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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% CI for the true proportion is (0.1116, 0.2884).

Step by step solution

01

Identify the Given Information

We are given that 11 out of 55 masks had lenses pop out at \( 250^{\circ} \)F. The sample size \( n \) is 55 and the number of successes (masks with popped lenses) \( x \) is 11.
02

Calculate the Sample Proportion

Compute the sample proportion \( \hat{p} \). It is given by \( \hat{p} = \frac{x}{n} = \frac{11}{55} \). Hence, \( \hat{p} \approx 0.2 \).
03

Determine the Critical Value

Using a standard Z-distribution table, for a 90% confidence interval, the critical value \( Z_{\alpha/2} \) corresponding to \( \alpha = 0.10 \) is approximately 1.645.
04

Calculate the Standard Error

The standard error (SE) is calculated using the formula: \( SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n}} \). Substituting the given values: \( SE = \sqrt{ \frac{0.2 \times 0.8}{55} } \approx 0.0538 \).
05

Construct the Confidence Interval

The 90% confidence interval is given by: \( \hat{p} \pm Z_{\alpha/2} \times SE \). Thus, it is \( 0.2 \pm 1.645 \times 0.0538 \). Calculating this gives the interval approximately \( (0.1116, 0.2884) \).
06

Interpret the Confidence Interval

The 90% confidence interval \( (0.1116, 0.2884) \) suggests that we are 90% confident that the true proportion of masks with lenses that would pop out at \( 250^{\circ} \)F lies between 11.16% and 28.84%.

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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 estimate of the true proportion in a population based on a sample. In our case of firefighter masks, it provides an initial guess of the proportion of masks that have a failure at high temperatures. To find the sample proportion, we use the formula:\[ \hat{p} = \frac{x}{n} \]where \( x \) is the number of successes (masks with popped lenses) and \( n \) is the total number of masks tested. Here, we calculated it as \( \hat{p} = \frac{11}{55} = 0.2 \). This means 20% of the sampled masks failed the test. The sample proportion is crucial because it sets the stage for constructing the confidence interval, helping us make inferences about the entire population of masks.
Standard Error
Standard error measures the variability or "spread" of the sample proportion from the true population proportion. Understanding it helps us determine how reliable our sample estimate is likely to be.To calculate the standard error of the sample proportion \( \hat{p} \), use the formula:\[ SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n}} \]This formula accounts for the proportion of successful outcomes \( \hat{p} \) and the probability of failure \( 1-\hat{p} \), scaled by the sample size \( n \). In our scenario:\[ SE = \sqrt{ \frac{0.2 \times 0.8}{55} } \approx 0.0538 \]A smaller standard error indicates that our sample proportion is more precise, which is comforting for making broader conclusions about mask reliability.
Critical Value
The critical value, derived from the chosen confidence level, determines the width of our confidence interval. For a 90% confidence interval, the critical value comes from the Z-distribution table (a standard table for normal distribution percentages).To find this value, we consider the complement of our confidence level, which is \( 1 - 0.90 = 0.10 \). Then, halve it to find the area in each tail, resulting in \( \alpha/2 = 0.05 \). The critical value \( Z_{\alpha/2} \) for \( 90\% \) is approximately 1.645. Both tales absorb the remaining 10%, each consisting of 5%. This critical factor is essential because it scales the confidence interval by representing the "distance" from the sample mean that captures the true mean.
Z-distribution
The Z-distribution, also known as the standard normal distribution, is a fundamental concept for conducting statistical analyses like creating confidence intervals. It has a mean of 0 and a standard deviation of 1. All values on a Z-distribution express how many standard deviations an element is from the mean. In the context of our mask reliability test, using a Z-distribution allows us to derive the critical value necessary to build a confidence interval. Since the Z-distribution is bell-shaped and symmetric, it helps define the "spread" or "range" where we expect to find the true population parameter. Each standardized value (Z-score) corresponds to a percentile in the distribution, which assists in determining how our sample's behavior compares to the expected probabilities in a true normal distribution. Thus, it is an invaluable tool for confirming the likelihood of our observations. By applying it correctly, we ensure that our confidence intervals are robust and reliable.

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Most popular questions from this chapter

The article "Extravisual Damage Detection? Defining the Standard Normal Tree" (Photogrammetric Engrg. Remote Sensing, 1981: 515-522) discusses the use of color infrared photography in identification of normal trees in Douglas fir stands. Among data reported were summary statistics for green-filter analytic optical densitometric measurements on samples of both healthy and diseased trees. For a sample of 69 healthy trees, the sample mean dye-layer density was \(1.028\), and the sample standard deviation was .163. a. Calculate a \(95 \%\) (two-sided) CI for the true average dye-layer density for all such trees. b. Suppose the investigators had made a rough guess of 16 for the value of \(s\) before collecting data. What sample size would be necessary to obtain an interval width of \(.05\) for a confidence level of \(95 \%\) ?

Determine the \(t\) critical value that will capture the desired \(t\) curve area in each of the following cases: a. Central area \(=.95\), df \(=10\) b. Central area \(=.95, \mathrm{df}=20\) c. Central area \(=99\), df \(=20\) d. Central area \(=.99, \mathrm{df}=50\) e. Upper-tail area \(=.01, \mathrm{df}=25\) f. Lower-tail area \(=.025\), df \(=5\)

Here are the names of 12 orchestra conductors and their performance times in minutes for Beethoven's Ninth Symphony: \(\begin{array}{llll}\text { Bermstein } & 71.03 & \text { Furtwängler } & 74.38 \\ \text { Leinsdorf } & 65.78 & \text { Ormandy } & 64.72 \\ \text { Solti } & 74.70 & \text { Szell } & 66.22 \\ \text { Bohm } & 72.68 & \text { Karajan } & 66.90 \\ \text { Masur } & 69.45 & \text { Rattle } & 69.93 \\\ \text { Steinberg } & 68.62 & \text { Tennstedt } & 68.40\end{array}\) a. Check to see that normality is a reasonable assumption for the performance time distribution. b. Compute a \(95 \%\) CI for the population standard deviation, and interpret the interval. c. Supposedly, classical music is \(100 \%\) determined by the composer's notation, including all timings. Based on your results, is this true or false?

A random sample of \(n=15\) heat pumps of a certain type yielded the following observations on lifetime (in years): \(\begin{array}{rrrrrrrr}2.0 & 1.3 & 6.0 & 1.9 & 5.1 & .4 & 1.0 & 5.3 \\ 15.7 & .7 & 4.8 & .9 & 12.2 & 5.3 & .6 & \end{array}\) a. Assume that the lifetime distribution is exponential and use an argument parallel to that of Example \(8.5\) to obtain a \(95 \%\) CI for expected (true average) lifetime. b. How should the interval of part (a) be altered to achieve a confidence level of \(99 \%\) ? c. What is a \(95 \%\) CI for the standard deviation of the lifetime distribution? [Hint: What is the standard deviation of an exponential random variable?]

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