Please help me with B **Flexural Strength Data for Concrete Beams and Cylinders****Concrete Beams:**The flexural strength (MPa) data for concrete beams is as follows:- 5.1, 7.2, 7.3, 6.3, 8.1, 6.8, 7.0, 7.2, 6.5, 7.0, 6.3, 7.9, 9.0- 8.4, 8.7, 7.8, 9.7, 7.4, 7.7, 8.0- 7.7, 8.0, 11.6, 11.3, 11.8, 10.7**Cylinders:**The accompanying strength observations for cylinders (MPa) are:- 6.5, 5.8, 7.8, 7.1, 7.2, 9.2, 6.6, 8.3, 7.0, 8.5- 7.2, 8.1, 7.4, 8.5, 8.9, 9.1- 14.1, 12.6, 11.4**Statistical Analysis of Flexural Strength:**Before collecting data, denote the beam strengths by \( X_1, \ldots, X_m \) and the cylinder strengths by \( Y_1, \ldots, Y_n \). Assume the \( X_i \)'s form a random sample from a distribution with mean \( \mu_1 \) and standard deviation \( \sigma_1 \), and the \( Y_j \)'s form a random sample from another distribution with mean \( \mu_2 \) and standard deviation \( \sigma_2 \).**Part (a): Expected Value Calculation**- Use rules of expected value to show that \( \overline{X} - \overline{Y} \) is an unbiased estimator of \( \mu_1 - \mu_2 \).The correct formula is:\[E(\overline{X} - \overline{Y}) = E(\overline{X}) - E(\overline{Y}) = \mu_1 - \mu_2 **Section (b): Variance and Standard Deviation of the Estimator**To obtain an expression for the variance and standard deviation (standard error) of the estimator from part (a), use the rules of variance:\[ V(\bar{X} - \bar{Y}) = V(\bar{X}) + V(\bar{Y}) = \frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2} \]**Identify the next step in this rule from the options below:**- \( \bigcirc \) \(\frac{\sigma^2_1}{n_1} - \frac{\sigma^2_2}{n_2}\) - \( \bullet \) \(\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2}\) ✔ - \( \bigcirc \) \(\frac{\sigma_1}{n_1} + \frac{\sigma_2}{n_2}\) - \( \bigcirc \) \(\frac{\sigma_1}{n_1} - \frac{\sigma_2}{n_2}\) **Since standard deviation is the square root of variance, it follows that:**\[ \sigma_{\bar{X} - \bar{Y}} = \sqrt{V(\bar{X} - \bar{Y})} \]**Options:**- \( \bigcirc \) \(\sqrt{\frac{\sigma_1}{n_1} - \frac{\sigma_2}{n_2}}\) - \( \bullet \) \(\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2}}\) ✔ - \( \bigcirc \) \(\sqrt{\frac{\sigma_1}{n_1} + \frac{\sigma_2}{n_2}}\) - \( \bigcirc \) \(\sqrt{\frac{\sigma^2_1}{n_1} - \frac{\sigma^2_2}{n_2}}\) **Compute the estimated standard error (in MPa), (Round your answer to three decimal places.)**\[ \_\_\_\_\_\_ \text{ MPa} \]**Section (c): Point Estimate

Answered: Image /qna-images/answer/e514ecfd-ba2b-4d82-8c74-0648480577ee.jpg

(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) = V(X) + (5 =ay² +0,² Identify the next step in this rule from the options below. OV(x-)-1_0₂² n₁ VX-1+%2² 0₁₂ ₂ OVX-1 + 2 1 7₂ O VX--1-02 10₂ Since standard deviation is the square root of variance, it follows that -7√x- V Oºx- ₂ Oox-V 01-02 n₁ 7₂ -₂ 0₂²2²0₂² +9 Oox-₁₂ 2 01+02 n₁ 0₂ Compute the estimated standard error (in MPa). (Round your answer to three decimal places.) MPa (c) Calculate a point estimate of the ratio ₁/₂ of the two standard deviations. (Round your answer 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) = V(X) + (5 =ay² +0,² Identify the next step in this rule from the options below. OV(x-)-1_0₂² n₁ VX-1+%2² 0₁₂ ₂ OVX-1 + 2 1 7₂ O VX--1-02 10₂ Since standard deviation is the square root of variance, it follows that -7√x- V Oºx- ₂ Oox-V 01-02 n₁ 7₂ -₂ 0₂²2²0₂² +9 Oox-₁₂ 2 01+02 n₁ 0₂ Compute the estimated standard error (in MPa). (Round your answer to three decimal places.) MPa (c) Calculate a point estimate of the ratio ₁/₂ of the two standard deviations. (Round your answer 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

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

Please help me with B

**Flexural Strength Data for Concrete Beams and Cylinders**

**Concrete Beams:**
The flexural strength (MPa) data for concrete beams is as follows:
- 5.1, 7.2, 7.3, 6.3, 8.1, 6.8, 7.0, 7.2, 6.5, 7.0, 6.3, 7.9, 9.0
- 8.4, 8.7, 7.8, 9.7, 7.4, 7.7, 8.0
- 7.7, 8.0, 11.6, 11.3, 11.8, 10.7

**Cylinders:**
The accompanying strength observations for cylinders (MPa) are:
- 6.5, 5.8, 7.8, 7.1, 7.2, 9.2, 6.6, 8.3, 7.0, 8.5
- 7.2, 8.1, 7.4, 8.5, 8.9, 9.1
- 14.1, 12.6, 11.4

**Statistical Analysis of Flexural Strength:**
Before collecting data, denote the beam strengths by \( X_1, \ldots, X_m \) and the cylinder strengths by \( Y_1, \ldots, Y_n \). Assume the \( X_i \)'s form a random sample from a distribution with mean \( \mu_1 \) and standard deviation \( \sigma_1 \), and the \( Y_j \)'s form a random sample from another distribution with mean \( \mu_2 \) and standard deviation \( \sigma_2 \).

**Part (a): Expected Value Calculation**
- Use rules of expected value to show that \( \overline{X} - \overline{Y} \) is an unbiased estimator of \( \mu_1 - \mu_2 \).

The correct formula is:
\[
E(\overline{X} - \overline{Y}) = E(\overline{X}) - E(\overline{Y}) = \mu_1 - \mu_2

**Section (b): Variance and Standard Deviation of the Estimator**

To obtain an expression for the variance and standard deviation (standard error) of the estimator from part (a), use the rules of variance:

\[ V(\bar{X} - \bar{Y}) = V(\bar{X}) + V(\bar{Y}) = \frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2} \]

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

- \( \bigcirc \) \(\frac{\sigma^2_1}{n_1} - \frac{\sigma^2_2}{n_2}\)
- \( \bullet \) \(\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2}\) ✔
- \( \bigcirc \) \(\frac{\sigma_1}{n_1} + \frac{\sigma_2}{n_2}\)
- \( \bigcirc \) \(\frac{\sigma_1}{n_1} - \frac{\sigma_2}{n_2}\)

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

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

**Options:**

- \( \bigcirc \) \(\sqrt{\frac{\sigma_1}{n_1} - \frac{\sigma_2}{n_2}}\)
- \( \bullet \) \(\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2}}\) ✔
- \( \bigcirc \) \(\sqrt{\frac{\sigma_1}{n_1} + \frac{\sigma_2}{n_2}}\)
- \( \bigcirc \) \(\sqrt{\frac{\sigma^2_1}{n_1} - \frac{\sigma^2_2}{n_2}}\)

**Compute the estimated standard error (in MPa), (Round your answer to three decimal places.)**

\[ \_\_\_\_\_\_ \text{ MPa} \]

**Section (c): Point Estimate

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