1. The problem involves comparing blood pressure measurements across two time periods or conditions with 28 participants. 2. Given values: initial mean pressure $\mu_1 = 145$ mmHg, final mean pressure $\mu_2 = 132$ mmHg, difference $d = 9$ mmHg, sample size $n = 28$. 3. Our objective is to assess if the change is statistically significant at the 5% significance level. 4. We assume the difference $d$ here is the sample mean difference between paired samples. 5. The standard error (SE) for the mean difference is $SE = \frac{s}{\sqrt{n}}$, where $s$ is the standard deviation of differences. 6. Since $s$ is not given, but if $9$ mmHg is the sample mean difference, we can't proceed with the t-test calculation without $s$. 7. To test significance, the t-statistic formula is $t = \frac{\bar{d} - 0}{SE}$. 8. Without $s$, we cannot compute exact t or p-value; additional data is needed. 9. Conclusion: More information such as standard deviation or standard error of the difference is required to perform the hypothesis test at 5% significance level.