## 29.6 Conclusions: Mean differences

The *one-tailed* \(P\)-value is \(0.0635\),
suggesting
only slight evidence supporting \(H_1\).
To write a conclusion,
an *answer to the RQ* is needed,
plus *evidence* leading to that conclusion; and
some *summary statistics*,
including a CI (indicating the precision of the statistic):

Slight evidence exists in the sample (paired \(t=1.68\); one-tailed \(P=0.0635\)) of a mean energy saving in the population (mean saving: 0.54 MWh; \(n=10\); 95% CI from \(-0.19\) to \(1.27\) MWh) after adding the insulation.

The wording implies the direction of the differences (by talking of ‘savings’). Of course, statistically validity shoud be checked; this was done in Sect. 23.9, but the validity conditions are given again in the next section, for completeness.

**Example 29.1 (COVID lockdown) **A study of \(n = 213\) Spanish health students (Romero-Blanco et al. 2020)
measured (among other things)
the number of minutes of vigorous physical activity (PA) performed by students
*before* and *during* the COVID-19 lockdown (from March to April 2020 in Spain).

These numerical summary of the data are shown in Example 23.1, so we do not repeat it here.
We define the *differences* as the number of minutes of vigorous PA *before* the COVID lockdown, minus the number of minutes of vigorous PA *during* the COVID lockdown.
A difference is computed for each participant, so the data are *paired*.

Using this definition,
a *positive* difference means the *Before* value is higher;
hence,
the differences tell us how much longer the student spent doing vigorous PA *before* the COVID lockdown.
Similarly,
a *negative * value means that the *During* value is higher.

The RQ is

For Spanish health students, is there a mean change in the amount of vigorous PA during and before the COVID lockdown?

In this situation, the *parameter* of interest is the population mean difference \(\mu_d\),
the mean amount that students spent in vigorous PA *before* the lockdown compared to *during* the lockdown.
The hypotheses are:

- \(H_0\): \(\mu_d = 0\)
- \(H_1\): \(\mu_d \ne 0\) (i.e., two-tailed)

The mean *difference* is \(\bar{d} = -2.68\) minutes, with a standard deviation of \(s_d = 51.30\) minutes.
However, we know that the sample mean difference could vary from sample to sample,
so has a standard error:
\[
\text{s.e.}(\bar{d}) = \frac{s_d}{n} = \frac{51.30}{\sqrt{213}} = 3.515018.
\]

The test statistic is

\[ t = \frac{\bar{d} - \mu_d}{\text{s.e.}(\bar{d})} = \frac{-2.68 - 0}{3.515018} = -0.76. \]

This is a very small value, so (using the 68-95-99.7 rule) the \(P\)-value will be very large: a sample mean difference of -2.68 minutes could easily have happened by chance even if the population mean difference was zero.

We write:

There is no evidence (paired \(t = -0.76\), \(P > 0.10\)) of a mean change in the amount of vigorous PA

beforeandduringlockdown (sample mean 2.68 minutes greater *during lockdown; standard deviation: 51.30 minutes).