Answered

The table shows data from local cable companies, representing the number of channels offered (x) and the monthly dollar cost of service (y).

[tex]\[
\begin{array}{|c|c|c|c|}
\hline
x & y & x^2 & xy \\
\hline
115 & 65 & 13,225 & 7,475 \\
\hline
122 & 72 & 14,884 & 8,784 \\
\hline
128 & 85 & 16,384 & 10,880 \\
\hline
132 & 87 & 17,424 & 11,484 \\
\hline
135 & 95 & 18,225 & 12,825 \\
\hline
\sum x = 632 & \sum y = 404 & \sum x^2 = 80,142 & \sum xy = 51,448 \\
\hline
\end{array}
\][/tex]

Which regression equation correctly models the data?



Answer :

GinnyAnswer
To determine the regression equation that models the given data, we can follow these steps:

1. Calculate the means of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:

[tex]\[ \bar{x} = \frac{\sum x}{n} \quad \text{and} \quad \bar{y} = \frac{\sum y}{n} \][/tex]

Given the values [tex]\(\sum x = 632\)[/tex], [tex]\(\sum y = 404\)[/tex], and [tex]\( n = 5 \)[/tex]:

[tex]\[ \bar{x} = \frac{632}{5} = 126.4 \quad \text{and} \quad \bar{y} = \frac{404}{5} = 80.8 \][/tex]

2. Calculate the slope ( [tex]\( b \)[/tex] ) of the regression line:

[tex]\[ b = \frac{\sum (xy) - n \cdot \bar{x} \cdot \bar{y}}{\sum (x^2) - n \cdot \bar{x}^2} \][/tex]

Using the given data [tex]\(\sum xy = 51448\)[/tex] and [tex]\(\sum x^2 = 80142\)[/tex]:

[tex]\[ b = \frac{51448 - 5 \cdot 126.4 \cdot 80.8}{80142 - 5 \cdot (126.4)^2} \][/tex]

Calculating the numerator and denominator separately:

[tex]\[ \text{Numerator} = 51448 - 5 \cdot 126.4 \cdot 80.8 \approx 51448 - 51065.6 = 382.4 \][/tex]

[tex]\[ \text{Denominator} = 80142 - 5 \cdot (126.4)^2 \approx 80142 - 80084.8 = 257.2 \][/tex]

So, the slope [tex]\( b \)[/tex] is:

[tex]\[ b = \frac{382.4}{257.2} \approx 1.48678 \][/tex]

3. Calculate the intercept ( [tex]\( a \)[/tex] ):

[tex]\[ a = \bar{y} - b \cdot \bar{x} \][/tex]

[tex]\[ a = 80.8 - 1.48678 \cdot 126.4 \approx 80.8 - 187.92908 = -107.12908 \][/tex]

4. Form the regression equation:

The regression equation is of the form [tex]\( y = a + bx \)[/tex].

Given the values calculated:

[tex]\[ y = -107.12908 + 1.48678x \][/tex]

So, the regression equation that correctly models the data is:

[tex]\[ \boxed{y = -107.12908 + 1.48678x} \][/tex]