91Ó°ÊÓ

The $95\%$ confidence interval was calculated using the one-proportion $z$ - interval technique.

According to the information, the $95\%$confidence level:

$\mathrm{Î\pm }=0.05$

Therefore,

As a result, the sample proportion would be:

$$\begin{gathered} \widehat{p}=\frac{x}{n} \\ =\frac{4}{103} \\ =0.0388 \end{gathered}$$

The confidence interval for $p$using the one-proportion $z$-interval technique is :

$\widehat{p}Â\pm z_{\mathrm{Î\pm }/2}â‹\dots \sqrt{\frac{\widehat{p}(1−\widehat{p})}{n}}$

$\begin{array}{r}\mathrm{CI}=0.0388Â\pm 1.96â‹\dots \sqrt{\frac{0.0388(1−0.0388)}{103}}\end{array}$

$$\begin{gathered} \mathrm{CI}=0.0388Â\pm 1.96×0.0190 \\ \mathrm{CI}=0.0388Â\pm 0.0373 \\ \mathrm{CI}=0.0015\text{to}0.0761 \end{gathered}$$

The 95% confidence interval was calculated using the one-proportion plus four $z$interval technique.

According to the information, the number of successes is:

$x=4$

$n=103$

In order to use the one-proportion plus four z-interval technique, the sample proportion would be:

$$\begin{gathered} \widehat{p}=\frac{x+2}{n+4} \\ =\frac{4+2}{103+4} \\ =0.056 \end{gathered}$$

The confidence interval for $p$using the one-proportion plus four $z$-interval technique is :

$\widehat{p}Â\pm z_{\mathrm{Î\pm }/2}â‹\dots \sqrt{\frac{\widehat{p}(1−\widehat{p})}{n+4}}$

$$\begin{gathered} \mathrm{CI}=0.056Â\pm 1.96â‹\dots \sqrt{\frac{0.056(1−0.056)}{103+4}} \\ \mathrm{CI}=0.056Â\pm 1.96×0.0222 \\ \mathrm{CI}=0.056Â\pm 0.0436 \\ \mathrm{CI}=0.0124\text{to}0.0995 \end{gathered}$$

To determine the cause of the substantial disparity between part (a) and part (b) results.

Only four out of $103$bulletproof jackets passed safety tests in a research.

This suggests that the probability of success is $x=4$, and the sample size is$n=103.$

When the confidence level is greater than $90\%$and the sample size is greater than ten, the one proportion plus four $z-$interval technique is utilised as a rule of thumb.

To determine which way of calculating the $95\%$confidence interval is optimal.

Only four out of $103$bulletproof jackets passed safety tests in a research.

This suggests that the probability of success is $x=4$, and the sample size is$n=103.$

When the confidence level is greater than $90\%$and the sample size is greater than ten, the one proportion plus four $z$-interval technique is utilised as a rule of thumb.

When we add two more successes to component (b), the total number of successes equals half of the original number. This is why we get a large discrepancy in the confidence interval when we calculate it.