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An electrical component has a time-to-failure (or lifetime) distribution that is exponential with parameter \(\lambda,\) so the mean lifetime is \(\mu=1 / \lambda .\) Suppose that a sample of \(n\) of these components is put on test, and let \(X_{i}\) be the observed lifetime of component \(i\). The test continues only until the \(r\) th unit fails, where \(r

Short Answer

Expert verified
The 95% confidence interval for the mean lifetime is approximately (298.33, 1144.77) hours.

Step by step solution

01

Understand the Distribution

The lifetime of the component follows an exponential distribution with parameter \( \lambda \), meaning the mean lifetime \( \mu \) is given by \( \mu = \frac{1}{\lambda} \). For our task, we're told that the statistic \( 2\lambda T_r \) has a chi-square distribution with \( 2r \) degrees of freedom, where \( T_r \) is the accumulated lifetime up until the \( r \text{th} \) failure, calculated by \( T_r = \sum_{i=1}^{r} X_i + (n-r)X_r \).
02

Develop the Confidence Interval Formula

Given that \( 2\lambda T_r \sim \chi^2(2r) \), the mean \( \mu = \frac{1}{\lambda} \). For a chi-square distribution, if \( Z = 2\lambda T_r \), then we can use the properties of the chi-square distribution to find \( \chi^2_{1-\alpha/2, 2r} \) and \( \chi^2_{\alpha/2, 2r} \). The confidence interval for \( 1/\lambda \) can be written as:\[\left( \frac{T_r}{\chi^2_{1-\alpha/2, 2r}/(2r)}, \frac{T_r}{\chi^2_{\alpha/2, 2r}/(2r)} \right)\]
03

Calculate Total Lifetime Accumulated \(T_r\)

For the problem, \( n = 20 \) and \( r = 10 \). The failure times are 15, 18, 19, 20, 21, 21, 22, 27, 28, and 29 hours. Calculate the sum of failure times \( T_r = 15 + 18 + 19 + 20 + 21 + 21 + 22 + 27 + 28 + 29 \) and add the surviving times factor: \( (n-r) \times X_r = (20 - 10) \times 29 \). Therefore, \( T_r = 220 + 290 = 510 \) hours.
04

Find Chi-Square Critical Values

For a 95% confidence interval, \( \alpha = 0.05 \), so we need the chi-square critical values \( \chi^2_{0.025, 20} \) and \( \chi^2_{0.975, 20} \), corresponding to \( 2r = 20 \) degrees of freedom. These values can be obtained from chi-square distribution tables or calculators: \( \chi^2_{0.025, 20} = 34.17 \) and \( \chi^2_{0.975, 20} = 8.91 \).
05

Calculate the Confidence Interval

The confidence interval for \( \mu = 1/\lambda \) is given by:\[\left( \frac{510}{34.17/20}, \frac{510}{8.91/20} \right)\]Calculate each boundary:\[\frac{510 \times 20}{34.17} \approx 298.33, \quad \frac{510 \times 20}{8.91} \approx 1144.77\]Thus, the 95% confidence interval for the mean lifetime is approximately \( (298.33, 1144.77) \) hours.

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Key Concepts

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

Exponential Distribution
In statistics, the exponential distribution is a continuous probability distribution used to model the time between events in a Poisson process. Simply put, it's used to understand how long you might wait before an event occurs. This distribution is characterized by the parameter \( \lambda \), which is the rate parameter.

The key properties of an exponential distribution include:
  • It is memoryless. This means that the probability of an event occurring in the next time period is independent of how much time has already elapsed. This property is unique to the exponential distribution.
  • The mean or expected value is given by \( \frac{1}{\lambda} \), so as \( \lambda \) increases, the mean time until the event decreases.
  • The distribution is defined for all non-negative values of time: \( x \geq 0 \).
It's often used in reliability engineering and survival analysis to model lifetimes of components or time-to-failure scenarios. For instance, it can describe how soon an electrical component will fail, helping in planning maintenance or warranty terms.
Chi-Square Distribution
The Chi-Square distribution is a continuous probability distribution that is used extensively in statistical tests, especially in hypothesis testing and constructing confidence intervals. This distribution is defined by its degrees of freedom, which often relate to the number of variables or samples in the analysis.

Some crucial elements about the Chi-Square distribution include:
  • It is positively skewed, especially for low degrees of freedom, but approaches a normal distribution as the degrees of freedom increase.
  • It's used in goodness-of-fit tests to see how well a theoretical distribution fits data, in assessing independence in contingency tables, and in estimating population variances.
  • In our context, \( 2 \lambda T_r \) follows a Chi-Square distribution with \( 2r \) degrees of freedom. This property derives from the exponential distribution and is crucial in constructing confidence intervals for the mean lifetime \( \mu \).
Understanding and applying the Chi-Square distribution is vital for working with data that has been gathered from various samples or tests.
Confidence Interval
A confidence interval is a range of values that estimate an unknown population parameter, in this case, the mean lifetime \( \mu \). Confidence intervals provide more information than point estimates because they demonstrate the precision and reliability of the estimation.

Key insights about confidence intervals:
  • The confidence level, commonly set at 95%, indicates that if the same population is sampled repeatedly, 95% of the calculated confidence intervals will contain the true mean.
  • For our exercise, we apply the Chi-Square distribution's properties to calculate the confidence interval for \( \mu = 1/\lambda \).
  • The width of the confidence interval depends on the sample size and variability within the data. Larger samples tend to produce narrower intervals, indicating more precise estimates.
Using confidence intervals helps in making predictions or conclusions about a population, providing a clear understanding of data reliability.
Mean Lifetime Estimation
Mean lifetime estimation refers to determining the average time until a particular event, such as failure or decay, occurs. It’s particularly significant in fields like reliability engineering and life testing.

In the context of the exponential distribution:
  • The mean \( \mu \) is simply \( \frac{1}{\lambda} \), where \( \lambda \) is the rate at which events occur.
  • For a censored life test, where testing stops after a certain number of failures, the total observed lifetime can be used to estimate the mean lifetime of components.
  • This estimation becomes useful in predicting future behaviors of processes or components, and in planning maintenance schedules to prevent unexpected failures.
By carefully calculating and interpreting the mean lifetime, we gain insights that are critical for optimizing and improving the reliability and performance of various systems.

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Most popular questions from this chapter

A healthcare provider monitors the number of CAT scans performed each month in each of its clinics. The most recent year of data for a particular clinic follows (the reported variable is the number of CAT scans each month expressed as the number of CAT scans per thousand members of the health plan): \(2.31,2.09,2.36,1.95,1.98,2.25,2.16,2.07,1.88,1.94,1.97,\) \(2.02 .\) (a) Find a \(95 \%\) two-sided CI on the mean number of CAT scans performed each month at this clinic. (b) Historically, the mean number of scans performed by all clinics in the system has been \(1.95 .\) If there any evidence that this particular clinic performs more CAT scans on average than the overall system average?

During the 1999 and 2000 baseball seasons, there was much speculation that the unusually large number of home runs hit was due at least in part to a livelier ball. One way to test the "liveliness" of a baseball is to launch the ball at a vertical surface with a known velocity \(V_{L}\) and measure the ratio of the outgoing velocity \(V_{0}\) of the ball to \(V_{L} .\) The ratio \(R=V_{0} / V_{L}\) is called the coefficient of restitution. Following are measurements of the coefficient of restitution for 40 randomly selected baseballs. The balls were thrown from a pitching machine at an oak surface. \(\begin{array}{llllll}0.6248 & 0.6237 & 0.6118 & 0.6159 & 0.6298 & 0.6192 \\\ 0.6520 & 0.6368 & 0.6220 & 0.6151 & 0.6121 & 0.6548 \\ 0.6226 & 0.6280 & 0.6096 & 0.6300 & 0.6107 & 0.6392 \\ 0.6230 & 0.6131 & 0.6223 & 0.6297 & 0.6435 & 0.5978 \\ 0.6351 & 0.6275 & 0.6261 & 0.6262 & 0.6262 & 0.6314 \\\ 0.6128 & 0.6403 & 0.6521 & 0.6049 & 0.6170 & \\ 0.6134 & 0.6310 & 0.6065 & 0.6214 & 0.6141 & \end{array}\) (a) Is there evidence to support the assumption that the coefficient of restitution is normally distributed? (b) Find a \(99 \%\) CI on the mean coefficient of restitution. (c) Find a \(99 \%\) prediction interval on the coefficient of restitution for the next baseball that will be tested. (d) Find an interval that will contain \(99 \%\) of the values of the coefficient of restitution with \(95 \%\) confidence. (e) Explain the difference in the three intervals computed in parts (b), (c), and (d).

A random sample has been taken from a normal distribution and the following confidence intervals constructed using the same data: (37.53,49.87) and (35.59,51.81) (a) What is the value of the sample mean? (b) One of these intervals is a \(99 \% \mathrm{CI}\) and the other is a \(95 \%\) CI. Which one is the \(95 \%\) CI and why?

Determine the \(t\) -percentile that is required to construct each of the following one-sided confidence intervals: (a) Confidence level \(=95 \%,\) degrees of freedom \(=14\) (b) Confidence level \(=99 \%,\) degrees of freedom \(=19\) (c) Confidence level \(=99.9 \%,\) degrees of freedom \(=24\)

The diameter of holes for a cable harness is known to have a normal distribution with \(\sigma=0.01\) inch. A random sample of size 10 yields an average diameter of 1.5045 inch. Find a \(99 \%\) two-sided confidence interval on the mean hole diameter.

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