I have been working on this practice problem for a while, and worked it out in two ways, and following from inputting the answers in the system, it notified me that the answers for both of them was wrong.
Could anyone help me figure out where I went wrong in my procedure? And help me work it out the correct way?
The question goes as follows:
The first method
The second method
Part (a)
$$E = \frac{0.2319-0.1270}{2}= 0.05245$$ $$p = 0.2314-0.05245 = 0.17945$$ $$\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.17945(1-0.17945)}{195}}= 0.0275$$
Therefore, the critical $z$ value is, $$z = \frac{E}{\sigma} = \frac{0.05245}{0.0275} = 1.9073 \approx 1.91$$
From the standard table, $$P ( -1.91 < z < 1.91) = 0.9719 $$
Confidence Level: 97.19%
Part (b)
Confidence Level computed as: $$\frac{19}{20}= 0.95$$
From the standard table, $$P(z < 1.645) = 0.95$$
For a point estimate of proportion here as $p = 0.17945$, the margin of error is, $$E = z \times \sqrt{\frac{p(1-p)}{n}}$$ $$n = \frac{z^2 \times p(1-p)}{E^2}= \frac{1.645^2 \times 0.17945(1-0.17945)}{0.05245^2}= 144.84$$
Required sample size: 145
You need to use a finite population correction factor for the interval estimate, because the batch has size $N = 2750$ and you are sampling $n = 195$ axles without replacement from that batch. So a $100(1-\alpha)\%$ asymptotic confidence interval will have the form $$\hat p \pm z_{\alpha/2}^* \sqrt{\frac{\hat p (1 - \hat p)}{n} \frac{N - n}{N - 1}},$$ where $\hat p$ is the point estimate of the proportion of failed axles, and $z_{\alpha/2}^*$ is the upper $\alpha/2$ quantile of the standard normal distribution, i.e. $\Pr[Z \le z_{\alpha/2}^*] = 1 - \alpha/2$. In your case, $$\hat p = \frac{0.1270 + 0.2319}{2},$$ hence the margin of error is $$ME = z_{\alpha/2}^* \sqrt{\frac{\hat p (1 - \hat p)}{n} \frac{N - n}{N - 1}} = \frac{0.2319 - 0.1270}{2}$$ and the critical value is $$z_{\alpha/2}^* = \frac{0.2319 - 0.1270}{2} \sqrt{\frac{195}{(0.17945)(1 - 0.17945)} \cdot \frac{2750 - 1}{2750 - 195}}.$$ This gives $1 - \alpha/2 = \Phi(1.97984)$ or $\alpha = 0.0477213$.
For the second part, what they are telling you is that the desired margin of error is $ME = 4\% = 0.04$, and the desired confidence level is $19/20 = 95\%$, corresponding to $\alpha/2 = 0.025$. If we assume the point estimate from part (a) stays the same, then the required sample size is $$n = \frac{m}{1 + \frac{m-1}{N}}, \quad m = \frac{(z_{\alpha/2}^*)^2 \hat p (1 - \hat p)}{ME^2}.$$ This gives $n = 314$, meaning $314 - 195 = 119$ more axles need to be sampled.