# How to do a hypothesis test for the mean with known standard deviation

## Description

Let’s say we are measuring a variable over a population, and we know its standard deviation $\sigma$ is known, and assume that the variable is normally distributed. We take a sample, $x_1, x_2, x_3, \ldots, x_k$, and compute its mean $\bar{x}$. We want to determine if the sample mean is significantly different from, greater than, or less than some hypothesized value, such as a hypothesized population mean. How do we formulate an appropriate hypothesis test?

Related tasks:

- How to compute a confidence interval for a population mean
- How to do a hypothesis test for a mean difference (matched pairs)
- How to do a hypothesis test for a population proportion
- How to do a hypothesis test for population variance
- How to do a hypothesis test for the difference between means when both population variances are known
- How to do a hypothesis test for the difference between two proportions
- How to do a hypothesis test for the ratio of two population variances
- How to do a hypothesis test of a coefficient’s significance
- How to do a one-sided hypothesis test for two sample means
- How to do a two-sided hypothesis test for a sample mean
- How to do a two-sided hypothesis test for two sample means

## Using SciPy, in Python

We will use the following fake data, but you can insert your actual data in its place. We have a sample of just 5 values and an assumed population standard deviation of 3.

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sample = [31, 44, 28, 25, 40] # sample data
pop_std = 3 # population standard deviation

We also choose a value $0 \le \alpha \le 1$ as our Type I error rate. We’ll let $\alpha$ be 0.05 here, but you can change that in the code below.

### Two-tailed test

In a two-tailed test, we compare the unknown population mean to a hypothesized value $m$ using the null hypothesis $H_0: \mu=m$. Here we’ll use $m=30$, but you can replace that value in the code below with the hypothesis relevant for your situation.

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from scipy import stats
import numpy as np
m = 30 # hypothesized mean
n = len(sample) # number of observations
xbar = np.mean(sample) # sample mean
test_stat = (xbar - m) / (pop_std/np.sqrt(n)) # test statistic
2*stats.norm.sf(abs(test_stat)) # two-tailed p-value

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0.007290358091535614

The $p$-value, 0.00729, is less than $\alpha$, so we have evidence to reject the null hypothesis and conclude that the actual population mean $\mu$ is significantly different from the hypothesized value of $m=30$.

### Right-tailed test

In a right-tailed hypothesis test, the null hypothesis is that the population mean is greater than or equal to a chosen value, $H_0: \mu \ge m$.

Most of the code below is the same as above, but we repeat it to make it easy to copy and paste. Only the computation of the $p$-value changes.

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from scipy import stats
import numpy as np
m = 30 # hypothesized mean
n = len(sample) # number of observations
xbar = np.mean(sample) # sample mean
test_stat = (xbar - m) / (pop_std/np.sqrt(n)) # test statistic
stats.norm.sf(abs(test_stat)) # right-tailed p-value

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0.003645179045767807

The $p$-value, 0.003645, is less than $\alpha$, so we have evidence to reject the null hypothesis and conclude that the actual population mean $\mu$ is significantly less than the hypothesized value of $m=30$.

### Left-tailed test

In a left-tailed hypothesis test, the null hypothesis is that the population mean is less than or equal to a chosen value, $H_0: \mu \le m$.

Most of the code below is the same as above, but we repeat it to make it easy to copy and paste. Only the computation of the $p$-value changes.

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from scipy import stats
import numpy as np
m = 30 # hypothesized mean
n = len(sample) # number of observations
xbar = np.mean(sample) # sample mean
test_stat = (xbar - m) / (pop_std/np.sqrt(n)) # test statistic
stats.norm.sf(-abs(test_stat)) # left-tailed p-value

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0.9963548209542322

The $p$-value, 0.99635, is greater than $\alpha$, so wwe do not have sufficient evidence to conclude that $\mu>m$ and should continue to accept the null hypothesis, that $\mu\le m$.

Content last modified on 24 July 2023.

See a problem? Tell us or edit the source.

## Solution, in R

We will use the following fake data, but you can insert your actual data in its place. We have a sample of just 5 values and an assumed population standard deviation of 3.

1
2

sample <- c(31, 44, 28, 25, 40) # sample data
pop.std <- 3 # population standard deviation

We also choose a value $0 \le \alpha \le 1$ as our Type I error rate. We’ll let $\alpha$ be 0.05 here, but you can change that in the code below.

### Two-tailed test

In a two-tailed test, we compare the unknown population mean to a hypothesized value $m$ using the null hypothesis $H_0: \mu=m$. Here we’ll use $m=30$, but you can replace that value in the code below with the hypothesis relevant for your situation.

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m <- 30 # hypothesized mean
n <- length(sample) # number of observations
xbar <- mean(sample) # sample mean
test.stat <- (xbar - m) / (pop.std/sqrt(n)) # test statistic
2*pnorm(abs(test.stat), 0, 1, lower.tail=FALSE) # two-tailed p-value

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[1] 0.007290358

The $p$-value, 0.00729, is less than $\alpha$, so we have evidence to reject the null hypothesis and conclude that the actual population mean $\mu$ is significantly different from the hypothesized value of $m=30$.

### Right-tailed test

In a right-tailed hypothesis test, the null hypothesis is that the population mean is greater than or equal to a chosen value, $H_0: \mu \ge m$.

Most of the code below is the same as above, but we repeat it to make it easy to copy and paste. Only the computation of the $p$-value changes.

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m <- 30 # hypothesized mean
n <- length(sample) # number of observations
xbar <- mean(sample) # sample mean
test.stat <- (xbar - m) / (pop.std/sqrt(n)) # test statistic
pnorm(test.stat, 0, 1, lower.tail=FALSE) # right-tailed p-value

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[1] 0.003645179

The $p$-value, 0.003645, is less than $\alpha$, so we have evidence to reject the null hypothesis and conclude that the actual population mean $\mu$ is significantly less than the hypothesized value of $m=30$.

### Left-tailed test

In a left-tailed hypothesis test, the null hypothesis is that the population mean is less than or equal to a chosen value, $H_0: \mu \le m$.

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m <- 30 # hypothesized mean
n <- length(sample) # number of observations
xbar <- mean(sample) # sample mean
test.stat <- (xbar - m) / (pop.std/sqrt(n)) # test statistic
pnorm(test.stat, 0, 1, lower.tail=TRUE) # left-tailed p-value

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[1] 0.9963548

The $p$-value, 0.99635, is greater than $\alpha$, so wwe do not have sufficient evidence to conclude that $\mu>m$ and should continue to accept the null hypothesis, that $\mu\le m$.

Content last modified on 24 July 2023.

See a problem? Tell us or edit the source.

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