3.2.4 Assessing the Fit of a Linear Model

Compare your answer: Your answer may vary, but here is a sample.

The line you drew should go through the middle of the data in your scatter plot.

Compare your answer:

$y=0.78x-9.44$

Compare your answer:

The slope is 0.78.

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This means that about 1 additional pound of ice cream is sold with every 0.78-degree increase in the outside temperature.

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The $y$-intercept is $(0,-9.44)$.

Compare your answer:

This means that no ice cream is sold when the temperature is about −9.44 degrees Celsius.

Compare your answer:

About −1.64 degrees Celsius.

Compare your answer:

Your answer may vary, but here is a sample. This may be a good estimate for the data since the points seem close to the line, but something different might happen when the temperature gets below freezing, so it’s possible that when only 10 pounds of ice cream are sold, the relationship with the weather may be different.

Compare your answer:

Card A: $y=1.1979x+1.3196$

Card B: $y=-0.4337x+12.288$

Card C: $y=0.9749x+4.6529$

Card D: $y=-2.063x+16.144$

Card E: $y=1.1979x+5.8196$

Compare your answer:

• Card A: $y=1.1979x+1.3196$. When $x$ is zero, $y$ is 1.3196. As $x$ increases by 1, $y$ increases, on average, by 1.1979.
• Card B: $y=-0.4337x+12.288$. When $x$ is zero, $y$ is 12.288. As $x$ increases by 1, $y$ decreases, on average, by 0.4337.
• Card C: $y=0.9749x+4.6529$. When $x$ is zero, $y$ is 4.6529. As $x$ increases by 1, $y$ increases, on average, by 0.9749.
• Card D: $y=-2.063x+16.144$. When $x$ is zero, $y$ is 16.144. As $x$ increases by 1, $y$ decreases, on average, by 2.063.
• Card E: $y=1.1979x+5.8196$. When $x$ is zero, $y$ is 5.8196. As $x$ increases by 1, $y$ increases, on average, by 1.1979.