To fit a straight line to the given data and estimate the value of $y$ at $x = 2.5$, we need to use the method of linear regression. A straight line is typically represented by the equation:

$y = mx + c$

where:

• $m$ is the slope of the line,
• $c$ is the y-intercept (value of $y$ when $x = 0$).

Step 1: Calculate the Slope ($m$)

The slope $m$ is calculated as:

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

where:

• $n$ is the number of data points,
• $\sum{x}$ is the sum of all $x$ values,
• $\sum{y}$ is the sum of all $y$ values,
• $\sum{xy}$ is the sum of the product of $x$ and $y$ values,
• $\sum{x^2}$ is the sum of squares of the $x$ values.

Step 2: Calculate the Y-Intercept ($c$)

Once we have the slope $m$, the y-intercept $c$ is given by:

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

Step 3: Predict the Value at $x = 2.5$

Once the linear equation is found, we can estimate the value of $y$ at $x = 2.5$ by substituting $x = 2.5$ into the equation $y = mx + c$.

Let me calculate these values for you using the data provided:

• $x: 0, 1, 2, 3, 4$
• $y: 1, 1.8, 3.3, 4.5, 6.3$

I'll perform the calculations now.The equation of the straight line that fits the given data is:

$y = 1.33x + 0.72$

Estimate for $y$ when $x = 2.5$:

Substituting $x = 2.5$ into the equation:

$y = 1.33(2.5) + 0.72 = 4.04$

Thus, the estimated value of $y$ at $x = 2.5$ is approximately 4.04.

Would you like more details on any of these steps or have any other questions?

Here are 5 related questions you could explore:

1. How do residuals help evaluate the accuracy of the fitted line?
2. How would the slope and intercept change if additional data points were included?
3. Can you fit a different type of curve (e.g., quadratic) to the data instead of a straight line?
4. What is the interpretation of the slope $m = 1.33$ in this context?
5. How does the least squares method minimize the error in linear regression?

Tip: When working with real-world data, always check if the linear model is appropriate by plotting the data and analyzing the residuals.