How to use t.ppf()? which are the arguments?

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I couldn't understand how to properly use t.ppf, could someone please explain it to me?

I have to use this information

  • scipy.stats.t
  • scipy.stats
  • a mean of 100
  • a standard deviation of 0.39
  • N = 851 (851 samples)

When I'm asked to calculate the (95%) margin of error using t.ppf() will the code look like below?

cutoff1 = t.ppf(0.05,100,0.36,850)

Can somebody help me, please?

distribution python
1 Answers

According to the reference docs, the arguments to t.ppf are q, df, loc, and scale. The df argument is degrees of freedom, which is usually the sample size minus 1 for a single population sampling problem. Since ppf calculates the inverse cumulative distribution function, by definition a result of x for a given q-value and df means P{T <= x} = q, i.e, there is probability q of getting outcomes less than or equal to x from a T distribution with the given loc and scale. The loc (mean) and scale (standard deviation) arguments are optional, and default to 0 and 1, respectively.

To get a 95% margin of error, you want 5% of the probability to be in the tails of the distribution. This is usually done symmetrically so that 2.5% is in each tail, so you would use q values of 0.025 and 0.975 for the lower and upper cutoff points respectively. For your particular problem, the code would look something like:

from scipy.stats import t

n = 851
mean = 100
std_dev = 0.39

lower_cutoff = t.ppf(0.025, n - 1, loc = mean, scale = std_dev)  # =>  99.23452406698323
upper_cutoff = t.ppf(0.975, n - 1, loc = mean, scale = std_dev)  # => 100.76547593301677
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