Solution

import scipy.stats
t = ((100-105)-(0))/(5**2/50+10**2/75)**.5
print(t)
print(scipy.stats.norm.cdf(t)*2)
print(scipy.stats.norm.cdf(t))

If we instead don’t know the standard deviations, the equation becomes….

When we compute this t value, the degrees of freedom equals min(N

₁

-1,N

₂

-1). We only need to do this with low sample sizes, otherwise the t score converges to the z score.

Let’s do an example…

dist1 = scipy.stats.norm(300,100)
dist2 = scipy.stats.norm(400,100)

pts1 = dist1.rvs(25,random_state=1)
pts2 = dist2.rvs(26,random_state=2)

So now we have two random samples, let’s get our t score.

import numpy as np
sd1 = np.std(pts1)
sd2 = np.std(pts2)
mean1 = np.mean(pts1)
mean2 = np.mean(pts2)
t = ((mean1-mean2)-(0))/(sd1**2/25+sd2**2/26)**.5
print(t)
print(scipy.stats.t.cdf(t,25-1)*2)

The chance that the two distributions are the same is small! We would expect this though, we set up two distributions that have a difference in means of 100! What if we had a hunch that our samples should have a difference of 100? This would meanH

₀

: μ

₁

+100 = μ

₂

.

import numpy as np
sd1 = np.std(pts1)
sd2 = np.std(pts2)
mean1 = np.mean(pts1)
mean2 = np.mean(pts2)
t = ((mean1-mean2)-(-40))/(sd1**2/25+sd2**2/26)**.5
print(t)
print(scipy.stats.t.cdf(t,25-1)*2)

Now, we do not have a significant value!