number of measurements ([tex]\( n \)[/tex]) = 9
- [tex]\( x_i \)[/tex] values = 1, 2, 3, ..., 9
- Variance ([tex]\( \sigma^2 \)[/tex]) = 45
- Desired range for slope estimate deviation = ±1.5 units
We need to find the probability that the estimated slope [tex]\( \hat{\beta_1} \)[/tex] will be within 1.5 units of the true slope [tex]\( \beta_1 \)[/tex].
Step-by-Step Solution:
1. Calculate the mean of the [tex]\( x_i \)[/tex] values ([tex]\( \bar{x} \)[/tex]):
[tex]\[ \bar{x} = \frac{1 + 2 + \cdots + 9}{9} = \frac{45}{9} = 5 \][/tex]
2. Calculate the sum of squared deviations of [tex]\( x_i \)[/tex]:
[tex]\[ \sum (x_i - \bar{x})^2 = \sum_{i=1}^{9} (x_i - 5)^2 \][/tex]
The values are:
[tex]\[ (1-5)^2 = 16, \ (2-5)^2 = 9, \ (3-5)^2 = 4, \ (4-5)^2 = 1, \ (5-5)^2 = 0, \ (6-5)^2 = 1, \ (7-5)^2 = 4, \ (8-5)^2 = 9, \ (9-5)^2 = 16 \][/tex]
Therefore,
[tex]\[ \sum (x_i - \bar{x})^2 = 16 + 9 + 4 + 1 + 0 + 1 + 4 + 9 + 16 = 60 \][/tex]
3. Calculate the standard error of the slope ([tex]\( \text{SE}(\hat{\beta_1}) \)[/tex]):
[tex]\[ \text{SE}(\hat{\beta_1}) = \frac{\sigma}{\sqrt{\sum (x_i - \bar{x})^2}} = \frac{\sqrt{45}}{\sqrt{60}} = \frac{6.7082}{7.745} \approx 0.866 \][/tex]
4. Calculate the z-scores for the lower and upper bounds:
- Lower bound deviation for the slope = -1.5
- Upper bound deviation for the slope = +1.5
[tex]\[ z_{\text{lower}} = \frac{-1.5}{\text{SE}(\hat{\beta_1})} \approx \frac{-1.5}{0.866} \approx -1.73 \][/tex]
[tex]\[ z_{\text{upper}} = \frac{1.5}{\text{SE}(\hat{\beta_1})} \approx \frac{1.5}{0.866} \approx 1.73 \][/tex]
5. Use the cumulative distribution function (CDF) for the normal distribution to find the probability:
[tex]\[ P(-1.73 \le z \le 1.73) \][/tex]
Using the standard normal distribution table or a tool like a cumulative distribution function:
[tex]\[ P(z \le 1.73) - P(z \le -1.73) \approx 0.9582 - 0.0418 \approx 0.9164 \][/tex]
Therefore, the probability that the estimated slope is within 1.5 units of the true slope is approximately 0.9164 or 91.64%.
### Problem 11.3.6: Prove the Useful Computing Formula
The formula to prove is:
[tex]\[ \sum_{i=1}^n \left(y_i - \hat{\beta}_0 - \hat{\beta}_1 x_i\right)^2 = \sum_{i=1}^n y_i^2 - \hat{\beta}_0 \sum_{i=1}^n y_i - \hat{\beta}_1 \sum_{i=1}^n x_i y_i \][/tex]
Proof:
1. Define the sum of squared errors:
[tex]\[ \text{SS}_\text{res} = \sum_{i=1}^n \left(y_i - \hat{y}_i\right)^2 = \sum_{i=1}^n (y_i - \hat{\beta}_0 - \hat{\beta}_1 x_i)^2 \][/tex]
2. By expanding:
[tex]\[ \text{SS}_\text{res} = \sum_{i=1}^n \left(y_i^2 - 2y_i(\hat{\beta}_0 + \hat{\beta}_1 x_i) + (\hat{\beta}_0 + \hat{\beta}_1 x_i)^2\right) \][/tex]
3. Separate and simplify each term:
[tex]\[ \text{SS}_\text{res} = \sum_{i=1}^n y_i^2 - 2 \hat{\beta}_0 \sum_{i=1}^n y_i - 2 \hat{\beta}_1 \sum_{i=1}^n x_i y_i + \hat{\beta}_0^2 \sum_{i=1}^n 1 + 2 \hat{\beta}_0 \hat{\beta}_1 \sum_{i=1}^n x_i + \hat{\beta}_1^2 \sum_{i=1}^n x_i^2 \][/tex]
4. Use the normal equations:
- Normal equation for [tex]\( \hat{\beta}_0 \)[/tex]:
[tex]\[ n \hat{\beta}_0 + \hat{\beta}_1 \sum_{i=1}^n x_i = \sum_{i=1}^n y_i \][/tex]
- Normal equation for [tex]\( \hat{\beta}_1 \)[/tex]:
[tex]\[ \hat{\beta}_0 \sum_{i=1}^n x_i + \hat{\beta}_1 \sum_{i=1}^n x_i^2 = \sum_{i=1}^n x_i y_i \][/tex]
5. Substitute these values back:
[tex]\[ \text{SS}_\text{res} = \sum_{i=1}^n y_i^2 - \hat{\beta}_0 \sum_{i=1}^n y_i - \hat{\beta}_1 \sum_{i=1}^n x_i y_i \][/tex]
Therefore, we have proved that:
[tex]\[ \sum_{i=1}^n \left(y_i - \hat{\beta}_0 - \hat{\beta}_1 x_i\right)^2=\sum_{i=1}^n y_i^2-\hat{\beta}_0 \sum_{i=1}^n y_i-\hat{\beta}_1 \sum_{i=1}^n x_i y_i \][/tex]
This completes the proof.
Given:
- Total number of measurements ([tex]\( n \)[/tex]) = 9
- [tex]\( x_i \)[/tex] values = 1, 2, 3, ..., 9
- Variance ([tex]\( \sigma^2 \)[/tex]) = 45
- Desired range for slope estimate deviation = ±1.5 units
We need to find the probability that the estimated slope [tex]\( \hat{\beta_1} \)[/tex] will be within 1.5 units of the true slope [tex]\( \beta_1 \)[/tex].
Step-by-Step Solution:
1. Calculate the mean of the [tex]\( x_i \)[/tex] values ([tex]\( \bar{x} \)[/tex]):
[tex]\[ \bar{x} = \frac{1 + 2 + \cdots + 9}{9} = \frac{45}{9} = 5 \][/tex]
2. Calculate the sum of squared deviations of [tex]\( x_i \)[/tex]:
[tex]\[ \sum (x_i - \bar{x})^2 = \sum_{i=1}^{9} (x_i - 5)^2 \][/tex]
The values are:
[tex]\[ (1-5)^2 = 16, \ (2-5)^2 = 9, \ (3-5)^2 = 4, \ (4-5)^2 = 1, \ (5-5)^2 = 0, \ (6-5)^2 = 1, \ (7-5)^2 = 4, \ (8-5)^2 = 9, \ (9-5)^2 = 16 \][/tex]
Therefore,
[tex]\[ \sum (x_i - \bar{x})^2 = 16 + 9 + 4 + 1 + 0 + 1 + 4 + 9 + 16 = 60 \][/tex]
3. Calculate the standard error of the slope ([tex]\( \text{SE}(\hat{\beta_1}) \)[/tex]):
[tex]\[ \text{SE}(\hat{\beta_1}) = \frac{\sigma}{\sqrt{\sum (x_i - \bar{x})^2}} = \frac{\sqrt{45}}{\sqrt{60}} = \frac{6.7082}{7.745} \approx 0.866 \][/tex]
4. Calculate the z-scores for the lower and upper bounds:
- Lower bound deviation for the slope = -1.5
- Upper bound deviation for the slope = +1.5
[tex]\[ z_{\text{lower}} = \frac{-1.5}{\text{SE}(\hat{\beta_1})} \approx \frac{-1.5}{0.866} \approx -1.73 \][/tex]
[tex]\[ z_{\text{upper}} = \frac{1.5}{\text{SE}(\hat{\beta_1})} \approx \frac{1.5}{0.866} \approx 1.73 \][/tex]
5. Use the cumulative distribution function (CDF) for the normal distribution to find the probability:
[tex]\[ P(-1.73 \le z \le 1.73) \][/tex]
Using the standard normal distribution table or a tool like a cumulative distribution function:
[tex]\[ P(z \le 1.73) - P(z \le -1.73) \approx 0.9582 - 0.0418 \approx 0.9164 \][/tex]
Therefore, the probability that the estimated slope is within 1.5 units of the true slope is approximately 0.9164 or 91.64%.
### Problem 11.3.6: Prove the Useful Computing Formula
The formula to prove is:
[tex]\[ \sum_{i=1}^n \left(y_i - \hat{\beta}_0 - \hat{\beta}_1 x_i\right)^2 = \sum_{i=1}^n y_i^2 - \hat{\beta}_0 \sum_{i=1}^n y_i - \hat{\beta}_1 \sum_{i=1}^n x_i y_i \][/tex]
Proof:
1. Define the sum of squared errors:
[tex]\[ \text{SS}_\text{res} = \sum_{i=1}^n \left(y_i - \hat{y}_i\right)^2 = \sum_{i=1}^n (y_i - \hat{\beta}_0 - \hat{\beta}_1 x_i)^2 \][/tex]
2. By expanding:
[tex]\[ \text{SS}_\text{res} = \sum_{i=1}^n \left(y_i^2 - 2y_i(\hat{\beta}_0 + \hat{\beta}_1 x_i) + (\hat{\beta}_0 + \hat{\beta}_1 x_i)^2\right) \][/tex]
3. Separate and simplify each term:
[tex]\[ \text{SS}_\text{res} = \sum_{i=1}^n y_i^2 - 2 \hat{\beta}_0 \sum_{i=1}^n y_i - 2 \hat{\beta}_1 \sum_{i=1}^n x_i y_i + \hat{\beta}_0^2 \sum_{i=1}^n 1 + 2 \hat{\beta}_0 \hat{\beta}_1 \sum_{i=1}^n x_i + \hat{\beta}_1^2 \sum_{i=1}^n x_i^2 \][/tex]
4. Use the normal equations:
- Normal equation for [tex]\( \hat{\beta}_0 \)[/tex]:
[tex]\[ n \hat{\beta}_0 + \hat{\beta}_1 \sum_{i=1}^n x_i = \sum_{i=1}^n y_i \][/tex]
- Normal equation for [tex]\( \hat{\beta}_1 \)[/tex]:
[tex]\[ \hat{\beta}_0 \sum_{i=1}^n x_i + \hat{\beta}_1 \sum_{i=1}^n x_i^2 = \sum_{i=1}^n x_i y_i \][/tex]
5. Substitute these values back:
[tex]\[ \text{SS}_\text{res} = \sum_{i=1}^n y_i^2 - \hat{\beta}_0 \sum_{i=1}^n y_i - \hat{\beta}_1 \sum_{i=1}^n x_i y_i \][/tex]
Therefore, we have proved that:
[tex]\[ \sum_{i=1}^n \left(y_i - \hat{\beta}_0 - \hat{\beta}_1 x_i\right)^2=\sum_{i=1}^n y_i^2-\hat{\beta}_0 \sum_{i=1}^n y_i-\hat{\beta}_1 \sum_{i=1}^n x_i y_i \][/tex]
This completes the proof.
