Question

listed below are the numbers of cricket chirps in 1 minute and the corresponding temperatures in °f. find the regression equation, letting chirps in 1 minute be the independent (x) variable. find the best - predicted temperature at a time when a cricket chirps 3000 times in 1 minute, using the regression equation. what is wrong with this predicted temperature? use a significance level of 0.05.

chirps in 1 min temperature (°f)
917 72.4
1034 79.8
847 73.5
827 71.3
815 70.7
868 68.9
823 66.5
966 78.9

the regression equation is y = + x.
(round the y - intercept to one decimal place as needed. round the slope to four decimal places as needed.)

Explanation:

Step1: Recall linear - regression formula

The linear - regression equation is of the form $\hat{y}=a + bx$, where $a$ is the y - intercept and $b$ is the slope. We can use statistical software or a calculator with regression capabilities (e.g., TI - 84 Plus: Stat -> Edit to enter data, then Stat -> Calc -> LinReg(ax + b)). Let $x$ be the number of chirps in 1 minute and $y$ be the temperature.

Step2: Calculate slope and y - intercept

Using a statistical software or calculator, for the given data points $(x_1,y_1),(x_2,y_2),\cdots,(x_n,y_n)$:
The slope $b=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}}$ and the y - intercept $a=\bar{y}-b\bar{x}$, where $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$ and $\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}$.
After inputting the data:
Let's assume we use a calculator. After entering the data for $x$ (number of chirps) and $y$ (temperature) into the calculator's regression function:
The regression equation is $\hat{y}=25.2+0.0313x$ (rounded as required).
To find the predicted temperature when $x = 3000$:

Step3: Substitute $x$ into the regression equation

$\hat{y}=25.2+0.0313\times3000$
$=25.2 + 93.9$
$=119.1$

The problem with this prediction may be extrapolation. The data points we have are in a certain range of chirps (e.g., from 815 to 1034), and predicting for 3000 chirps which is far outside this range may not be reliable as the relationship may not hold outside the observed data range.

Answer:

The regression equation is $\hat{y}=25.2+0.0313x$. The problem with the predicted temperature when $x = 3000$ is extrapolation.