The table compares the average daily temperature and ice cream sales each day.
\begin{tabular}{|l|l|}
\hline Temperature [tex]$\left({ }^{\circ} F \right)$[/tex] & Ice Cream Sales \\
\hline 56.1 & [tex]$\$[/tex] 107$ \\
\hline 60.3 & [tex]$\$[/tex] 116$ \\
\hline 64.2 & [tex]$\$[/tex] 120$ \\
\hline 66.4 & [tex]$\$[/tex] 122$ \\
\hline 71.3 & [tex]$\$[/tex] 125$ \\
\hline 73.9 & [tex]$\$[/tex] 130$ \\
\hline 74.2 & [tex]$\$[/tex] 132$ \\
\hline 75.2 & [tex]$\$[/tex] 135$ \\
\hline 80.1 & [tex]$\$[/tex] 141$ \\
\hline 84.8 & [tex]$\$[/tex] 152$ \\
\hline
\end{tabular}

What is the slope of the line of best fit, where [tex]$x$[/tex] represents the average daily temperature and [tex]$y$[/tex] represents the total ice cream sales? (Round your answer to one decimal place.)

A. 6.3
B. 4.2
C. 2.8
D. 1.4



Answer :

To find the slope of the line of best fit, we need to perform a linear regression analysis on the given temperature (\( x \)) and ice cream sales data (\( y \)). Here are the steps involved in calculating the slope:

1. Calculate the means of \( x \) and \( y \):
- Mean of temperature (\( \bar{x} \)):
[tex]\[ \bar{x} = 70.65 \][/tex]
- Mean of ice cream sales (\( \bar{y} \)):
[tex]\[ \bar{y} = 128.0 \][/tex]

2. Calculate the intermediate sums:
- Sum of the products of corresponding \( x \) and \( y \) values (\( \sum xy \)):
[tex]\[ \sum xy = 91451.9 \][/tex]
- Sum of the \( x \) values (\( \sum x \)):
[tex]\[ \sum x = 706.5 \][/tex]
- Sum of the \( y \) values (\( \sum y \)):
[tex]\[ \sum y = 1280 \][/tex]
- Sum of the squares of the \( x \) values (\( \sum x^2 \)):
[tex]\[ \sum x^2 = 50626.53 \][/tex]

3. Calculate the slope:
- \( n \) is the number of data points, which is 10 in this case.
- The slope (\( m \)) of the line of best fit is given by the formula:
[tex]\[ m = \frac{n \sum xy - (\sum x)(\sum y)}{n \sum x^2 - (\sum x)^2} \][/tex]
Plugging in the values, we get:
[tex]\[ m = \frac{10 \cdot 91451.9 - (706.5 \cdot 1280)}{10 \cdot 50626.53 - (706.5)^2} \][/tex]
Simplifying this, we find:
[tex]\[ m = 1.4 \][/tex]

Therefore, the slope of the line of best fit, rounded to one decimal place, is [tex]\( \boxed{1.4} \)[/tex].