The data show systolic and diastolic blood pressure of certain people. Find the regression​ equation, letting the systolic reading be the independent​ (x) variable. If one of these people has a systolic blood pressure of 118 mm​ Hg, what is the best predicted diastolic blood​ pressure?
Systolic
150
124
149
120
114
135
136
116

Diastolic
109
91
110
85
82
74
72
87
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Part 1
What is the regression​ equation?
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enter your response herex​ (Round to two decimal places as​ needed.)

To find the regression equation, we first need to calculate the slope (b) and the intercept (a) of the line. The regression equation can be written as:

\[
\hat{y} = a + bx
\]

Where:
- \( \hat{y} \) is the predicted diastolic blood pressure,
- \( x \) is the systolic blood pressure,
- \( a \) is the intercept,
- \( b \) is the slope of the regression line.

### Step 1: Calculate the Mean of X and Y
Let's denote:
- \( x \) as systolic blood pressure,
- \( y \) as diastolic blood pressure.

Given dataTo find the regression equation, we first need to calculate the slope (b) and the intercept (a) of the line. The regression equation can be written as:

\[
\hat{y} = a + bx
\]

Where:
- \( \hat{y} \) is the predicted diastolic blood pressure,
- \( x \) is the systolic blood pressure,
- \( a \) is the intercept,
- \( b \) is the slope of the regression line.

### Step 1: Calculate the Mean of X and Y
Let's denote:
- \( x \) as systolic blood pressure,
- \( y \) as diastolic blood pressure.

Given 
- Systolic (X): 150, 124, 149, 120, 114, 135, 136, 116
- Diastolic (Y): 109, 91, 110, 85, 82, 74, 72, 87

Calculate the means:
\[
\bar{x} = \frac{\sum x}{n}, \quad \bar{y} = \frac{\sum y}{n}
\]

### Step 2: Calculate the Slope (b)
The slope \( b \) is given by:

\[
b = \frac{\sum (x - \bar{x})(y - \bar{y})}{\sum (x - \bar{x})^2}
\]

### Step 3: Calculate the Intercept (a)
The intercept \( a \) is given by:

\[
a = \bar{y} - b\bar{x}
\]

Let's calculate these values.The regression equation, with systolic blood pressure as the independent variable (x), is:

\[
\hat{y} = 15.44 + 0.56x
\]

### Step 4: Predict the Diastolic Blood Pressure
To predict the diastolic blood pressure when the systolic blood pressure is 118 mm Hg, substitute \( x = 118 \) into the regression equation:

\[
\hat{y} = 15.44 + 0.56(118)
\] 

Let's calculate this value.The best predicted diastolic blood pressure for a systolic reading of 118 mm Hg is approximately **81.73 mm Hg**.

Would you like any more details or have any questions? Here are a few related questions you might find useful:

1. How is the Pearson correlation coefficient calculated?
2. What does the slope of the regression line represent in this context?
3. How can we assess the accuracy of this regression model?
4. What is the significance of the intercept in the regression equation?
5. How would the prediction change if the systolic value was different?

**Tip:** Always check the residuals (differences between actual and predicted values) to understand how well your regression model fits the data.

The data show systolic and diastolic blood pressure of certain people. Find the regression equation, letting the systolic reading be the independent (x) variable. If one of these people has a systolic blood pressure of 118 mm Hg, what is the best predicted diastolic blood pressure? Systolic: 150, 124, 149, 120, 114, 135, 136, 116. Diastolic: 109, 91, 110, 85, 82, 74, 72, 87.

Learn how to calculate the linear regression equation to predict diastolic blood pressure from systolic readings. This example uses data points for systolic blood pressure and diastolic pressure, with detailed step-by-step calculations to determine the slope, intercept, and prediction for a systolic value of 118 mm Hg. Ideal for Grades 10-12 students studying statistics.

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