How to do a hypothesis test for the mean with known standard deviation (in R)

Task

Let’s say we are measuring a variable over a population, and we know its standard deviation $σ$ 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 $\bar{x}$ . 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?

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Solution

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.

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

We also choose a value $0 \leq α \leq 1$ as our Type I error rate. We’ll let $α$ 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 $H_{0} : μ = m$ . Here we’ll use $m = 30$ , but you can replace that value in the code below with the hypothesis relevant for your situation.

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

[1] 0.007290358

The $p$ -value, 0.00729, is less than $α$ , so we have evidence to reject the null hypothesis and conclude that the actual population mean $μ$ 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, $H_{0} : μ \geq m$ .

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.

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

[1] 0.003645179

The $p$ -value, 0.003645, is less than $α$ , so we have evidence to reject the null hypothesis and conclude that the actual population mean $μ$ 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, $H_{0} : μ \leq m$ .

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.

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

[1] 0.9963548

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

Content last modified on 24 July 2023.

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Contributed by:

Elizabeth Czarniak (CZARNIA_ELIZ@bentley.edu)
Nathan Carter (ncarter@bentley.edu)