## 30.8 Example: Health Promotion services

A study
(Becker et al. 1991)
compared the access to health promotion (HP) services
for people with and without a disability.
(This study was seen in
Sect. 24.10.)
Access was measured using the
*Barriers to Health Promoting Activities for Disabled Persons*
BHADP scale,
where higher scores mean greater barriers.
The RQ is:

Is the mean BHADP score the same for people with and without a disability?

The parameter is \(\mu_D - \mu_N\), the difference between the population mean BHADP score (people with disabilities, minus people without disabilities).

In this case, only numerical summary data is available (Table 30.3), not the original data. (An appropriate graphical summary, an error bar chart, can be constructed from the summary information (Fig. 24.12, though a boxplot cannot be constructed from the information.) Denoting those with and without a disability with subscripts \(D\) and \(N\) respectively, the hypotheses are:

- \(H_0\): \(\mu_D - \mu_N = 0\): There
*is no*difference in the population mean BHADP scores - \(H_1\): \(\mu_D - \mu_N \ne 0\): There
*is a*difference in the population mean BHADP scores

Sample mean | Std deviation | Sample size | Std error | |
---|---|---|---|---|

Disability | 31.83 | 7.73 | 132 | 0.6728 |

No disability | 25.07 | 4.8 | 137 | 0.4101 |

Difference | 6.76 | 0.80285 |

The best estimate of the difference in *population* means
is the difference between the *sample* means:
\((\bar{x}_D - \bar{x}_{ND}) = 6.76\).
The table also gives
the standard error for estimating this *difference* as
\(\text{s.e.}(\bar{x}_D - \bar{x}_{ND}) = 0.80285\)
(as *given* in the article).

*standard error is given here*; you

**cannot**easily calculate this from the given information. You are not expected to do so.

Using the summary information in Table 30.3, the \(t\)-score is computed using Equation (27.1):

\[
t = \frac{6.76 - 0}{0.80285} = 8.42.
\]
(Recall that \(\mu_D - \mu_N = 0\) from the null hypothesis.)
Using the 68–95–99.7 rule,
this *very* large \(t\)-score implies
the \(P\)-value will be *very* small.
We conclude:

Strong evidence exists in the sample (\(t = 8.42\); two-tailed \(P < 0.001\)) that people with a disability (mean: 31.83; \(n = 132\); standard deviation: \(7.73\)) and people without a disability (mean: 25.07; \(n = 137\); standard deviation: \(4.80\)) have different population mean access to health promotion services (95% CI for the difference: 5.17 to 8.35).