9 the data table below shows water temperatures at various depths in an ocean. water depth (x) (meters) temperature (y) (°c) 50 18 75 15 100 12 150 7 200 1 write the linear regression equation for this set of data, rounding all values to the nearest thousandth. using this equation, predict the temperature (°c), to the nearest integer, at a water depth of 255 meters.

Question

Accepted Answer

# Explanation:
## Step1: Calculate the means of x and y
Let $x_i$ be the water - depth values and $y_i$ be the temperature values.
$n = 5$
$\bar{x}=\frac{50 + 75+100 + 150+200}{5}=\frac{575}{5}=115$
$\bar{y}=\frac{18 + 15+12 + 7+1}{5}=\frac{53}{5}=10.6$
## Step2: Calculate the slope $m$
$$
\begin{align*}
m&=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2}\
\sum_{i = 1}^{5}(x_i - 115)(y_i - 10.6)&=(50 - 115)(18 - 10.6)+(75 - 115)(15 - 10.6)+(100 - 115)(12 - 10.6)+(150 - 115)(7 - 10.6)+(200 - 115)(1 - 10.6)\
&=(- 65)\times7.4+(-40)\times4.4+(-15)\times1.4+(35)\times(-3.6)+(85)\times(-9.6)\
&=-481-176 - 21-126 - 816\
&=-1620
\end{align*}
$$
$$
\begin{align*}
\sum_{i = 1}^{5}(x_i - 115)^2&=(50 - 115)^2+(75 - 115)^2+(100 - 115)^2+(150 - 115)^2+(200 - 115)^2\
&=(-65)^2+(-40)^2+(-15)^2+(35)^2+(85)^2\
&=4225 + 1600+225+1225+7225\
&=14500
\end{align*}
$$
$m=\frac{-1620}{14500}\approx - 0.112$
## Step3: Calculate the y - intercept $b$
$b=\bar{y}-m\bar{x}=10.6-(-0.112)\times115=10.6 + 12.88=23.480$
The linear - regression equation is $y=-0.112x + 23.480$
## Step4: Predict the temperature at $x = 255$
Substitute $x = 255$ into the equation $y=-0.112\times255+23.480$
$y=-28.56+23.480=-5.08\approx - 5$

# Answer:
The linear regression equation is $y=-0.112x + 23.480$ and the predicted temperature at a water depth of 255 meters is $-5^{\circ}C$

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the means of x and y

Step2: Calculate the slope \(m\)

Step3: Calculate the y - intercept \(b\)

Step4: Predict the temperature at \(x = 255\)

Answer: