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Chapter 7: Problem 54

It is important that face masks used by firefighters be able to withstand high temperatures because firefighters commonly work in temperatures of \(200-500^{\circ} \mathrm{F}\). In a test of one type of mask, 11 of 55 masks had lenses pop out at \(250^{\circ}\). Construct a \(90 \%\) CI for the true proportion of masks of this type whose lenses would pop out at \(250^{\circ}\).

Short Answer

Expert verified
The 90% confidence interval is approximately (0.112, 0.288).

Step by step solution

01

Define the sample data

We know that out of 55 masks tested, 11 masks had lenses pop out. Thus, the sample proportion \( \hat{p} \) is calculated as follows:\[ \hat{p} = \frac{11}{55} = 0.2 \]
02

Identify the confidence level and find the z-score

The problem asks for a 90% confidence interval, which correlates to a 5% significance level at each tail (since it is a two-tailed test). The critical z-value for a 90% confidence interval is approximately 1.645.
03

Calculate the standard error

The standard error of the sample proportion is calculated using the formula:\[ SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \]Where \( n \) is the sample size. Substituting the values we have:\[ SE = \sqrt{ \frac{0.2 \times 0.8}{55} } \approx 0.05366 \]
04

Compute the margin of error

The margin of error \( ME \) is calculated by multiplying the z-score with the standard error:\[ ME = z \times SE = 1.645 \times 0.05366 \approx 0.08826 \]
05

Determine the confidence interval

The confidence interval can now be computed using the formula:\[ \text{CI} = (\hat{p} - ME, \hat{p} + ME) \]Plugging in the values, we get:\[ (0.2 - 0.08826, 0.2 + 0.08826) \approx (0.11174, 0.28826) \]
06

Conclusion: Interpret the confidence interval

With 90% confidence, we can say that the true proportion of masks that would have lenses pop out at \(250^{\circ}\) lies between approximately 0.112 and 0.288.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
The sample proportion is an important statistical measure that represents the proportion of a certain outcome in a sample. In this exercise, our focus is on the proportion of firefighter masks that failed the performance test at a specified temperature.

To calculate the sample proportion, you use the formula:
  • Divide the number of successful outcomes by the total number of observations.
  • For this exercise, 11 masks failed out of 55, so the sample proportion \( \hat{p} \) is calculated as \( \hat{p} = \frac{11}{55} = 0.2 \).
This means 20% of the masks tested failed.

Sample proportion helps provide an estimate of the proportion in the entire population based on the sample data. This will be used to further understand other concepts such as standard error and confidence intervals.
Standard Error
The standard error gives us an idea of the variability or spread of the sample proportion. It helps to understand how the sample proportion might differ if we were to take multiple samples. This measure is crucial because it plays a key role in calculating confidence intervals.

In our exercise, the formula for standard error of the sample proportion is:
  • \( SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \)
  • With a sample proportion \( \hat{p} = 0.2 \) and total sample size \( n = 55 \), the standard error calculates to \( SE \approx 0.05366 \).
This value tells us about the possible extent of deviation from the true population proportion that our sample proportion could have.

The standard error is what we multiply with the z-score to determine the margin of error.
Margin of Error
The margin of error is an essential component when constructing a confidence interval. It quantifies the uncertainty associated with the sample proportion and signifies how much the sample results may vary from the true population parameter.

To compute the margin of error, the standard error \( SE \) is multiplied by the critical value \( z \) derived from the confidence level:
  • \( ME = z \times SE \)
  • For a 90% confidence level with a \( z \)-value of 1.645 and a standard error of 0.05366, the margin of error is approximately 0.08826.
This means our observed sample proportion could vary by about 8.8% from the true population proportion.

In practical terms, the margin of error gives us the range above and below the sample proportion to predict where the true population proportion most likely falls.
Z-score
In statistics, the z-score plays a vital role in constructing confidence intervals by determining how many standard deviations away from the mean a data point is. It gives us confidence in estimating the population parameter based on sample statistics.

For confidence intervals, the z-score corresponds to the confidence level desired. For example, a 90% confidence interval uses a standard z-score of approximately 1.645.

The role of the z-score includes:
  • Converting the standard error into the margin of error by providing the number of standard errors to move away from the sample proportion.
  • Ensuring the covering of a particular range of the probability distribution based on the desired confidence level.
When applied in our exercise, it demonstrates a balance between confidence and an acceptable range of error, ensuring the results are reliable enough for interpretation.

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Most popular questions from this chapter

The article "Limited Yield Estimation for Visual Defect Sources" (IEEE Trans. on Semiconductor Manuf., 1997: 17-23) reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 201 of these passed the probe. Assuming a stable process, calculate a \(95 \%\) (two-sided) confidence interval for the proportion of all dies that pass the probe.

On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with \(\sigma=100\). The composition of the bar has been slightly modified, but the modification is not believed to have affected either the normality or the value of \(\sigma\). a. Assuming this to be the case, if a sample of 25 modified bars resulted in a sample average yield point of \(8439 \mathrm{lb}\), compute a \(90 \%\) CI for the true average yield point of the modified bar. b. How would you modify the interval in part (a) to obtain a confidence level of \(92 \%\) ?

Consider a normal population distribution with the value of \(\sigma\) known. a. What is the confidence level for the interval \(\bar{x} \pm\) \(2.81 \sigma / \sqrt{n}\) ? b. What is the confidence level for the interval \(\bar{x} \pm\) \(1.44 \sigma / \sqrt{n} ?\) c. What value of \(z_{\mathrm{a} / 2}\) in the CI formula (7.5) results in a confidence level of \(99.7 \%\) ? d. Answer the question posed in part (c) for a confidence level of \(75 \%\).

Determine the following: a. The 95 th percentile of the chi-squared distribution with \(v=10\) b. The 5 th percentile of the chi-squared distribution with \(v=10\) c. \(P\left(10.98 \leq \chi^{2} \leq 36.78\right)\), where \(\chi^{2}\) is a chi- squared rv with \(v=22\) d. \(P\left(\chi^{2}<14.611\right.\) or \(\left.\chi^{2}>37.652\right)\), where \(\chi^{2}\) is a chisquared rv with \(v=25\)

The following observations were made on fracture toughness of a base plate of \(18 \%\) nickel maraging steel ["Fracture Testing of Weldments," ASTM Special Publ. No. 381, 1965: \(328-356\) (in ksi \(\sqrt{\text { in., }}\) given in increasing order)]: \(\begin{array}{llllllll}69.5 & 71.9 & 72.6 & 73.1 & 73.3 & 73.5 & 75.5 & 75.7 \\\ 75.8 & 76.1 & 76.2 & 76.2 & 77.0 & 77.9 & 78.1 & 79.6 \\ 79.7 & 79.9 & 80.1 & 82.2 & 83.7 & 93.7 & & \end{array}\) Calculate a \(99 \%\) CI for the standard deviation of the fracture toughness distribution. Is this interval valid whatever the nature of the distribution? Explain.
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