How to do a two-sided hypothesis test for a sample mean

Description

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

Related tasks:

• How to compute a confidence interval for a population mean
• How to do a two-sided hypothesis test for two sample means
• How to do a one-sided hypothesis test for two sample means
• How to do a hypothesis test for a mean difference (matched pairs)
• How to do a hypothesis test for a population proportion

Solution, in Julia

View this solution alone.

This is a two-sided test with the null hypothesis $H_0:\mu =\overline{x}$. We choose a value $0\le \alpha \le 1$ 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, $\mu =\overline{x}$.

When you are using the most common value for $\alpha$, which is $0.05$ for the $95\mathrm{\%}$ 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.

See a problem? Tell us or edit the source.

Using SciPy, in Python

View this solution alone.

This is a two-sided test with the null hypothesis $H_0:\mu =\overline{x}$. We choose a value $0\le \alpha \le 1$ 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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from scipy import stats

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

# Run a one-sample t-test and print out alpha, the p value,
# and whether the comparison says to reject the null hypothesis.
t_statistic, p_value = stats.ttest_1samp( 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 sample does not give us enough information to reject the null hypothesis. We would continue to assume that the sample is like the population, $\mu =\overline{x}$.

Content last modified on 24 July 2023.

See a problem? Tell us or edit the source.

Solution, in R

View this solution alone.

This is a two-sided test with the null hypothesis $H_0:\mu =\overline{x}$. We choose a value $0\le \alpha \le 1$ 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 <- c( 9, 12, 14, 8, 13 )

# Run a one-sample t-test and print out alpha, the p value,
# and whether the comparison says to reject the null hypothesis.
t.test( sample, mu=pop.mean, conf.level=1-alpha )

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	One Sample t-test

data:  sample
t = 1.0366, df = 4, p-value = 0.3585
alternative hypothesis: true mean is not equal to 10
95 percent confidence interval:
  7.986032 14.413968
sample estimates:
mean of x 
     11.2 

Although we can deduce the answer to our question from the above output, by comparing the $p$ value with $\alpha$ manually, we can also ask R to do it.

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# Is there enough evidence to reject the null hypothesis?
result <- t.test( sample, mu=pop.mean, conf.level=1-alpha )
result$p.value < alpha

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[1] FALSE

In this case, 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, $\mu =\overline{x}$.

Content last modified on 24 July 2023.

See a problem? Tell us or edit the source.

Topics that include this task

• Bentley University GB213
• Bentley University GR521

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