Incorrect y intercept using fitlm

Hi and thanks in advance.

I'm currently working on creating basic plots and obtaining the equation of the best-fit line using 'fitlm' in MATLAB. However, I've observed that the y-intercept of the equation doesn't seem to match the value displayed on the graph. The best-fit line visually crosses around approximately 1 on the y-axis, but the displayed value is approximately -0.21 from the equation. Is there a way to correct this discrepancy to match the displayed value, or is there something else happening that I am unaware of

thanks

% Example data (replace with your actual data)
xData = [1, 2, 3, 4, 5, 6, 7];
yData = [1.1, 1.5, 4, 3.8, 6, 6.2, 8];

% Fit a linear model to the data
mdl = fitlm(xData, yData);

% Plotting the data and the best fit line
figure;
scatter(xData, yData, 'filled', 'o'); % Scatter plot of data points
hold on;
plot(mdl); % Plotting the best fit line
hold off;
xlabel('X Data');
ylabel('Y Data');
title('Best Fit Line');
grid on;

% Display the equation of the line
disp(['Equation of the best fit line: y = ' num2str(mdl.Coefficients.Estimate(2)) 'x + ' num2str(mdl.Coefficients.Estimate(1))]);

[Stephen23]

" The best-fit line visually crosses around approximately 1 on the y-axis"

Yes, but at x=1, so it has nothing to do with the y-intercept (which is defined as being at x=0).

"...but the displayed value is approximately -0.21 from the equation."

Which seems about correct.

"Is there a way to correct this discrepancy to match the displayed value"

The only discrepancy that I can see is that you are confusing x=1 for x=0.

"or is there something else happening that I am unaware of"

Plot the line of best fit at x=0 and see what value it has.

[Sarvesh]

Silly overlook from my end! thanks @Stephen23

Answer 1

[Sam Chak]

Extend the line, and you will see the y-intercept at .

% Example data (replace with your actual data)
xData = [1, 2, 3, 4, 5, 6, 7];
yData = [1.1, 1.5, 4, 3.8, 6, 6.2, 8];

% Fit a linear model to the data
mdl = fitlm(xData, yData)

mdl = 
Linear regression model:
    y ~ 1 + x1

Estimated Coefficients:
                   Estimate      SE        tStat        pValue  
                   ________    _______    ________    __________

    (Intercept)    -0.21429    0.50341    -0.42567       0.68805
    x1               1.1464    0.11257      10.185    0.00015658


Number of observations: 7, Error degrees of freedom: 5
Root Mean Squared Error: 0.596
R-squared: 0.954,  Adjusted R-Squared: 0.945
F-statistic vs. constant model: 104, p-value = 0.000157

% Plotting the data and the best fit line
figure;
scatter(xData, yData, 'filled', 'o'); % Scatter plot of data points
hold on;
plot(mdl); % Plotting the best fit line
xlabel('X Data');
ylabel('Y Data');
title('Best Fit Line');
grid on;

%% ---- plot the extension ----
m   = mdl.Coefficients.Estimate(2);
c   = mdl.Coefficients.Estimate(1);
x   = linspace(0, 1, 101);
y   = m*x + c;
plot(x, y, '--')
xlim([0 7])
hold off;

% Display the equation of the line
disp(['Equation of the best fit line: y = ' num2str(mdl.Coefficients.Estimate(2)) 'x + ' num2str(mdl.Coefficients.Estimate(1))]);

Equation of the best fit line: y = 1.1464x + -0.21429

[Sarvesh]

thanks @Sam Chak. just realized the mistake a while ago.