A lifetime test is carried out on a particular type of component

A total of 10 components are observed, with failure times as follows:

7.2, 5.6, 4.7, 8.3, 9.5, 1.7, 0.1, 9.2, 8.8, 4.6.

The failure times are assumed to be iid Exponential with hazard rate λ.

Determine the maximum likelihood estimate of  λ.

(Give answers to 3 decimal places)

To determine the maximum likelihood estimate (MLE) of the hazard rate \( \lambda \) for a set of independent and identically distributed (iid) failure times that follow an Exponential distribution, we can follow these steps:

### 1. Understanding the Exponential Distribution
The probability density function (PDF) of an Exponential distribution with hazard rate \( \lambda \) is:
\[
f(t; \lambda) = \lambda e^{-\lambda t}
\]
where \( t \geq 0 \) and \( \lambda > 0 \).

### 2. Write the Likelihood Function
Given the failure times \( t_1, t_2, \dots, t_n \), the likelihood function \( L(\lambda) \) is:
\[
L(\lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda t_i} = \lambda^n \exp\left(-\lambda \sum_{i=1}^{n} t_i\right)
\]

### 3. Log-Likelihood Function
Taking the natural logarithm of the likelihood function gives the log-likelihood function:
\[
\ln L(\lambda) = n \ln \lambda - \lambda \sum_{i=1}^{n} t_i
\]

### 4. Derive the MLE for \( \lambda \)
To find the MLE of \( \lambda \), differentiate the log-likelihood function with respect to \( \lambda \) and set it equal to zero:
\[
\frac{d}{d\lambda} \ln L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^{n} t_i = 0
\]

Solving for \( \lambda \) gives:
\[
\hat{\lambda} = \frac{n}{\sum_{i=1}^{n} t_i}
\]

### 5. Calculate the MLE
Here, \( n = 10 \) and the failure times are \( t_1 = 7.2 \), \( t_2 = 5.6 \), \( t_3 = 4.7 \), \( t_4 = 8.3 \), \( t_5 = 9.5 \), \( t_6 = 1.7 \), \( t_7 = 0.1 \), \( t_8 = 9.2 \), \( t_9 = 8.8 \), \( t_{10} = 4.6 \).

First, sum the failure times:
\[
\sum_{i=1}^{10} t_i = 7.2 + 5.6 + 4.7 + 8.3 + 9.5 + 1.7 + 0.1 + 9.2 + 8.8 + 4.6 = 59.7
\]

Now, calculate \( \hat{\lambda} \):
\[
\hat{\lambda} = \frac{10}{59.7} \approx 0.168
\]

### Final Answer:
The maximum likelihood estimate of \( \lambda \) is **0.168** (correct to 3 decimal places).

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Would you like further details or have any questions?

Here are five related questions to explore further:

1. How does the MLE for \( \lambda \) change if the sample size \( n \) increases?
2. What are the assumptions underlying the use of the Exponential distribution in this context?
3. How would the estimate of \( \lambda \) change if one failure time was significantly larger than the others?
4. Can you explain how the hazard rate \( \lambda \) relates to the mean of the Exponential distribution?
5. How would you compute a confidence interval for \( \lambda \) based on this estimate?

**Tip:** In lifetime analysis, always check whether the chosen distribution (like Exponential) fits the data well by examining the nature of the data and performing goodness-of-fit tests.

Learn how to determine the maximum likelihood estimate (MLE) of the hazard rate \( \lambda \) for a set of failure times assumed to follow an Exponential distribution. Includes step-by-step calculation and formula derivations.

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