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

A total of 10 components are observed, with failure times observed from t = 0 until t = 50.

Ten components fail during this interval.

The respective failure times are as follows:

6.8, 6.7, 6.9, 0.7, 5.6, 3.8, 6.1, 3.8, 1.8, 4.1.

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

Determine the maximum likelihood estimate lambda.

(Give answers to 3 decimal places)

To determine the maximum likelihood estimate (MLE) for the hazard rate \(\lambda\) of the exponential distribution given the failure times, we can follow these steps:

### Step 1: Understand the Problem
The failure times are assumed to follow an Exponential distribution with hazard rate \(\lambda\). The probability density function (PDF) for an exponential distribution is given by:

\[
f(t|\lambda) = \lambda e^{-\lambda t}
\]

Given the dataTo determine the maximum likelihood estimate (MLE) for the hazard rate \(\lambda\) of the exponential distribution given the failure times, we can follow these steps:

### Step 1: Understand the Problem
The failure times are assumed to follow an Exponential distribution with hazard rate \(\lambda\). The probability density function (PDF) for an exponential distribution is given by:

\[
f(t|\lambda) = \lambda e^{-\lambda t}
\]

Given the 

\[
\text{Failure times: } 6.8, 6.7, 6.9, 0.7, 5.6, 3.8, 6.1, 3.8, 1.8, 4.1
\]

### Step 2: Write the Likelihood Function
The likelihood function for the given data is the product of the individual densities:

\[
L(\lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda t_i}
\]

where \(t_i\) are the observed failure times and \(n\) is the total number of observations (here \(n = 10\)).

### Step 3: Simplify the Likelihood Function
The likelihood function simplifies to:

\[
L(\lambda) = \lambda^n e^{-\lambda \sum_{i=1}^{n} t_i}
\]

### Step 4: Take the Log-Likelihood
Taking the natural logarithm of the likelihood function gives the log-likelihood function:

\[
\ln L(\lambda) = n \ln \lambda - \lambda \sum_{i=1}^{n} t_i
\]

### Step 5: Differentiate and Solve for \(\lambda\)
To find the MLE, we differentiate the log-likelihood with respect to \(\lambda\) and set it to zero:

\[
\frac{d}{d\lambda} \ln L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^{n} t_i = 0
\]

Solving for \(\lambda\):

\[
\hat{\lambda} = \frac{n}{\sum_{i=1}^{n} t_i}
\]

### Step 6: Calculate the MLE
Substitute the given failure times into the formula:

\[
\hat{\lambda} = \frac{10}{6.8 + 6.7 + 6.9 + 0.7 + 5.6 + 3.8 + 6.1 + 3.8 + 1.8 + 4.1}
\]

Now, let's calculate the sum of the failure times and then the MLE for \(\lambda\).

\[
\sum_{i=1}^{n} t_i = 46.3
\]

Thus:

\[
\hat{\lambda} = \frac{10}{46.3} \approx 0.216
\]

### Final Answer
The maximum likelihood estimate (MLE) for \(\lambda\) is:

\[
\hat{\lambda} \approx 0.216 \text{ (to three decimal places)}
\]

Would you like more details on any part of this solution, or have any further questions?

**Related Questions:**
1. What is the significance of the hazard rate \(\lambda\) in the context of an exponential distribution?
2. How does the assumption of independent and identically distributed (iid) failures influence the MLE calculation?
3. What are some real-world applications of estimating the hazard rate for components?
4. How would the MLE for \(\lambda\) change if one additional component was observed without failure?
5. How does the exponential distribution compare to other distributions commonly used in survival analysis?

**Tip:** When dealing with exponential distributions in reliability studies, the hazard rate \(\lambda\) is the reciprocal of the mean time to failure.

Learn how to determine the maximum likelihood estimate (MLE) for the hazard rate λ of the exponential distribution given failure times observed over a specific interval. Includes step-by-step calculation with example data.

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