The regression equation is correct ...help me solve the second question and the answer is NOT 66.1 and 161.6 ....is there a way I could do the second part on a ti84 plus # Systolic Blood Pressure Measurement Analysis## Data DescriptionListed below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. The goal is to find the regression equation, letting the right arm blood pressure be the predictor (x) variable. Additionally, we need to find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 95 mm Hg. The significance level used for this analysis is 0.05.## Data Table| Measurement Site | Systolic Blood Pressure (mm Hg) || ---------------- | ------------------------------- || **Right Arm** | 103, 102, 94, 76, 75 || **Left Arm** | 177, 172, 145, 145, 144 |## Calculation Steps1. Click the icon to view the critical values of the Pearson correlation coefficient r.2. Determine the regression equation using the right arm blood pressure as the predictor variable.### Regression EquationThe regression equation is given by:\[ \hat{y} = \beta_0 + \beta_1 x \]For this data set, the regression equation is:\[ \hat{y} = 66 + 1.0x \]*Note:* The values are rounded to one decimal place as needed.### Predicted ValueGiven that the systolic blood pressure in the right arm is 95 mm Hg, the best predicted systolic blood pressure in the left arm can be calculated using the regression equation:\[ \hat{y} = 66 + 1.0 \times 95 \]\[ \hat{y} = 161 \text{ mm Hg} \]For educational purposes, the predicted systolic blood pressure in the left arm, given that the right arm measurement is 95 mm Hg, is **161 mm Hg**. This value was rounded to one decimal place as needed.# Further ExplorationTo delve deeper into the calculation process, including finding the Pearson correlation coefficient and the steps to derive the regression equation, students are encouraged to use statistical software or manual computation techniques according to the standard statistical methods provided in the course materials.

Let X denote the right arm blood pressure Let Y denote the left arm blood pressure The regression…

Listed below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. Find the regression equation, letting the right arm blood pressure be the predictor (x) variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 95 mm Hg. Use a significance level of 0.05. Right Arm Left Arm BEE Click the icon 103 102 94 76 177 172 145 145 o view the critical values of the Pearson correlation coefficient r The regression equation is y= 66 +1.0 x (Round to one decimal place as needed.) 75 144 COLLE Given that the systolic blood pressure in the right arm is 95 mm Hg, the best predicted systolic blood pressure in the left arm is (Round to one decimal place as needed.) mm Hg.

Listed below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. Find the regression equation, letting the right arm blood pressure be the predictor (x) variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 95 mm Hg. Use a significance level of 0.05. Right Arm Left Arm BEE Click the icon 103 102 94 76 177 172 145 145 o view the critical values of the Pearson correlation coefficient r The regression equation is y= 66 +1.0 x (Round to one decimal place as needed.) 75 144 COLLE Given that the systolic blood pressure in the right arm is 95 mm Hg, the best predicted systolic blood pressure in the left arm is (Round to one decimal place as needed.) mm Hg.

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

The regression equation is correct ...help me solve the second question and the answer is NOT 66.1 and 161.6 ^....is there a way I could do the second part on a ti84 plus

# Systolic Blood Pressure Measurement Analysis

## Data Description
Listed below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. The goal is to find the regression equation, letting the right arm blood pressure be the predictor (x) variable. Additionally, we need to find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 95 mm Hg. The significance level used for this analysis is 0.05.

## Data Table
| Measurement Site | Systolic Blood Pressure (mm Hg) |
| ---------------- | ------------------------------- |
| **Right Arm** | 103, 102, 94, 76, 75 |
| **Left Arm** | 177, 172, 145, 145, 144 |

## Calculation Steps

1. Click the icon to view the critical values of the Pearson correlation coefficient r.
2. Determine the regression equation using the right arm blood pressure as the predictor variable.

### Regression Equation
The regression equation is given by:
\[ \hat{y} = \beta_0 + \beta_1 x \]

For this data set, the regression equation is:
\[ \hat{y} = 66 + 1.0x \]

*Note:* The values are rounded to one decimal place as needed.

### Predicted Value
Given that the systolic blood pressure in the right arm is 95 mm Hg, the best predicted systolic blood pressure in the left arm can be calculated using the regression equation:
\[ \hat{y} = 66 + 1.0 \times 95 \]
\[ \hat{y} = 161 \text{ mm Hg} \]

For educational purposes, the predicted systolic blood pressure in the left arm, given that the right arm measurement is 95 mm Hg, is **161 mm Hg**. This value was rounded to one decimal place as needed.

# Further Exploration
To delve deeper into the calculation process, including finding the Pearson correlation coefficient and the steps to derive the regression equation, students are encouraged to use statistical software or manual computation techniques according to the standard statistical methods provided in the course materials.

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