Prediction from an exponential regression

Enter the four points into a graphing calculator and run the exponential regression ($y=a{b}^x$). It fits a straight line to $\ln y$ against $x$ and converts back, which gives approximately $$a\approx 24.076,b\approx 1.5568$$ So the model is $y\approx 24.076(1.5568{)}^x$.

Evaluate at $x=1.5$: $$y\approx 24.076(1.5568{)}^{1.5}\approx 46.767$$ That is choice A.

Keep the full stored values of $a$ and $b$ rather than rounded ones; rounding them early can move the third decimal place. The regression curve does not pass exactly through any of the four data points, which is why its prediction differs from simply reading between the table values at $x=1$ and $x=2$.

Each choice

• A (correct). Running the exponential regression gives roughly $24.076(1.5568{)}^x$, and at $x=1.5$ that is about 46.767.
• B. This is what a quadratic regression predicts at 1.5. The question names an exponential model, so the calculator's exponential regression is the one to run.
• C. A linear regression gives 47.800 at $x=1.5$. Choosing the wrong regression command on the calculator is the likely slip.
• D. This is the cubic regression's value. With four points a cubic passes through every one of them exactly, but it is not the model the question specifies.