I want to ask about the confidence interval in NIST SP 800 22 so I can make sure I got it right. When evaluating the randomness of many binary strings, for example 1000 binary strings. In the NIST SP 800 22 document section 4.2.1, it is recommended to calculate the confidence interval given by the formula: $\hat{p} \pm 3 \sqrt{\frac{\hat{p}(1-\hat{p})}{m}} $, where $\hat{p}=1-\alpha$ with $\alpha=0.01$, and $m$ is the sample size.
Suppose i need to test 1000 binary strings where all binary strings have $P-value \geq0.01$, then the proportion is $1000/1000=1$, the confidence interval is $0.99 \pm 3 \sqrt{\frac{0.99(0.01)}{1000}} = 0.99 \pm 0.0094392$, means the proportion should above 0.9805607 and below 0.9994393, but proportion 1 falls outside of this interval so these 1000 binary strings are not random.
Is there something wrong here or am I misunderstanding it? 1000 binary strings all pass but the conclusion is not random because it is not in the confidence interval ?