Question

$103$Bulletproof Vests. In the New York Times article "A Common Police Vest Fails the Bulletproof Test," E. Lichtblau reported on U.S. Department of Justice study of $103$bulletproof vests containing a fiber known as Zylon. In ballistics tests, only $4$of these vests produced acceptable safety outcomes (and resulted in immediate changes in federal safety guidelines). Find a $95\%$confidence interval for the proportion of all such vests that would produce acceptable safety outcomes by using the

a. one-proportion $z$-interval procedure.

b. one-proportion plus-four $z$-interval procedure. (See page 462 for the details of this procedure.)

c. Explain the large discrepancy between the two methods.

d. Which confidence interval would you use? Explain your answer.

Short answer

1. Using the one-proportion $z$- interval procedure, the $95\%$confidence interval is $0.0015$to $0.0761$
2. Using the one-proportion plus four $z$- interval technique, the $95\%$confidence interval is $0.0124$to $0.0995$.
3. 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.
4. The one-proportion $z$ - interval technique is optimal for (a), (b) and (c)

Step-by-step solution

Part (a) Step 1: Given Information

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

Part (a) Step 2: Explanation

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

$\alpha =0.05$

Therefore,

$$\begin{gathered} z_{\alpha /2}=z_{0.025} \\ =1.96 \end{gathered}$$

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 :

$\widehat{p}\pm z_{\alpha /2}\cdot \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

$\begin{array}{r}\mathrm{CI}=0.0388\pm 1.96\cdot \sqrt{\frac{0.0388(1-0.0388)}{103}}\end{array}$

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

Part  (b) Step 1: Given Information

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

Part (b) Step 2: Explanation 

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 :

$\widehat{p}\pm z_{\alpha /2}\cdot \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n+4}}$

$$\begin{gathered} \mathrm{CI}=0.056\pm 1.96\cdot \sqrt{\frac{0.056(1-0.056)}{103+4}} \\ \mathrm{CI}=0.056\pm 1.96\times 0.0222 \\ \mathrm{CI}=0.056\pm 0.0436 \\ \mathrm{CI}=0.0124\text{to}0.0995 \end{gathered}$$

Part (c) Step 1: Given Information

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

Part (c) Step 2: Explanation

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.

Part (d) Step 1: Given Information

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

Part (d) Step 2: Explanation

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.