To calculate the linear regression line (also called the line of best fit), we will use the formula:

$y = mx + b$

where:

• $m$ is the slope of the line.
• $b$ is the y-intercept.

Step 1: Calculate the slope ($m$)

The formula for the slope $m$ is:

$m = \frac{n \sum xy - \sum x \sum y}{n \sum x^2 - (\sum x)^2}$

where:

• $n$ is the number of data points,
• $\sum xy$ is the sum of the products of $x$ and $y$,
• $\sum x$ is the sum of the $x$-values,
• $\sum y$ is the sum of the $y$-values,
• $\sum x^2$ is the sum of the squares of the $x$-values.

Step 2: Compute the required sums

$x$	$y$	$x \cdot y$	$x^2$
1	22.18	22.18	1
2	22.66	45.32	4
3	26.54	79.62	9
4	28.42	113.68	16
5	28.20	141.00	25
6	29.68	178.08	36
7	31.26	218.82	49
8	34.74	277.92	64
9	35.52	319.68	81
10	37.40	374.00	100
11	40.68	447.48	121
12	44.16	529.92	144

$\sum x = 78, \quad \sum y = 381.44, \quad \sum xy = 2747.70, \quad \sum x^2 = 650, \quad n = 12$

Step 3: Calculate the slope ($m$)

$m = \frac{12(2747.70) - (78)(381.44)}{12(650) - (78)^2}$

First, simplify the numerator and denominator:

$12(2747.70) = 32972.4, \quad (78)(381.44) = 29752.32$ $12(650) = 7800, \quad (78)^2 = 6084$

Now:

$m = \frac{32972.4 - 29752.32}{7800 - 6084} = \frac{3220.08}{1716} \approx 1.88$

Step 4: Calculate the y-intercept ($b$)

The formula for the y-intercept is:

$b = \frac{\sum y - m \sum x}{n}$

Substitute the values:

$b = \frac{381.44 - 1.88(78)}{12}$

$1.88(78) = 146.64$

$b = \frac{381.44 - 146.64}{12} = \frac{234.80}{12} \approx 19.57$

Step 5: Write the regression equation

Now that we have the slope and intercept:

$y = 1.88x + 19.57$

Final Answer

The regression line is:

$y = 1.88x + 19.57$

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Do you want a step-by-step graph or any further breakdown of the calculations?

Here are five related questions to explore next:

1. How do we interpret the slope in a regression line?
2. How can we measure the goodness-of-fit for this regression line?
3. What is the predicted value of $y$ when $x = 15$?
4. How do outliers affect the regression line?
5. How would adding more data points change the slope and intercept?

Tip: The closer the points are to the regression line, the better the line represents the relationship between $x$ and $y$.