## 27.4 The test statistic and \(t\)-scores: One mean

The sampling distributions
describes what to **expect** from the sample mean,
**assuming** \(\mu = 37.0^\circ\text{C}\).
The value of \(\bar{x}\) that is **observed**,
however,
is \(\bar{x}=36.8051^\circ\)
How likely is it that such a value could occur by chance?

The value of the observed sample mean can be located the picture of the sampling distribution (Fig. 27.3). The value \(\bar{x}=36.8051^\circ\text{C}\) is unusually small. About how many standard deviations is \(\bar{x}\) away from \(\mu=37\)? A lot…

Relatively speaking,
the *distance* that the observed sample mean (of \(\bar{x}=36.8051\))
is from
the mean of the sampling distribution
(Fig. 27.3).
is found by computing *how many* standard deviations
the value of \(\bar{x}\) is from the mean of the distribution;
that is, computing something like a \(z\)-score.
(Remember that the standard deviation in
Fig. 27.3
is the the *standard error*: the amount of variation in the sample means.)

Since the mean and standard deviation (i.e., the *standard error*)
of this
normal distribution are known,
the number of standard deviations
that \(\bar{x} = 36.8051\) is from the mean is

\[
\frac{36.8051 - 37.0}{0.035724} = -5.453.
\]
This value is like a \(z\)-score.
However, this is actually called a \(t\)-score
because it has been computed when
*the population standard deviation is unknown*,
and the best estimate (the *sample* standard deviation) is used
when \(\text{s.e.}(\bar{x})\) was computed.

Both \(t\) and \(z\) scores measure
*the number of standard deviations
that an observation is from the mean*:
\(z\)-scores use \(\sigma\) and \(t\)-scores use \(s\).
Here,
the distribution of the *sample statistic* is relevant,
so the appropriate standard deviation is the standard deviation
of the sampling distribution: the *standard error*.

Like \(z\)-scores, \(t\)-scores measure the
number of standard deviations that an observation is from the mean.
\(z\)-scores are calculated using the *population* standard deviation,
and \(t\)-scores are calculated using the *sample* standard deviation.

In hypothesis testing, \(t\)-scores are more commonly used than \(z\)-scores, because almost always the population standard deviation is unknown, and the sample standard deviation is used instead.

In this course, it is sufficient to think of \(z\)-scores and \(t\)-scores as approximately the same. Unless sample sizes are small, this is a reasonable approximation.So the calculation is:

\[
t = \frac{36.8051 - 37.0}{0.035724} = -5.453;
\]
the observed sample mean
is *more than five standard deviation below the population mean*.
This is *highly* unusual
based on the
68–95–99.7 rule,
as seen in
Fig. 27.3.

In general, a \(t\)-score in hypothesis testing is

\[\begin{equation} t = \frac{\text{sample statistic} - \text{assumed population parameter}} {\text{standard error of the sample statistic}}. \tag{27.1} \end{equation}\]