Statistical inference on bolt machine On the average about 3% of the bolts produced by a company are defective. To maintain this quality of performance, a sample of 200 bolts produced is examined every 4 hours.Determine the (a0)99% and (b)95% control limits for the number of defective bolts in each sample.
The textbook answer is (a)6 and (b)4.
I can calculate $\sigma = \sqrt{\frac{pq}{n}}=\sqrt{\frac{0.03*0.97}{200}}=0.01206$
However I not sure how to get to six bolts from there
99%: 
0.03 $\pm$ 2.576*0.01206
0.03 $\pm$ 0.0311
95%: 
0.03 $\pm$ 1.96*0.01206
0.03 $\pm$ 0.0236
When multiplied by n=200 this becomes:
(a) 6 $\pm$ 6.22 bolts
(b) 6 $\pm$ 4.72 bolts
It seems the answer ignores the mean value of 6 and only uses the integer part from the volatility part is this correct?
 A: To be precise, if the hypothesis is $$H_0 : p = p_0 = 0.03 \quad \text{vs.} \quad H_a : p > 0.03,$$ where $p$ is the true defect rate, then at a significance level of $\alpha$, the exact test statistic is the random variable $$X \mid H_0 \sim \operatorname{Binomial}(n = 200, p = p_0 = 0.03)$$ that counts the random number of observed defects.  The test will reject $H_0$ in favor of $H_a$ if $X$ is "too large," i.e., we want to find the boundary $x$ of the rejection region such that the Type I error is at most $\alpha$;  $$\Pr[X \ge x \mid H_0] \le \alpha.$$  Since $p$ is so small, most of the probability mass of $X$ will be concentrated at small $X$, thus we can compute 
$$\begin{align*}
\Pr[X = 0] &= \binom{200}{0}(0.03)^0 (1-0.03)^{200} \approx 0.00226124 \\
\Pr[X \le 1] &= \Pr[X = 0] + \binom{200}{1}(0.03)^1 (1-0.03)^{199} \approx 0.0162483 \\
\Pr[X \le 2] &= \Pr[X \le 1] + \binom{200}{2}(0.03)^2 (1-0.03)^{198} \approx 0.0592909 \\
&\,\,\vdots \\
\Pr[X \le 9] &\approx 0.919221 \\
\Pr[X \le 10] &\approx 0.959872 \\
\Pr[X \le 11] &\approx 0.981589 \\
\Pr[X \le 12] &\approx 0.992167.
\end{align*}$$
This tells us that at a significance level of $\color{red}{\alpha = 0.01}$, the rejection region is $\color{red}{X \ge 13}$:  if there are at least $13$ defects in the lot of $200$, we are at least $99\%$ confident that the true mean defect rate exceeds $3\%$.  It is not $X \ge 12$ because then the Type I error is not strictly controlled at $\alpha = 0.01$; it is $$1 - 0.981589 = 0.0184113 > 0.01.$$  In other words, if you observed only $X = 12$ defects in the batch, your chance of erroneously concluding that the true defect rate exceeds $3\%$ is almost twice the claimed Type I error--you can only be $98.1\%$ confident, not $99\%$ confident, that your conclusion is valid.
Similarly, at a significance level of $\color{blue}{\alpha = 0.05}$, the rejection region is $\color{blue}{X \ge 11}$.  This means if you observe at least $11$ defects in the batch, you can be at least $95\%$ confident that the true mean defect rate exceeds $3\%$.
