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(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)=√(x+vn - 0x² +0,² Identify the next step in this rule from the options below. x-5-₂²_0₂² ₁ 2 VX-1 + 2 0₂ OV--1+%2 n1 na OV--1-%2 n1 2 Since standard deviation is the square root of variance, it follows that 01 √22-9/322 n₁ 0,² 0,² 01+02 Ox-√₂ Compute the estimated standard error (in MPa). (Round your answer to three decimal places.) MPa (c) Calculate a point estimate of the ratio a/a₂ of the two standard deviations. (Round your answer to three decimal places.) 4 (d) Suppose a single beam and a single cylinder are randomly selected. Calculate a point estimate (in MPa) of the variance of the difference X - Y between beam strength and cylinder strength. (Round your answer to two decimal places.) MPa2

(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)=√(x+vn - 0x² +0,² Identify the next step in this rule from the options below. x-5-₂²_0₂² ₁ 2 VX-1 + 2 0₂ OV--1+%2 n1 na OV--1-%2 n1 2 Since standard deviation is the square root of variance, it follows that 01 √22-9/322 n₁ 0,² 0,² 01+02 Ox-√₂ Compute the estimated standard error (in MPa). (Round your answer to three decimal places.) MPa (c) Calculate a point estimate of the ratio a/a₂ of the two standard deviations. (Round your answer to three decimal places.) 4 (d) Suppose a single beam and a single cylinder are randomly selected. Calculate a point estimate (in MPa) of the variance of the difference X - Y between beam strength and cylinder strength. (Round your answer to two decimal places.) MPa2

MATLAB: An Introduction with Applications
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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Can u help find the answer for b and explain d?
The image contains a detailed explanation and steps for deriving an expression for the variance and standard deviation of an estimator, using variance rules. --- **(b) Use rules of variance to obtain an expression for the variance and standard deviation (standard error) of the estimator in part (a):** \[ V(\overline{X} - \overline{Y}) = V(\overline{X}) + V(\overline{Y}) = \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} \] Identify the next step in this rule from the options below: - \[ V(\overline{X} - \overline{Y}) = \frac{\sigma_1^2}{n_1^2} + \frac{\sigma_2^2}{n_2^2} \] - \[ V(\overline{X} - \overline{Y}) = \frac{\sigma_1^2}{n_1^2} + \frac{\sigma_2^2}{n_2} \] - \[ V(\overline{X} - \overline{Y}) = \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} \] (Selected option) - \[ V(\overline{X} - \overline{Y}) = \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2^2} \] Since standard deviation is the square root of variance, it follows that: \[ \sigma_{\overline{X} - \overline{Y}} = \sqrt{V(\overline{X} - \overline{Y})} \] Options for the expression: - \[ \sigma_{\overline{X} - \overline{Y}} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \] - \[ \sigma_{\overline{X} - \overline{Y}} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}^2} \] - \[ \sigma_{\overline{
Transcribed Image Text:The image contains a detailed explanation and steps for deriving an expression for the variance and standard deviation of an estimator, using variance rules. --- **(b) Use rules of variance to obtain an expression for the variance and standard deviation (standard error) of the estimator in part (a):** \[ V(\overline{X} - \overline{Y}) = V(\overline{X}) + V(\overline{Y}) = \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} \] Identify the next step in this rule from the options below: - \[ V(\overline{X} - \overline{Y}) = \frac{\sigma_1^2}{n_1^2} + \frac{\sigma_2^2}{n_2^2} \] - \[ V(\overline{X} - \overline{Y}) = \frac{\sigma_1^2}{n_1^2} + \frac{\sigma_2^2}{n_2} \] - \[ V(\overline{X} - \overline{Y}) = \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} \] (Selected option) - \[ V(\overline{X} - \overline{Y}) = \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2^2} \] Since standard deviation is the square root of variance, it follows that: \[ \sigma_{\overline{X} - \overline{Y}} = \sqrt{V(\overline{X} - \overline{Y})} \] Options for the expression: - \[ \sigma_{\overline{X} - \overline{Y}} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \] - \[ \sigma_{\overline{X} - \overline{Y}} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}^2} \] - \[ \sigma_{\overline{
**Educational Content on Flexural Strength Estimation** ### Data on Flexural Strength for Concrete Beams and Cylinders #### Concrete Beams Flexural Strength (MPa) - 5.1, 7.2, 7.3, 6.8, 6.5, 7.0, 6.3, 7.9, 9.0 - 8.4, 8.7, 7.8, 9.7, 7.4, 7.7, 8.0, 11.6, 11.3, 11.8, 10.7 #### Cylinders Strength Observations - 6.5, 5.8, 7.8, 7.1, 7.2, 9.2, 6.6, 8.5, 7.0, 8.5 - 7.2, 8.1, 7.4, 8.5, 8.9, 9.9, 12.6, 11.2, 6.9, 11.4 ### Prior Considerations for Data Analysis - **Beam Strengths**: Denoted by X₁, ..., Xₘ - **Cylinder Strengths**: Denoted by Y₁, ..., Yₙ - Assume X's are a random sample from a distribution with mean μ₁ and standard deviation σ₁. - Assume Y's are a random sample from another distribution with mean μ₂ and standard deviation σ₂. ### Analysis #### Task (a): Demonstrate that X̄ – Ȳ is an unbiased estimator of μ₁ – μ₂ Using rules of expected value: Options: 1. \(E(X̄ - Ȳ) = (E(X) - E(Y)) = \mu_1 + \mu_2\) 2. \(E(X̄ - Ȳ) = (E(X) - E(Y))^2 = (\mu_1 - \mu_2)\) 3. \(E(X̄ - Ȳ) = (E(X) - E(Y)) = \frac{\mu_1 - \mu_2}{nm}\) 4. **\(E(X̄ - Ȳ) = (E(X) - E(Y
Transcribed Image Text:**Educational Content on Flexural Strength Estimation** ### Data on Flexural Strength for Concrete Beams and Cylinders #### Concrete Beams Flexural Strength (MPa) - 5.1, 7.2, 7.3, 6.8, 6.5, 7.0, 6.3, 7.9, 9.0 - 8.4, 8.7, 7.8, 9.7, 7.4, 7.7, 8.0, 11.6, 11.3, 11.8, 10.7 #### Cylinders Strength Observations - 6.5, 5.8, 7.8, 7.1, 7.2, 9.2, 6.6, 8.5, 7.0, 8.5 - 7.2, 8.1, 7.4, 8.5, 8.9, 9.9, 12.6, 11.2, 6.9, 11.4 ### Prior Considerations for Data Analysis - **Beam Strengths**: Denoted by X₁, ..., Xₘ - **Cylinder Strengths**: Denoted by Y₁, ..., Yₙ - Assume X's are a random sample from a distribution with mean μ₁ and standard deviation σ₁. - Assume Y's are a random sample from another distribution with mean μ₂ and standard deviation σ₂. ### Analysis #### Task (a): Demonstrate that X̄ – Ȳ is an unbiased estimator of μ₁ – μ₂ Using rules of expected value: Options: 1. \(E(X̄ - Ȳ) = (E(X) - E(Y)) = \mu_1 + \mu_2\) 2. \(E(X̄ - Ȳ) = (E(X) - E(Y))^2 = (\mu_1 - \mu_2)\) 3. \(E(X̄ - Ȳ) = (E(X) - E(Y)) = \frac{\mu_1 - \mu_2}{nm}\) 4. **\(E(X̄ - Ȳ) = (E(X) - E(Y
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