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, $x_1,x_2,x_3,\dots ,x_k$, and compute its mean $\overline{x}$. 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

View this solution alone.

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

View this solution alone.

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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Copy
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, $H_0:\mu \le m$.

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=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.

Topics that include this task

• Bentley University MA214

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