the owner of an ice cream stand kept track of the number of ice cream cones that were sold each day of the first week in june. she compared the ice cream sales to the average daily temperature. the data are shown in the table below. state the linear regression equation for these data, rounding all values to the nearest hundredth.

Question

Accepted Answer

# Explanation:
## Step1: Calculate the means of $x$ and $y$
The mean of $x$ values ($\bar{x}$):
$$
\bar{x}=\frac{72 + 75+81+78+77+76+80}{7}=\frac{539}{7} = 77
$$
The mean of $y$ values ($\bar{y}$):
$$
\bar{y}=\frac{126+183 + 263+229+200+185+249}{7}=\frac{1435}{7}=205
$$
## Step2: Calculate the slope $m$
$$
m=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}
$$
$$
\begin{align*}
\sum_{i = 1}^{7}(x_{i}-\bar{x})(y_{i}-\bar{y})&=(72 - 77)(126-205)+(75 - 77)(183 - 205)+(81-77)(263 - 205)+(78-77)(229 - 205)+(77-77)(200 - 205)+(76-77)(185 - 205)+(80-77)(249 - 205)\
&=(- 5)(-79)+(-2)(-22)+4\times58+1\times24+0\times(-5)+(-1)(-20)+3\times44\
&=395 + 44+232+24+0 + 20+132\
&=847
\end{align*}
$$
$$
\begin{align*}
\sum_{i = 1}^{7}(x_{i}-\bar{x})^{2}&=(72 - 77)^{2}+(75 - 77)^{2}+(81-77)^{2}+(78-77)^{2}+(77-77)^{2}+(76-77)^{2}+(80-77)^{2}\
&=(-5)^{2}+(-2)^{2}+4^{2}+1^{2}+0^{2}+(-1)^{2}+3^{2}\
&=25+4 + 16+1+0+1+9\
&=56
\end{align*}
$$
$$
m=\frac{847}{56}\approx15.125\approx15.13
$$
## Step3: Calculate the $y$ - intercept $b$
Using the formula $y=mx + b$, substitute $m = 15.13$, $x=\bar{x}=77$, $y=\bar{y}=205$
$$
205=15.13\times77+b
$$
$$
205 = 1165.01+b
$$
$$
b=205-1165.01=-960.01
$$

# Answer:
The linear regression equation is $y = 15.13x-960.01$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Calculate the means of \(x\) and \(y\)

Step2: Calculate the slope \(m\)

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

Answer: