t-Test for One Sample

The $t$-test is the most powerful parametric test for calculating the significance of a small sample mean. A one sample $t$-test has the null hypothesis, or $H_0$, that the population mean equals the hypothesized value. Expressed formally:

$H_0: \, \mu = c$

where the Greek letter $\mu$ represents the population mean and $c$ represents its assumed (hypothesized) value. The $t$-test is the small sample analog of the $z$-test, which is suitable for large samples. A small sample is generally regarded as one of size $n < 30$.

In order to perform a $t$-test, one first has to calculate the degrees of freedom. This quantity takes into account the sample size and the number of parameters that are being estimated. Here, the population parameter $\mu$ is being estimated by the sample statistic $\bar { X }$, the mean of the sample data. For a $t$-test the degrees of freedom of the single mean is $n-1$. This is because only one population parameter (the population mean) is being estimated by a sample statistic (the sample mean).

Example

A college professor wants to compare her students' scores with the national average. She chooses a simple random sample of $20$ students who score an average of $50.2$ on a standardized test. Their scores have a standard deviation of $2.5$. The national average on the test is a $60$. She wants to know if her students scored significantly lower than the national average.

1. First, state the problem in terms of a distribution and identify the parameters of interest. Mention the sample. We will assume that the scores ( $\bar{X}$) of the students in the professor's class are approximately normally distributed with unknown parameters $\mu$ and $\sigma$.

2. State the hypotheses in symbols and words:

${ H }_{ 0 }:\, \mu =60$

i.e.: The null hypothesis is that her students scored on par with the national average.

${ H }_{ A }:\, \mu <60$

i.e.: The alternative hypothesis is that her students scored lower than the national average.

3. Identify the appropriate test to use. Since we have a simple random sample of small size and do not know the standard deviation of the population, we will use a one-sample $t$-test. The formula for the $t$-statistic $T$ for a one-sample test is as follows:

$T=\dfrac { \bar { X } -60 }{ S/\sqrt { 20 } }$,

where $\bar { X }$ is the sample mean and $S$ is the sample standard deviation. The standard deviation of the sample divided by the square root of the sample size is known as the "standard error" of the sample.

4. State the distribution of the test statistic under the null hypothesis. Under $H_0$ the statistic $T$ will follow a Student's distribution with $19$ degrees of freedom: $T\sim \tau \cdot (20-1)$.

5. Compute the observed value $t$ of the test statistic $T$, by entering the values, as follows:

$t=\dfrac { \bar { x } -60 }{ s/\sqrt { 20 } } =\dfrac { 50.2-60 }{ 2.5/\sqrt { 20 } } =\dfrac { -9.8 }{ 0.559 } =-17.5$

6. Determine the so-called $p$-value of the value $t$ of the test statistic $T$. We will reject the null hypothesis for too-small values of $T$, so we compute the left $p$-value:

$p = P\left( T\le t;{ H }_{ 0 } \right) =P\left( T\left( 19 \right) \le -17.5 \right) \approx 0$

The Student's distribution gives $T\left( 19 \right) =1.729$ at probabilities $0.95$ and degrees of freedom $19$. The $p$-value is approximated at $1.777$.

7. Lastly, interpret the results in the context of the problem. The $p$-value indicates that the results almost certainly did not happen by chance and we have sufficient evidence to reject the null hypothesis. This is to say, the professor's students did score significantly lower than the national average.