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I assume the plots follow common frequentist methodology. The numbers of sigma are found by Building a likelihood function $$\mathcal{L}(x) \equiv \max_\text{other params} p(D|x)$$ where $D$ is the data. NB that this is not a PDF of the variable $x$. The PDF $p(x|D)$ is never calculated in frequentist methods. Finding a test-statistic \lambda = -2\ln ... 1 I think what you have described is not the correct method to estimate a 68% confidence interval in one parameter of interest. The error is in freezing the other parameter when minimising chi-squared. A better procedure to evaluate the uncertainty in parameter 1, is to evaluate the minimum chi-squared for a set of values of parameter 1, whilst allowing ... 1 In the case of a 2-parameter fit, the 68% confidence space is typically an ellipse, not necessarily aligned with the axes. If you want to figure out the size of the ellipse, you should find the orientation of the ellipse, not just the intercepts of the ellipse with a horizontal and vertical line through its center. Example: fity=a+b x$for a large set of ... 0 if you mean the statistical methods used to evaluate the significance in scientific papers. I should say there is no differences in methods, only the sample sizes and significance value can change depending on the subject. 2 During the acceleration phase the object's movement can be modeled with the quadratic curve $$x=x_0 + v_0t+\frac{1}{2}at^2 \qquad\text{where } x_0 \text{ is the initial position, and }v_0 \text{ is the initial velocity}$$ During the constant velocity phase, the object's movement can be modeled with the linear equation $$x=x_1 + v_1(t-t_1)$$ where$x_1$, ... 0 The average value of acceleration should just depend on the initial and final velocity and the time interval between them. Since the average value of a function over the interval a to b is the integral of the function from a to b divided by (b-a), and since the integral of acceleration gives you velocity then if the limits are$t_1$and$t_2\$ the average ...