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