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Permutation test on frog data

The average strike force of Frog A was 0.71 Newtons (N), and that of Frog B was 0.42 N for a difference of 0.29 N. It is possible the frogs strike with the same force and this observed difference was by chance. You will compute the probability of getting at least a 0.29 N difference in mean strike force under the hypothesis that the distributions of strike forces for the two frogs are identical. We use a permutation test with a test statistic of the difference of means to test this hypothesis.

For your convenience, the data has been stored in the arrays force_a and force_b.

Este ejercicio forma parte del curso

Statistical Thinking in Python (Part 2)

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Instrucciones de ejercicio

  • Define a function with call signature diff_of_means(data_1, data_2) that returns the differences in means between two data sets, mean of data_1 minus mean of data_2.
  • Use this function to compute the empirical difference of means that was observed in the frogs.
  • Draw 10,000 permutation replicates of the difference of means.
  • Compute the p-value.
  • Print the p-value.

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def diff_of_means(data_1, data_2):
    """Difference in means of two arrays."""

    # The difference of means of data_1, data_2: diff
    diff = ____

    return diff

# Compute difference of mean impact force from experiment: empirical_diff_means
empirical_diff_means = ____

# Draw 10,000 permutation replicates: perm_replicates
perm_replicates = draw_perm_reps(____, ____,
                                 ____, size=10000)

# Compute p-value: p
p = np.sum(____ >= ____) / len(____)

# Print the result
print('p-value =', p)
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