The common cricket can be used as a crude thermometer. The colder the temperature, the slower the rate of chirping. The table below shows the average chirp rate of a cricket at various temperatures. $\begin{array} { c c }
\hline \begin{array} { c }
\text { Chirp Rate } \
\text { (chirps/second) }
\end{array} & \begin{array} { c }
\text { Temperature } \
\left( { } ^ { \circ } \mathrm { F } \right)
\end{array} \
\hline 3.7 & 67.8 \
2.6 & 64.6 \
1.9 & 55.1 \
1.8 & 44.3 \
3.9 & 71 \
2.8 & 67.8 \
\hline
\end{array}$
Compute the least-squares regression line for predicting the temperature from the chirp rate.
A) y=9.9492+34.0748 x
B) y=12.7143+34.0748 x
C) y=34.0748+12.7143 x
D) y=34.0748+9.9492 x

Question

The common cricket can be used as a crude thermometer. The colder the temperature, the slower the rate of chirping. The table below shows the average chirp rate of a cricket at various temperatures. $\begin{array} { c c } 
\hline \begin{array} { c } 
\text { Chirp Rate } \
\text { (chirps/second) }
\end{array} & \begin{array} { c } 
\text { Temperature } \
\left( { } ^ { \circ } \mathrm { F } \right)
\end{array} \
\hline 3.7 & 67.8 \
2.6 & 64.6 \
1.9 & 55.1 \
1.8 & 44.3 \
3.9 & 71 \
2.8 & 67.8 \
\hline
\end{array}$
 Compute the least-squares regression line for predicting the temperature from the chirp rate. 
A) y=9.9492+34.0748 x
B) y=12.7143+34.0748 x
C) y=34.0748+12.7143 x
D) y=34.0748+9.9492 x

Quizplus · Accepted Answer

To find the least-squares regression line, we need to calculate the slope and y-intercept using the formulas:
$$b=\frac{\sum\left(x_{i}-\bar{x}\right)\left(y_{i}-\bar{y}\right)}{\sum\left(x_{i}-\bar{x}\right)^{2}}$$
$$a=\bar{y}-b \bar{x}$$
where
$$\bar{x}=\frac{\sum x_{i}}{n}$$
$$\bar{y}=\frac{\sum y_{i}}{n}$$
and n is the sample size.

Using these formulas with the given data, we get:
$$\bar{x}=2.833$$
$$\bar{y}=62.775$$
$$b=\frac{(67.8-62.775)(3.7-2.833)+(64.6-62.775)(2.6-2.833)+(55.1-62.775)(1.9-2.833)+(44.3-62.775)(1.8-2.833)+(71-62.775)(3.9-2.833)+(67.8-62.775)(2.8-2.833)}{(3.7-2.833)^2+(2.6-2.833)^2+(1.9-2.833)^2+(1.8-2.833)^2+(3.9-2.833)^2+(2.8-2.833)^2}=9.9492$$
$$a=\bar{y}-b \bar{x}=62.775-9.9492(2.833)=34.0748$$

Therefore, the least-squares regression line for predicting temperature from chirp rate is:
$$y=34.0748+9.9492x$$
or
$$y=9.9492x+34.0748$$

The answer is D.

The Common Cricket Can Be Used as a Crude Thermometer