isted 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) variabl 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 101 100 94 79 78 D 175 168 145 145 145 E Click the icon to view the critical values of the Pearson correlation coefficient r Data table The regression equation is y x Round to one decimal place as needed.) Critical Values of the n a= 0.05 Siven 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.) 4 0.950 mm Hg. 0.878 6 0 811

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Below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. The goal is to find the regression equation, using the right arm's blood pressure as the predictor (x) variable. The task is also to determine the best predicted systolic blood pressure in the left arm when the systolic blood pressure in the right arm is 95 mm Hg. A significance level of 0.05 is used.

**Blood Pressure Measurements:**

- **Right Arm:** 101, 100, 94, 79, 78
- **Left Arm:** 175, 168, 145, 145, 145

**Regression Calculation:**
- Click the icon to view the critical values of the Pearson correlation coefficient \( r \).

The regression equation is given as:
\[ \hat{y} = [\text{value}] + [\text{value}]x \]
(Round to one decimal place as needed.)

For a systolic blood pressure in the right arm of 95 mm Hg, the best predicted systolic blood pressure in the left arm is:
\[ \text{value} \text{ mm Hg} \]
(Round to one decimal place as needed.)

**Data Table:**

The table provides critical values of the Pearson correlation coefficient \( r \) for different sample sizes \( n \), with significance levels \( \alpha = 0.05 \) and \( \alpha = 0.01 \).

| \( n \) | \( \alpha = 0.05 \) | \( \alpha = 0.01 \) |
|---------|---------------------|---------------------|
| 3       | 0.950               | 0.990               |
| 4       | 0.878               | 0.959               |
| 5       | 0.811               | 0.917               |
| 6       | 0.754               | 0.875               |
| 7       | 0.707               | 0.834               |
| 8       | 0.666               | 0.798               |
| 9       | 0.632               | 0.764               |
| 10      | 0.602               | 0.735               |
| 11      | 0.576               | 0.708               |
| 12      | 0.553               | 0
Transcribed Image Text:Below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. The goal is to find the regression equation, using the right arm's blood pressure as the predictor (x) variable. The task is also to determine the best predicted systolic blood pressure in the left arm when the systolic blood pressure in the right arm is 95 mm Hg. A significance level of 0.05 is used. **Blood Pressure Measurements:** - **Right Arm:** 101, 100, 94, 79, 78 - **Left Arm:** 175, 168, 145, 145, 145 **Regression Calculation:** - Click the icon to view the critical values of the Pearson correlation coefficient \( r \). The regression equation is given as: \[ \hat{y} = [\text{value}] + [\text{value}]x \] (Round to one decimal place as needed.) For a systolic blood pressure in the right arm of 95 mm Hg, the best predicted systolic blood pressure in the left arm is: \[ \text{value} \text{ mm Hg} \] (Round to one decimal place as needed.) **Data Table:** The table provides critical values of the Pearson correlation coefficient \( r \) for different sample sizes \( n \), with significance levels \( \alpha = 0.05 \) and \( \alpha = 0.01 \). | \( n \) | \( \alpha = 0.05 \) | \( \alpha = 0.01 \) | |---------|---------------------|---------------------| | 3 | 0.950 | 0.990 | | 4 | 0.878 | 0.959 | | 5 | 0.811 | 0.917 | | 6 | 0.754 | 0.875 | | 7 | 0.707 | 0.834 | | 8 | 0.666 | 0.798 | | 9 | 0.632 | 0.764 | | 10 | 0.602 | 0.735 | | 11 | 0.576 | 0.708 | | 12 | 0.553 | 0
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