Question

find the equation for the least squares regression line of the data described below.
jaylen has noticed that his bike ride to work takes longer on some days than others. he is curious to see how the morning temperature is related to the duration of his commute.
for the past several mornings, he measured the temperature (in celsius), x, and the time his commute had taken (in minutes), y.
temperature (degrees celsius) | commute time (minutes)
-4.47 | 58.58
0.97 | 59.10
1.61 | 44.39
1.92 | 39.91
3.59 | 46.44
round your answers to the nearest thousandth.
y = x +

Explanation:

Step1: Calculate necessary sums

First, we list the data points:
\( (x_1, y_1) = (-4.47, 58.58) \)
\( (x_2, y_2) = (0.97, 59.10) \)
\( (x_3, y_3) = (1.61, 44.39) \)
\( (x_4, y_4) = (1.92, 39.91) \)
\( (x_5, y_5) = (3.59, 46.44) \)

Calculate \( \sum x \), \( \sum y \), \( \sum xy \), \( \sum x^2 \):

\( \sum x = -4.47 + 0.97 + 1.61 + 1.92 + 3.59 = 3.62 \)

\( \sum y = 58.58 + 59.10 + 44.39 + 39.91 + 46.44 = 228.42 \)

\( \sum xy = (-4.47)(58.58) + (0.97)(59.10) + (1.61)(44.39) + (1.92)(39.91) + (3.59)(46.44) \)
\( = -261.8526 + 57.327 + 71.4679 + 76.6272 + 166.7196 = 110.2891 \)

\( \sum x^2 = (-4.47)^2 + (0.97)^2 + (1.61)^2 + (1.92)^2 + (3.59)^2 \)
\( = 19.9809 + 0.9409 + 2.5921 + 3.6864 + 12.8881 = 39.0884 \)

Step2: Calculate slope \( m \)

The formula for the slope \( m \) of the least squares regression line is:
\( m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} \)
where \( n = 5 \) (number of data points).

Substitute the values:
\( m = \frac{5(110.2891) - (3.62)(228.42)}{5(39.0884) - (3.62)^2} \)
\( = \frac{551.4455 - 826.8804}{195.442 - 13.1044} \)
\( = \frac{-275.4349}{182.3376} \approx -1.510 \)

Step3: Calculate y-intercept \( b \)

The formula for the y-intercept \( b \) is:
\( b = \frac{\sum y - m\sum x}{n} \)

Substitute the values:
\( b = \frac{228.42 - (-1.510)(3.62)}{5} \)
\( = \frac{228.42 + 5.4662}{5} \)
\( = \frac{233.8862}{5} \approx 46.777 \)

Answer:

\( y = -1.510x + 46.777 \)