96 Statistical Test of the Population Mean with unknown Variance

96.1 Theory

96.1.1 Case 1 (denominator \(n\) convention)

Assume that

\[ \begin{cases} U = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \sim \text{N}(0,1) \\ V = \frac{ns^2}{\sigma^2} \sim \chi_{n-1}^2\end{cases} \]

where

\[ \bar{x} \sim \text{N} \left( \mu, \frac{\sigma^2}{n} \right) \]

and

\[ \begin{cases} s^2 = \frac{\sum_{i=1}^{n}\left( x_i - \bar{x} \right)^2}{n} \\ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i\end{cases} \]

and assume that \(U\) and \(V\) are independent.

Since the t-density is defined as

\[ \frac{U}{\sqrt{\frac{V}{n-1}}} = \frac{\text{N}(0,1)}{\sqrt{\frac{\chi_{n-1}^2}{n-1}}} \sim t_{n-1} \]

it follows that

\[ \begin{align*}\frac{\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}}{\sqrt{\frac{\frac{ns^2}{\sigma^2}}{n-1}}} &= \frac{\bar{x} -\mu}{\frac{\sigma}{\sqrt{n}}} \times \frac{1}{\frac{s}{\sigma}\sqrt{\frac{n}{n-1}}} \\&= \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n-1}}} \sim t_{n-1}\end{align*} \]

96.1.2 Case 2 (standard unbiased sample variance)

Case 2 uses the usual unbiased estimator of the variance (denominator \(n-1\)), which is the convention used in most textbooks and software (including t.test() in R).

Assume that

\[ \begin{cases}U = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \sim \text{N}(0,1) \\V = \frac{(n-1)s^2}{\sigma^2} \sim \chi_{n-1}^2\end{cases} \]

where

\[ \bar{x} \sim \text{N} \left( \mu, \frac{\sigma^2}{n} \right) \]

and

\[ \begin{cases}s^2 = \frac{\sum_{i=1}^{n}\left( x_i - \bar{x} \right)^2}{n-1} \\\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i\end{cases} \]

and assume that \(U\) and \(V\) are independent.

Since the t-density is defined as

\[ \frac{U}{\sqrt{\frac{V}{n-1}}} = \frac{\text{N}(0,1)}{\sqrt{\frac{\chi_{n-1}^2}{n-1}}} \sim t_{n-1} \]

it follows that

\[ \begin{align*}\frac{\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}}{\sqrt{\frac{\frac{(n-1)s^2}{\sigma^2}}{n-1}}} &= \frac{\bar{x} -\mu}{\frac{\sigma}{\sqrt{n}}} \times \frac{1}{\frac{s}{\sigma}} \\&= \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \sim t_{n-1}\end{align*} \]

96.2 Software

There are two R modules available to preform the one sample t-test. These are the URLs on the public website:

One Sample T-Test using Confidence Intervals:
https://compute.wessa.net/rwasp_hypothesismeanu.wasp
One Sample T-Test using p-values:
https://compute.wessa.net/rwasp_onesampletests_mean.wasp

In RFC these R modules can be found under the “Hypotheses / Empirical Tests” menu item.

To compute the One Sample T-Test on your local machine, the following script can be used in the R console:

set.seed(123)
x <- runif(25, 20, 40)
# compute confidence intervals
par1 = 0.95 #Confidence
par2 = 30 #Null Hypothesis
len <- length(x)
df <- len - 1
sd <- sd(x)
mx <- mean(x)
delta2 <- abs(qt((1-par1)/2,df)) * sd / sqrt(len)
delta1 <- abs(qt((1-par1),df)) * sd / sqrt(len)
#Sample size
len
#Sample standard deviation
sd
#Sample Mean
mx
#2-sided Confidence Interval
dum <- paste('[',mx-delta2)
dum <- paste(dum,',')
dum <- paste(dum,mx+delta2)
dum <- paste(dum,']')
dum
#Left-sided Confidence Interval
dum <- paste('[',mx-delta1)
dum <- paste(dum,', +inf ]')
dum
#Right-sided Confidence Interval
dum <- paste('[ -inf,',mx+delta1)
dum <- paste(dum,']')
dum
# compute two-sided interval and p-value
par1 = 'two.sided'
par2 = 0.95 #Confidence
par3 = 20 #Null Hypothesis
(tt <- t.test(x,mu=par3,alternative=par1,conf.level=par2))

[1] 25
[1] 6.010489
[1] 31.91119
[1] "[ 29.4301862828312 , 34.3922024969133 ]"
[1] "[ 29.8545466508001 , +inf ]"
[1] "[ -inf, 33.9678421289444 ]"

    One Sample t-test

data:  x
t = 9.9087, df = 24, p-value = 5.878e-10
alternative hypothesis: true mean is not equal to 20
95 percent confidence interval:
 29.43019 34.39220
sample estimates:
mean of x 
 31.91119

96.3 Practical Example

A sample of intrinsic motivation scores was obtained from students in a statistics course. We wish to test the following hypothesis:

\[ \begin{cases}\text{H}_0: \mu = 19.8 \\\text{H}_A: \mu \neq 19.8\end{cases} \]

The sample data and computational results are available in the R module shown below. Do we have to accept or reject the Null Hypothesis if we choose a type I error of 3%?

This is a two-sided test because the alternative is pre-specified as \(\text{H}_A:\mu \neq 19.8\).

Interactive Shiny app (click to load).

Open in new tab

Answer: the sample mean \(\bar{x} \simeq 20.06778\) is significantly different from \(\mu_0 = 19.8\) because the p-value is \(0.02471 < 0.03\).