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Chapter 8: Problem 10

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?]

Short Answer

Expert verified
a. 95% CI for mean: (2.61, 8.54). b. 99% CI: (2.40, 9.23). c. 95% CI for SD: (2.61, 8.54).

Step by step solution

01

Understand the Exponential Distribution

For an exponential distribution with rate parameter \( \lambda \), the mean is given by \( \frac{1}{\lambda} \). Thus, the expected lifetime \( \mu \) in this context is \( \mu = \frac{1}{\lambda} \). The variance is \( \frac{1}{\lambda^2} \). We want to calculate a 95% confidence interval for \( \mu \), which is equivalent to finding a confidence interval for \( \frac{1}{\lambda} \).
02

Calculate the Sample Mean

First, we need to calculate the sample mean of the provided lifetimes. This is done by summing all the lifetimes and dividing by the number of observations \( n \).\[\bar{x} = \frac{2.0 + 1.3 + 6.0 + 1.9 + 5.1 + 0.4 + 1.0 + 5.3 + 15.7 + 0.7 + 4.8 + 0.9 + 12.2 + 5.3 + 0.6}{15} = 4.08 \]
03

Determine the 95% Confidence Interval

For an exponential distribution, the mean has a confidence interval: \( \left(\frac{2n\bar{x}}{\chi^2_{1-\alpha/2, 2n}}, \frac{2n\bar{x}}{\chi^2_{\alpha/2, 2n}}\right) \). For a 95% CI with \( n = 15 \), \( 2n = 30 \), and \( \alpha = 0.05 \). Using \( \chi^2_{0.975, 30} = 16.047 \) and \( \chi^2_{0.025, 30} = 46.979 \):\[\left(\frac{2 \times 15 \times 4.08}{46.979}, \frac{2 \times 15 \times 4.08}{16.047}\right) = (2.61, 8.54) \]
04

Convert the Interval to 99% CI

To achieve a 99% confidence interval, we change \( \alpha \) to 0.01, therefore using \( \chi^2_{0.995, 30} = 13.787 \) and \( \chi^2_{0.005, 30} = 50.892 \):\[\left(\frac{2 \times 15 \times 4.08}{50.892}, \frac{2 \times 15 \times 4.08}{13.787}\right) = (2.40, 9.23) \]
05

Calculate the 95% CI for the Standard Deviation

The standard deviation \( \sigma \) of an exponential distribution is the same as the mean, \( \sigma = \frac{1}{\lambda} = \mu \). Thus, the 95% CI for the standard deviation is the same as the CI for the mean: \( (2.61, 8.54) \).

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Key Concepts

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

Exponential Distribution
The exponential distribution is a common probability distribution used to model time to an event, such as the lifetime of a product. It is defined by a single parameter, \( \lambda \), known as the rate parameter. This distribution is particularly suited for modeling the time intervals between events in a Poisson process, which is why it's often used in reliability testing, life data analysis, and queuing models.

In this distribution, the mean or expected value is given by the formula \( \mu = \frac{1}{\lambda} \). This means that the average time or duration of the event we're examining is inversely proportional to the rate \( \lambda \). If \( \lambda \) is high, events happen frequently, so the mean time between them is short.

A key characteristic of the exponential distribution is the memoryless property. The probability that an event lasts a time \( t + s \) given that it has already lasted \( t \) is the same as the probability of lasting time \( s \). This property makes the exponential distribution unique in modeling lifetime data.
Sample Mean
The sample mean is an important statistical measure that represents the average value of a set of numbers. It is computed by summing up all the individual measurements or observations and dividing by the total number of observations.

For our exercise, the sample mean is calculated from 15 observations of heat pump lifetimes. By summing all values and dividing by 15, we get a mean lifetime of \( \bar{x} = 4.08 \) years. This sample mean serves as an estimate for the true population mean, which is an important step in constructing confidence intervals.

The sample mean is a critical component in statistical inference. In the context of confidence intervals, it helps estimate the range within which the true population mean is likely to fall, thus giving us insights into the central tendency of our data.
Standard Deviation
The standard deviation measures the amount of variation or dispersion of a set of values. In simpler terms, it shows how much the data deviates from the mean, indicating the spread of data points.

In an exponential distribution, the standard deviation happens to be the same as the mean, \( \sigma = \frac{1}{\lambda} \). This property implies that both the expected value and the standard deviation are equal, highlighting the simplicity and special characteristics of this distribution.

To find a confidence interval for the standard deviation of an exponential distribution, we use the same interval as for the mean because of this equality. The 95% confidence interval for both the mean and the standard deviation in our problem is \((2.61, 8.54)\) years. This interval suggests that the true standard deviation and mean of the population from which the sample was drawn is likely to fall within this range.

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Most popular questions from this chapter

Here is a sample of ACT scores (average of the Math, English, Social Science, and Natural Science scores) for students taking college freshman calculus: \(\begin{array}{lllllll}24.00 & 28.00 & 27.75 & 27.00 & 24.25 & 23.50 & 26.25 \\\ 24.00 & 25.00 & 30.00 & 23.25 & 26.25 & 21.50 & 26.00 \\ 28.00 & 24.50 & 22.50 & 28.25 & 21.25 & 19.75 & \end{array}\) a. Using an appropriate graph, see if it is plausible that the observations were selected from a normal distribution. b. Calculate a two-sided \(95 \%\) confidence interval for the population mean. c. The university ACT average for entering freshmen that year was about 21. Are the calculus students better than average, as measured by the ACT?

According to the article "Fatigue Testing of Condoms" (Polymer Testing, 2009: 567-571), "tests currently used for condoms are surrogates for the challenges they face in use", including a test for holes, an inflation test, a package seal test, and tests of dimensions and lubricant quality (all fertile territory for the use of statistical methodology!). The investigators developed a new test that adds cyclic strain to a level well below breakage and determines the number of cycles to break. A sample of 20 condoms of one particular type resulted in a sample mean number of 1584 and a sample standard deviation of 607 . Calculate and interpret a confidence interval at the \(99 \%\) confidence level for the true average number of cycles to break. [Note: The article presented the results of hypothesis tests based on the \(t\) distribution; the validity of these depends on assuming normal population distributions.]

Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let \(\mu\) denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting \(95 \%\) confidence interval is \((7.8,9.4)\). a. Would a \(90 \%\) confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning. b. Consider the following statement There is a \(95 \%\) chance that \(\mu\) is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? c. Consider the following statement: We can be highly confident that \(95 \%\) of all bottles of this type of cough syrup have an alcohol content that is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? d. Consider the following statement: If the process of selecting a sample of size 50 and then computing the corresponding \(95 \%\) interval is repeated 100 times, 95 of the resulting intervals will include \(\mu\). Is this statement correct? Why or why not?

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 chisquared rv with \(v=22\) d. \(P\left(x^{2}<14.611\right.\) or \(\left.\chi^{2}>37.652\right)\), where \(x^{2}\) is a chi-squared rv with \(v=25\)

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a uniform distribution on the interval \([0, \theta]\), so that $$ f(x)= \begin{cases}\frac{1}{\theta} & 0 \leq x \leq \theta \\ 0 & \text { otherwise }\end{cases} $$ Then if \(Y=\max \left(X_{i}\right)\), by the first proposition in Section \(5.5, U=Y / \theta\) has density function $$ f_{U}(u)=\left\\{\begin{array}{cl} n u^{n-1} & 0 \leq u \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Use \(f_{U}(u)\) to verify that $$ P\left[(\alpha / 2)^{1 / n} \leq \frac{Y}{\theta} \leq(1-\alpha / 2)^{1 / n}\right]=1-\alpha $$ and use this to derive a \(100(1-\alpha) \%\) CI for \(\theta\). b. Verify that \(P\left(\alpha^{1 / n} \leq Y / \theta \leq 1\right)=1-\alpha\), and derive a \(100(1-\alpha) \%\) CI for \(\theta\) based on this probability statement. c. Which of the two intervals derived previously is shorter? If your waiting time for a morning bus is uniformly distributed and observed waiting times are \(x_{1}=4.2, x_{2}=3.5, x_{3}=1.7\), \(x_{4}=1.2\), and \(x_{5}=2.4\), derive a 95\% CI for \(\theta\) by using the shorter of the two intervals.
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