## 31.4 The test statistic: Comparing odds

The decision-making process
compares what is *expected* from the sample statistic
if the null hypothesis about the parameter is true
(Table 31.3)
to what is **observe** in the sample
(Table 31.1).
Previously,
when the summary statistics were means,
\(t\)-tests were used.
However,
these data are not summarised by means,
and a different test statistic is used.

Rather than using a \(t\)-score as
the *test-statistic*,
the test-statistic here is a ‘chi-squared’ statistic,
written \(\chi^2\).
A \(\chi^2\) statistic measures the overall size
of the differences between the expected counts
and observed counts, over the entire table.

The Greek letter \(\chi\) is pronounced ‘ki,’ as in **ki**te.

From the software
(jamovi: Fig. 31.1;
SPSS: Fig. 31.2),
\(\chi^2=6.934\).
In a \(2\times 2\) table of counts
(when the ‘degrees of freedom,’ or `df`

, is equal to 1,
as shown in the computer output),
the *square root* of the \(\chi^2\) value is
approximately equivalent to a \(z\)-score.
So here,
the equivalent \(z\)-score is about \(\sqrt{6.934} = 2.63\),
which is fairly large:
a small \(P\)-value is expected.

More generally, for two-way tables of any size,

\[
\sqrt{\frac{\chi^2}{\text{df}}}
\]
is like a \(z\)-score,
where df is the ‘degrees of freedom’
(related to the size of the table^{14}),
as shown in the software output.
This allows a \(P\)-value to be estimated using the
68–95–99.7 rule
from the value of the \(\chi^2\) statistic.

`df`

in the software output),
the value of
\[
\sqrt{\frac{\chi^2}{\text{df}}}
\]
is like a \(z\)-score.
This allows the \(P\)-value to be estimated using the
68–95–99.7 rule.
For those who want to know: the degrees of freedom in a two-way table is the number of rows of data minus one, times the number of columns of data minus one. So, for a two-way table, the degrees of freedom is \((2-1) \times (2-1) = 1\).↩︎