Question

of a regression line
for each of three scenarios, use a graphing utility to create a scatterplot of the data, and then answer the questions that follow.

1. the national weather service uses a chart to determine the wind chill factor at various wind speeds. wind chill is the cooling effect caused by wind blowing on a surface. this phenomenon explains why you can feel colder on a windy day than on a calm day, even if both days have the same air temperature. the following table shows the air temperatures in degrees fahrenheit (°f) and the corresponding wind chill factor (the temperature in degrees fahrenheit that it feels like) for a wind speed of 20 miles per hour (mph):

air temperature (°f)	40	35	30	25	20	15
wind chill factor (°f)	30	24	17	11	4	-2

(a) determine the equation of the regression line.
(b) what is the slope of the regression line and what does it mean in the context of the problem?
(c) what is the dependent variable axis intercept and what does it mean in the context of the problem?
(d) use the equation of the regression line to estimate the wind chill factor for a 32°f day with a wind speed of 20 mph. show the work that leads to your answer.

Explanation:

Part (a)

Step 1: Define Variables

Let \( x \) be the Air Temperature (\(^\circ\text{F}\)) and \( y \) be the Wind Chill Factor (\(^\circ\text{F}\)). The data points are \((40, 30)\), \((35, 24)\), \((30, 17)\), \((25, 11)\), \((20, 4)\), \((15, -2)\).

Step 2: Calculate Necessary Sums

• \( n = 6 \) (number of data points)
• \( \sum x = 40 + 35 + 30 + 25 + 20 + 15 = 165 \)
• \( \sum y = 30 + 24 + 17 + 11 + 4 + (-2) = 84 \)
• \( \sum xy = (40 \times 30) + (35 \times 24) + (30 \times 17) + (25 \times 11) + (20 \times 4) + (15 \times (-2)) = 1200 + 840 + 510 + 275 + 80 - 30 = 2875 \)
• \( \sum x^2 = 40^2 + 35^2 + 30^2 + 25^2 + 20^2 + 15^2 = 1600 + 1225 + 900 + 625 + 400 + 225 = 4975 \)

Step 3: Calculate Slope (\(m\))

The formula for the slope of the regression line is:
\[
m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}
\]
Substitute the values:
\[
m = \frac{6 \times 2875 - 165 \times 84}{6 \times 4975 - 165^2}
\]
\[
m = \frac{17250 - 13860}{29850 - 27225}
\]
\[
m = \frac{3390}{2625} \approx 1.291
\]

Step 4: Calculate Intercept (\(b\))

The formula for the intercept is:
\[
b = \frac{\sum y - m\sum x}{n}
\]
Substitute the values:
\[
b = \frac{84 - 1.291 \times 165}{6}
\]
\[
b = \frac{84 - 213.015}{6}
\]
\[
b = \frac{-129.015}{6} \approx -21.5025
\]

Step 5: Regression Line Equation

The equation of the regression line is \( \hat{y} = mx + b \), so:
\[
\hat{y} \approx 1.291x - 21.5025
\]

Step 1: Identify the Slope

From part (a), the slope \( m \approx 1.291 \).

Step 2: Interpret the Slope

In the context of the problem, the slope represents the change in the wind chill factor (in degrees Fahrenheit) for a one - degree Fahrenheit increase in the air temperature, when the wind speed is 20 mph. So, for every 1°F increase in air temperature, the wind chill factor is expected to increase by approximately 1.29°F (when wind speed is 20 mph).

Step 1: Identify the Intercept

From part (a), the intercept \( b \approx - 21.5025 \).

Step 2: Interpret the Intercept

The intercept is the value of the wind chill factor when the air temperature \( x = 0^\circ\text{F}\) (and wind speed is 20 mph). However, we need to be cautious as an air temperature of \( 0^\circ\text{F}\) may be outside the range of our data (our data starts at \( 15^\circ\text{F}\)), but in the context of the regression model, it represents the predicted wind chill factor when the air temperature is \( 0^\circ\text{F}\) (with a wind speed of 20 mph).

Answer:

The equation of the regression line is approximately \( \hat{y} = 1.29x - 21.50 \) (or more precisely \( \hat{y} \approx 1.291x - 21.50 \)).

Part (b)