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write a power regression equation for the following data, rounding all …

Question

write a power regression equation for the following data, rounding all coefficients to the nearest hundredth.

xy
46140
86910
97075
127726
148366
178200

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Explanation:

Step1: Recall Power Regression Form

A power regression equation has the form \( y = ax^{b} \), where \( a \) and \( b \) are constants to be determined. To find \( a \) and \( b \), we can take the natural logarithm of both sides: \( \ln(y)=\ln(a)+b\ln(x) \). This transforms the power regression into a linear regression problem of the form \( Y = A + bX \), where \( Y=\ln(y) \), \( A = \ln(a) \), and \( X=\ln(x) \).

Step2: Calculate Logarithms of Data

First, we calculate \( \ln(x) \) and \( \ln(y) \) for each data point:

  • For \( x = 2, y = 5399 \): \( \ln(2)\approx0.6931, \ln(5399)\approx8.5933 \)
  • For \( x = 4, y = 6140 \): \( \ln(4)\approx1.3863, \ln(6140)\approx8.7232 \)
  • For \( x = 8, y = 6910 \): \( \ln(8)\approx2.0794, \ln(6910)\approx8.8473 \)
  • For \( x = 9, y = 7075 \): \( \ln(9)\approx2.1972, \ln(7075)\approx8.8633 \)
  • For \( x = 12, y = 7726 \): \( \ln(12)\approx2.4849, \ln(7726)\approx8.9503 \)
  • For \( x = 14, y = 8366 \): \( \ln(14)\approx2.6391, \ln(8366)\approx9.0333 \)
  • For \( x = 17, y = 8200 \): \( \ln(17)\approx2.8332, \ln(8200)\approx9.0150 \)

Step3: Perform Linear Regression on Transformed Data

We now perform linear regression on the data \( (X, Y) \) where \( X=\ln(x) \) and \( Y=\ln(y) \). Using a calculator or software (e.g., linear regression function), we find the slope \( b \) and the intercept \( A \) of the line \( Y = A + bX \).

After performing the linear regression, we find that \( b\approx0.17 \) and \( A\approx\ln(a)\approx8.525 \).

Step4: Solve for \( a \)

Since \( A = \ln(a) \), we solve for \( a \) by exponentiating \( A \): \( a = e^{A} \). Substituting \( A\approx8.525 \), we get \( a = e^{8.525}\approx5050.67 \).

Step5: Write the Power Regression Equation

Substituting \( a\approx5050.67 \) and \( b\approx0.17 \) into the power regression form \( y = ax^{b} \), we get \( y = 5050.67x^{0.17} \).

Answer:

\( y = 5050.67x^{0.17} \)