9.1 Normal Model - Inference about Means

Example 1 Iris petal lengths - Best estimate

**Table 9.1.** Iris petal lengths, in centimeters, for Iris type *Setosa*.
1.4	1.4	1.3	1.5	1.4
1.7	1.4	1.5	1.4	1.5
1.5	1.6	1.4	1.1	1.2
1.5	1.3	1.4	1.7	1.5
1.7	1.5	1.0	1.7	1.9
1.6	1.6	1.5	1.4	1.6
1.6	1.5	1.5	1.4	1.5
1.2	1.3	1.5	1.3	1.5
1.3	1.3	1.3	1.6	1.9
1.4	1.6	1.4	1.5	1.4

Table 9.1 shows data for the lengths (in centimeters) of the petals of one species of Iris flower¹. If we want to estimate the “true” length of the the petal for this species, given all of these examples, we would apply the following model of the data:

$\begin{matrix} d a t a = true value + Normal(mean=0,known σ) \end{matrix}$

or equivalently

$d a t a equals Normal(mean= true value,known σ )$

The resulting distribution for the “true value”, $μ$ , is also a Normal distribution (Section 7.3),

$\begin{matrix} P (μ | d a t a, σ) = N o r m a l (\bar{x}, σ / \sqrt{N}) \end{matrix}$

where the best estimate of the true value, $μ$ is the sample mean, $\bar{x}$ , and the uncertainty is related to the sample deviation (which we're going to take as the “known” deviation, $σ ∼ 0.174$ ) in this case. Thus,

$\begin{matrix} \hat{μ} = \bar{x} = \frac{1.4 + 1.4 + 1.3 + 1.5 + \dots + 1.6 + 1.4 + 1.5 + 1.4}{50} \\ = 1.464 \end{matrix}$

and the full answer, with uncertainty, is

$\begin{matrix} \hat{μ} = 1.464 [c m] \pm \frac{0.174}{\sqrt{50}} [c m] \\ = 1.464 [c m] \pm 0.025 [c m] \end{matrix}$

Example 2 Iris petal lengths - A different species?

Show solution

Here we apply the $z$ -test (Section $z$-test on page 159) to a new observation to see if there is reason to believe it to be a different species. Imagine we have a single observation of another iris with petal length 2.5 [cm]. Is this likely to be the same type as the Setosa type above? As outlined in Section 7.2, we get the best estimate for the difference as:

$\begin{matrix} μ_{d i f f} = 2.5 - 1.464 = 1.036 \end{matrix}$

with uncertainty the same as the uncertainty of the Setosa type, so the final estimate with uncertainty is:

$1.036 open bracket c m close bracket ± 0.025 open bracket c m close bracket$

which is

$\begin{matrix} \frac{1.036 [c m]}{0.025 [c m]} = 41 deviations away from zero! \end{matrix}$

which makes it virtually certain to be a different type (see Table 8.1).

Adapted from Statistical Inference for Everyone, by Brian Blais (Bryant University), licensed under CC BY-SA 4.0 (dual-licensed under the GNU FDL 1.2 or later; this adaptation uses the CC BY-SA grant). Changes were made; this adaptation is distributed under the same license. License: CC-BY-SA-4.0.

1.4	1.4	1.3	1.5	1.4
1.7	1.4	1.5	1.4	1.5
1.5	1.6	1.4	1.1	1.2
1.5	1.3	1.4	1.7	1.5
1.7	1.5	1.0	1.7	1.9
1.6	1.6	1.5	1.4	1.6
1.6	1.5	1.5	1.4	1.5
1.2	1.3	1.5	1.3	1.5
1.3	1.3	1.3	1.6	1.9
1.4	1.6	1.4	1.5	1.4

1.4	1.4	1.3	1.5	1.4
1.7	1.4	1.5	1.4	1.5
1.5	1.6	1.4	1.1	1.2
1.5	1.3	1.4	1.7	1.5
1.7	1.5	1.0	1.7	1.9
1.6	1.6	1.5	1.4	1.6
1.6	1.5	1.5	1.4	1.5
1.2	1.3	1.5	1.3	1.5
1.3	1.3	1.3	1.6	1.9
1.4	1.6	1.4	1.5	1.4

1.4	1.4	1.3	1.5	1.4
1.7	1.4	1.5	1.4	1.5
1.5	1.6	1.4	1.1	1.2
1.5	1.3	1.4	1.7	1.5
1.7	1.5	1.0	1.7	1.9
1.6	1.6	1.5	1.4	1.6
1.6	1.5	1.5	1.4	1.5
1.2	1.3	1.5	1.3	1.5
1.3	1.3	1.3	1.6	1.9
1.4	1.6	1.4	1.5	1.4