This test can also be used for continuous data if the data are first summarised in a frequency table. The range of possible values of the distribution is partitioned into classes such as "10 ≤ X < 11", then the frequencies are the numbers of values in these classes.

Body temperature

In a study to determine the "normal" body temperature of healthy adults, body temperatures were found from 130 adults. The following frequency table summarises the data.

Temperature, x Frequency
\(X \lt 96.0\) 0
\(96.0 \le X \lt 96.5\) 2
\(96.5 \le X \lt 97.0\) 4
\(97.0 \le X \lt 97.5\) 13
\(97.5 \le X \lt 98.0\) 21
\(98.0 \le X \lt 98.5\) 38
\(98.5 \le X \lt 99.0\) 33
\(99.0 \le X \lt 99.5\) 15
\(99.5 \le X \lt 100.0\) 2
\(100.0 \le X \lt 100.5\) 1
\(100.5 \le X \lt 101.0\) 1
\(X \ge 101.0\) 0

Could this be a random sample from a normal distribution?

We first find the method of moments estimates of the normal distribution's parameters,

\[ \hat{\mu} \;=\; \overline{x} \;=\; 98.25 \spaced{and} \hat{\sigma} \;=\; s \;=\; 0.7332 \]

A histogram of the data and the best-fitting normal distribution seem similar in shape, but we will formally test the normal model with a chi-squared goodness-of-fit test.

The probabilities for values within these classes were found from the best-fitting \(\NormalDistn(\mu=98.25, \sigma = 0.7332)\) distribution, then multiplied by the number of values, 130, to get expected counts. However since several expected counts are low, classes must be combined before calculating the chi-squared goodness-of-fit statistic.

Temperature
x
Observed count,
O
Expected count,
E
\(X \lt 97.0\) 6 5.61
\(97.0 \le X \lt 97.5\) 13 14.20
\(97.5 \le X \lt 98.0\) 21 27.76
\(98.0 \le X \lt 98.5\) 38 34.69
\(98.5 \le X \lt 99.0\) 33 27.72
\(99.0 \le X \lt 99.5\) 15 14.16
\(X \ge 99.5\) 4 5.71

The test statistic is

\[ X^2 \;=\; \sum_{x} {\frac{\left(O_x - E_x\right)^2}{E_x}} \;=\; 3.657 \]

Since there are 7 counts in the combined frequency table and 2 estimated parameters, the test statistic should be compared to the chi-squared distribution with \((7-2-1) = 4\) degrees of freedom. The p-value is the probability of a value from the \(\ChiSqrDistn(4 \text{ df})\) distribution as high as 3.657 and can be found (e.g. using Excel) to be 0.4545.

Since there would be almost 50% probability of getting observed counts as far from those expected from a normal distribution if the data did come from a normal distribution, we conclude that the data are consistent with coming from a normal distribution.