Use linear regression to find the equation for the linear function that best fits this data. Round both numbers to two decimal places. Write your final answer in a form of an equation
x	1	2	3	4	5	6
y	86	108	126	160	181	193

To find the equation of the linear function that best fits the given data using linear regression, we will calculate the slope (\(m\)) and the y-intercept (\(b\)) of the line in the form \(y = mx + b\).

Given data points:
\[
x: 1, 2, 3, 4, 5, 6
\]
\[
y: 86, 108, 126, 160, 181, 193
\]

### Step 1: Calculate the means of \(x\) and \(y\)
\[
\bar{x} = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = 3.5
\]
\[
\bar{y} = \frac{86 + 108 + 126 + 160 + 181 + 193}{6} = \frac{854}{6} \approx 142.33
\]

### Step 2: Calculate the slope (\(m\))
The formula for the slope \(m\) is:
\[
m = \frac{\sum{(x_i - \bar{x})(y_i - \bar{y})}}{\sum{(x_i - \bar{x})^2}}
\]

Let's compute the sums needed:
\[
\sum{(x_i - \bar{x})(y_i - \bar{y})} = (1-3.5)(86-142.33) + (2-3.5)(108-142.33) + \dots + (6-3.5)(193-142.33)
\]
\[
= (-2.5)(-56.33) + (-1.5)(-34.33) + (-0.5)(-16.33) + (0.5)(17.67) + (1.5)(38.67) + (2.5)(50.67)
\]
\[
= 140.825 + 51.495 + 8.165 + 8.835 + 58.005 + 126.675 = 394.0
\]

\[
\sum{(x_i - \bar{x})^2} = (1-3.5)^2 + (2-3.5)^2 + \dots + (6-3.5)^2
\]
\[
= (-2.5)^2 + (-1.5)^2 + (-0.5)^2 + (0.5)^2 + (1.5)^2 + (2.5)^2
\]
\[
= 6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25 = 17.5
\]

\[
m = \frac{394.0}{17.5} \approx 22.51
\]

### Step 3: Calculate the y-intercept (\(b\))
The y-intercept \(b\) is calculated using the formula:
\[
b = \bar{y} - m\bar{x}
\]
\[
b = 142.33 - 22.51 \times 3.5 \approx 142.33 - 78.785 \approx 63.54
\]

### Step 4: Write the final equation
The linear equation that best fits the data is:
\[
y = 22.51x + 63.54
\]

Would you like more details or have any questions?

Here are 5 related questions you might find interesting:
1. How would the linear regression change if an outlier was added to the data set?
2. How can you determine the strength of the fit in a linear regression?
3. What are some real-world applications of linear regression?
4. How does linear regression differ from multiple regression?
5. Can linear regression be used to predict values outside the given range of data?

**Tip:** In linear regression, it's important to check if the relationship between the variables is actually linear, as using a linear model for nonlinear data can lead to inaccurate predictions.

Learn how to use linear regression to find the equation of the best-fit line for given data points (x: 1, 2, 3, 4, 5, 6; y: 86, 108, 126, 160, 181, 193). Step-by-step solution provided, rounding to two decimal places. Suitable for high school students studying statistics and linear regression.

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