CODEWITHSUNDEEP
scipystatisticsconfidence-interval
Eric Kim
Eric Kim

Reputation: 2698

Confidence interval of mean - scipy implementation doesn't agree with mathematic formula

Confidence interval of mean has the following analytical solution:

enter image description here

Assuming that my dataset is normally distributed, and I do not know the population standard deviation, I can use t-score to compute the CI of mean. So I did:

from scipy import stats
import numpy as np

arr = np.array([4, 4, 1, 6, 6, 8, 1, 2, 3, 2, 2, 3, 4, 7, 6, 8, 0, 2, 8, 6, 5])

alpha = 0.05                       # significance level = 5%
df = len(arr) - 1                  # degress of freedom = 20
t = stats.t.ppf(1 - alpha/2, df)   # 95% confidence t-score = 2.086
s = np.std(arr, ddof=1)            # sample standard deviation = 2.502
n = len(arr)

lower = np.mean(arr) - (t * s / np.sqrt(n))
upper = np.mean(arr) + (t * s / np.sqrt(n))

print((lower, upper))
>>> (3.0514065531195387, 5.329545827832843)

print(stats.t.interval(1 - alpha/2, df, loc=np.mean(arr), scale=s / np.sqrt(n)))
>>> (2.8672993716475763, 5.513653009304806)

And the interval I manually calculated using the equation does not agree with the scipy implementation of the CI. Where is this error coming from?

Upvotes: 1

Views: 1091

Answers (1)

Warren Weckesser
Warren Weckesser

Reputation: 114831

Your signifance level is 0.05, so the confidence level is 0.95. Pass that value to stats.t.interval. Don't divide by 2; the function does that for you:

In [62]: print(stats.t.interval(1 - alpha, df, loc=np.mean(arr), scale=s / np.sqrt(n)))              
(3.0514065531195387, 5.329545827832843)

Upvotes: 3

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