Math Problem Statement

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 137 mmHg, what is the best predicted diastolic blood pressure?

Solution

To find the regression equation and use it to predict the diastolic blood pressure for a systolic reading of 137 mmHg, follow these steps:

Step 1: Calculate the Regression Equation

The regression equation is in the form: y^=a+bx\hat{y} = a + b x where:

  • b=Covariance of x and yVariance of xb = \frac{\text{Covariance of } x \text{ and } y}{\text{Variance of } x}
  • a=yˉbxˉa = \bar{y} - b \bar{x}

From the table:

  • xx: Systolic values: 129, 119, 126, 145, 126, 125, 142, 110
  • yy: Diastolic values: 94, 64, 68, 84, 93, 91, 104, 79

Step 2: Compute Means (xˉ\bar{x} and yˉ\bar{y})

xˉ=xn,yˉ=yn\bar{x} = \frac{\sum x}{n}, \quad \bar{y} = \frac{\sum y}{n}

Step 3: Compute Slope (bb) and Intercept (aa)

Use the formulas: b=(xxˉ)(yyˉ)(xxˉ)2b = \frac{\sum (x - \bar{x})(y - \bar{y})}{\sum (x - \bar{x})^2} a=yˉbxˉa = \bar{y} - b \bar{x}

Step 4: Predict Diastolic Blood Pressure

Substitute x=137x = 137 into the regression equation.

I'll calculate as np

Data

systolic = np.array([129, 119, 126, 145, 126, 125, 142, 110]) diastolic = np.array([94, 64, 68, 84, 93, 91, 104, 79])

Calculate means

mean_systolic = np.mean(systolic) mean_diastolic = np.mean(diastolic)

Calculate slope (b)

numerator = np.sum((systolic - mean_systolic) * (diastolic - mean_diastolic)) denominator = np.sum((systolic - mean_systolic) ** 2) b = numerator / denominator

Calculate intercept (a)

a = mean_diastolic - b * mean_systolic

Regression equation: y = a + b * x

Predict diastolic blood pressure for systolic = 137

predicted_diastolic = a + b * 137

a, b, predicted_diastThe regression equation is:

y^=6.60+0.61x\hat{y} = 6.60 + 0.61x

Using this equation, the predicted diastolic blood pressure for a systolic reading of 137 mmHg is approximately:

y^=90.27mmHg\hat{y} = 90.27 \, \text{mmHg}

Would you like a detailed breakdown of these calculations or further assistance? Here are 5 questions to expand on this:

  1. How are the slope (bb) and intercept (aa) derived in regression analysis?
  2. What does the Pearson correlation coefficient signify for this data?
  3. Can we assess the accuracy of this regression equation using residual analysis?
  4. How can we calculate the coefficient of determination (R2R^2) for this dataset?
  5. What are some potential sources of error in this prediction?

Tip: Always plot the data to visually inspect the relationship before using regression for predictions

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Math Problem Analysis

Mathematical Concepts

Regression Analysis
Linear Regression
Pearson Correlation Coefficient

Formulas

ŷ = a + bx
b = Σ((x - x̄)(y - ȳ)) / Σ((x - x̄)^2)
a = ȳ - b * x̄
Pearson correlation coefficient formula

Theorems

Linear regression prediction
Least-squares estimation

Suitable Grade Level

Grades 10-12, College Introductory Statistics