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How to do a two-sided hypothesis test for a sample mean (in Julia)

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Task

Say we have a population whose mean $\mu$ is known. We take a sample $x_1,\ldots,x_n$ and compute its mean, $\bar x$. We then ask whether this sample is significantly different from the population at large, that is, is $\mu=\bar x$?

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Solution

This is a two-sided test with the null hypothesis $H_0:\mu=\bar x$. We choose a value $0\leq\alpha\leq1$ as the probability of a Type I error (false positive, finding we should reject $H_0$ when it’s actually true).

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# Replace these first three lines with the values from your situation.
alpha = 0.05
pop_mean = 10
sample = [ 9, 12, 14, 8, 13 ]

# The following code runs the test for your chosen alpha:
using HypothesisTests
p_value = pvalue( OneSampleTTest( sample, pop_mean ) )
reject_H0 = p_value < alpha
alpha, p_value, reject_H0
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(0.05, 0.35845634462296455, false)

In this case, the $p$-value was larger than $\alpha$, so the sample does not give us enough information to reject the null hypothesis. We would continue to assume that the sample is like the population, $\mu=\bar x$.

When you are using the most common value for $\alpha$, which is $0.05$ for the $95\%$ confidence interval, you can simply print out the test itself and get a detailed printout with all the information you need, thus saving a few lines of code.

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OneSampleTTest( sample, pop_mean )
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One sample t-test
-----------------
Population details:
    parameter of interest:   Mean
    value under h_0:         10
    point estimate:          11.2
    95% confidence interval: (7.986, 14.41)

Test summary:
    outcome with 95% confidence: fail to reject h_0
    two-sided p-value:           0.3585

Details:
    number of observations:   5
    t-statistic:              1.0366421106976316
    degrees of freedom:       4
    empirical standard error: 1.1575836902790224

Content last modified on 24 July 2023.

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Contributed by Nathan Carter (ncarter@bentley.edu)