(b) Use rules of variance to obtain an expression for the variance and standard deviation (standard error) of the estimator in part (a). v(x-7)= v(x+v(n = 0,² +0,² Identify the next step in this rule from the options below. 2 0 x-5-² X-5-02-02 --- --- Since standard deviation is the square root of variance, it follows that 0₂

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### Statistical Analysis: Variance and Standard Deviation Computation

#### (b) Variance and Standard Deviation (Standard Error) Computation

Use the rules of variance to obtain an expression for the variance and standard deviation (standard error) of the estimator in part (a):

\[ V(\overline{X} - \overline{Y}) = V(\overline{X}) + V(\overline{Y}) = \frac{\sigma_{X}^{2}}{n_{1}} + \frac{\sigma_{Y}^{2}}{n_{2}} \]

Identify the next step in this rule from the options below:

- \( \bigcirc \) \( V(\overline{X} - \overline{Y}) = \frac{\sigma_{1}^{2}}{n_{1}} + \frac{\sigma_{2}^{2}}{n_{2}} \)
- \( \phantom{\bigcirc} \) \( V(\overline{X} - \overline{Y}) = \frac{\sigma_{1}^{2}}{n_{1}} - \frac{\sigma_{2}^{2}}{n_{2}} \)
- \( \phantom{\bigcirc} \) \( V(\overline{X} - \overline{Y}) = \frac{\sigma_{1}}{n_{1}} + \frac{\sigma_{2}}{n_{2}} \)
- \( \phantom{\bigcirc} \) \( V(\overline{X} - \overline{Y}) = \frac{\sigma_{1}}{n_{1}} - \frac{\sigma_{2}}{n_{2}} \)

Since standard deviation is the square root of variance, it follows that:

\[ \sigma_{\overline{X} - \overline{Y}} = \sqrt{V(\overline{X} - \overline{Y})} \]

- \( \phantom{\bigcirc} \) \( \sigma_{\overline{X} - \overline{Y}} = \sqrt{\frac{\sigma_{1}^{2}}{n_{1}} - \frac{\sigma_{2}^{2}}{n_{2}}} \)
- \( \bigcirc \) \( \sigma_{\overline{X} - \overline{Y}} = \sqrt{\frac{\sigma_{1}^{2}}
Transcribed Image Text:### Statistical Analysis: Variance and Standard Deviation Computation #### (b) Variance and Standard Deviation (Standard Error) Computation Use the rules of variance to obtain an expression for the variance and standard deviation (standard error) of the estimator in part (a): \[ V(\overline{X} - \overline{Y}) = V(\overline{X}) + V(\overline{Y}) = \frac{\sigma_{X}^{2}}{n_{1}} + \frac{\sigma_{Y}^{2}}{n_{2}} \] Identify the next step in this rule from the options below: - \( \bigcirc \) \( V(\overline{X} - \overline{Y}) = \frac{\sigma_{1}^{2}}{n_{1}} + \frac{\sigma_{2}^{2}}{n_{2}} \) - \( \phantom{\bigcirc} \) \( V(\overline{X} - \overline{Y}) = \frac{\sigma_{1}^{2}}{n_{1}} - \frac{\sigma_{2}^{2}}{n_{2}} \) - \( \phantom{\bigcirc} \) \( V(\overline{X} - \overline{Y}) = \frac{\sigma_{1}}{n_{1}} + \frac{\sigma_{2}}{n_{2}} \) - \( \phantom{\bigcirc} \) \( V(\overline{X} - \overline{Y}) = \frac{\sigma_{1}}{n_{1}} - \frac{\sigma_{2}}{n_{2}} \) Since standard deviation is the square root of variance, it follows that: \[ \sigma_{\overline{X} - \overline{Y}} = \sqrt{V(\overline{X} - \overline{Y})} \] - \( \phantom{\bigcirc} \) \( \sigma_{\overline{X} - \overline{Y}} = \sqrt{\frac{\sigma_{1}^{2}}{n_{1}} - \frac{\sigma_{2}^{2}}{n_{2}}} \) - \( \bigcirc \) \( \sigma_{\overline{X} - \overline{Y}} = \sqrt{\frac{\sigma_{1}^{2}}
### Comparing Flexural Strength of Concrete Beams and Cylinders

#### Flexural Strength Data of Concrete Beams
The flexural strength (MPa) for concrete beams of a certain type is given in the following table:

| 5.3 | 7.2 | 7.3 | 6.3 | 8.1 | 6.8 | 7.0 | 7.4 | 8.6 | 6.5 | 7.0 | 6.3 | 7.9 | 9.0 | 8.6 | 8.7 | 7.8 | 7.4 | 7.7 | 7.7 | 9.7 | 7.6 | 8.6 | 7.5 | 9.7 | 8.6 | 7.8 | 7.0 | 11.4 | 7.1 | 8.6 | 8.6 | 8.6 | 10.4 | 11.6 | 11.3 | 11.8 | 11.8 | 10.7 |

#### Strength Observations for Cylinders
The strength for cylinders is recorded in the following table:

| 6.7 | 5.8 | 7.8 | 7.1 | 7.2 | 9.2 | 6.6 | 8.3 | 7.0 | 8.1 | 7.2 | 8.1 | 7.4 | 8.5 | 8.9 | 9.8 | 9.4 | 7.1 | 12.1 | 12.6 | 11.0 |

### Statistical Analysis

#### Part (a): Bias and Estimation
- **Objective**: Determine if \(\bar{X} - \bar{Y}\) is an unbiased estimator of \(\mu_1 - \mu_2\).
  
Let \(\bar{X}\) denote the mean strength for beams and \(\bar{Y}\) for cylinders, where \(\{X_i\}\) is a random sample from distribution with mean \(\mu_1\) and standard deviation \(\sigma_1\), and \(\{Y_i\}\) is from another distribution with mean \(\mu_2\) and
Transcribed Image Text:### Comparing Flexural Strength of Concrete Beams and Cylinders #### Flexural Strength Data of Concrete Beams The flexural strength (MPa) for concrete beams of a certain type is given in the following table: | 5.3 | 7.2 | 7.3 | 6.3 | 8.1 | 6.8 | 7.0 | 7.4 | 8.6 | 6.5 | 7.0 | 6.3 | 7.9 | 9.0 | 8.6 | 8.7 | 7.8 | 7.4 | 7.7 | 7.7 | 9.7 | 7.6 | 8.6 | 7.5 | 9.7 | 8.6 | 7.8 | 7.0 | 11.4 | 7.1 | 8.6 | 8.6 | 8.6 | 10.4 | 11.6 | 11.3 | 11.8 | 11.8 | 10.7 | #### Strength Observations for Cylinders The strength for cylinders is recorded in the following table: | 6.7 | 5.8 | 7.8 | 7.1 | 7.2 | 9.2 | 6.6 | 8.3 | 7.0 | 8.1 | 7.2 | 8.1 | 7.4 | 8.5 | 8.9 | 9.8 | 9.4 | 7.1 | 12.1 | 12.6 | 11.0 | ### Statistical Analysis #### Part (a): Bias and Estimation - **Objective**: Determine if \(\bar{X} - \bar{Y}\) is an unbiased estimator of \(\mu_1 - \mu_2\). Let \(\bar{X}\) denote the mean strength for beams and \(\bar{Y}\) for cylinders, where \(\{X_i\}\) is a random sample from distribution with mean \(\mu_1\) and standard deviation \(\sigma_1\), and \(\{Y_i\}\) is from another distribution with mean \(\mu_2\) and
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