There are several ways to accomplish what you asking for: Using only matplotlib from matplotlib import pyplot as plt import numpy as np #some example data x = np.linspace(0.1, 9.9, 20) y = 3.0 * x #some confidence interval ci = 1.96 * np.std(y)/np.sqrt(len(x)) fig, ax = plt.subplots() ax.plot(x,y) ax.fill_between(x, (y-ci), (y+ci), color=”b”, alpha=.1) fill_between … Read more
Have you tried scipy? You will need to installl the scipy library…more about installing it here: http://www.scipy.org/install.html Once installed, you can replicate the Excel functionality like such: from scipy import stats #Studnt, n=999, p<0.05, 2-tail #equivalent to Excel TINV(0.05,999) print stats.t.ppf(1-0.025, 999) #Studnt, n=999, p<0.05%, Single tail #equivalent to Excel TINV(2*0.05,999) print stats.t.ppf(1-0.05, 999) You … Read more
You can bootstrap the ROC computations (sample with replacement new versions of y_true / y_pred out of the original y_true / y_pred and recompute a new value for roc_curve each time) and the estimate a confidence interval this way. To take the variability induced by the train test split into account, you can also use … Read more
The 68% confidence interval for a single draw from a normal distribution with mean mu and std deviation sigma is stats.norm.interval(0.68, loc=mu, scale=sigma) The 68% confidence interval for the mean of N draws from a normal distribution with mean mu and std deviation sigma is stats.norm.interval(0.68, loc=mu, scale=sigma/sqrt(N)) Intuitively, these formulas make sense, since if … Read more
The usual way is to compute a confidence interval on the scale of the linear predictor, where things will be more normal (Gaussian) and then apply the inverse of the link function to map the confidence interval from the linear predictor scale to the response scale. To do this you need two things; call predict() … Read more
import numpy as np import scipy.stats def mean_confidence_interval(data, confidence=0.95): a = 1.0 * np.array(data) n = len(a) m, se = np.mean(a), scipy.stats.sem(a) h = se * scipy.stats.t.ppf((1 + confidence) / 2., n-1) return m, m-h, m+h You can calculate like this.