To compute the least-squares regression equation for the given dataset, we need to find the equation of the line in the form:

$y = mx + b$

where $m$ is the slope, and $b$ is the y-intercept.

Steps:

1. Calculate the mean of x and y values:

$\bar{x} = \frac{1 + 2 + 5 + 6 + 7}{5} = \frac{21}{5} = 4.2$

$\bar{y} = \frac{1 + 5 + 4 + 2 + 3}{5} = \frac{15}{5} = 3$

2. Calculate the slope $m$:

$m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}$

We will first compute the necessary sums.

• $x_i - \bar{x}$:

  • For $x_1 = 1$, $1 - 4.2 = -3.2$
  • For $x_2 = 2$, $2 - 4.2 = -2.2$
  • For $x_3 = 5$, $5 - 4.2 = 0.8$
  • For $x_4 = 6$, $6 - 4.2 = 1.8$
  • For $x_5 = 7$, $7 - 4.2 = 2.8$

• $y_i - \bar{y}$:

  • For $y_1 = 1$, $1 - 3 = -2$
  • For $y_2 = 5$, $5 - 3 = 2$
  • For $y_3 = 4$, $4 - 3 = 1$
  • For $y_4 = 2$, $2 - 3 = -1$
  • For $y_5 = 3$, $3 - 3 = 0$

• Compute $\sum (x_i - \bar{x})(y_i - \bar{y})$: $(-3.2)(-2) + (-2.2)(2) + (0.8)(1) + (1.8)(-1) + (2.8)(0) = 6.4 - 4.4 + 0.8 - 1.8 + 0 = 1$

• Compute $\sum (x_i - \bar{x})^2$: $(-3.2)^2 + (-2.2)^2 + (0.8)^2 + (1.8)^2 + (2.8)^2 = 10.24 + 4.84 + 0.64 + 3.24 + 7.84 = 26.8$

Now calculate the slope $m$:

$m = \frac{1}{26.8} \approx 0.0373$

3. Calculate the y-intercept $b$:

$b = \bar{y} - m\bar{x}$

$b = 3 - (0.0373 \times 4.2) = 3 - 0.15666 \approx 2.843$

Final regression equation:

$y = 0.0373x + 2.843$

Let me know if you would like more details or have any questions.

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Here are five questions related to least-squares regression:

1. How do you interpret the slope and intercept of a regression line?
2. What does it mean if the slope of the regression line is zero?
3. Can you calculate the correlation coefficient from this dataset?
4. What are residuals in the context of regression analysis?
5. How does the least-squares method minimize error in the regression line?

Tip: The slope in a least-squares regression equation represents the average change in $y$ for each unit increase in $x$.