91Ó°ÊÓ

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from an exponential distribution with parameter \(\lambda\). Then it can be shown that \(2 \lambda \sum X_{i}\) has a chi-squared distribution with \(\nu=2 n\) (by first showing that \(2 \lambda X_{i}\) has a chi-squared distribution with \(\nu=2\) ). a. Use this fact to obtain a test statistic and rejection region that together specify a level \(\alpha\) test for \(H_{0}: \mu=\mu_{0}\) versus each of the three commonly encountered alternatives. [Hint: \(E\left(X_{i}\right)=\mu=1 / \lambda\), so \(\mu=\mu_{0}\) is equivalent to \(\lambda=1 / \mu_{0}\). b. Suppose that ten identical components, each having exponentially distributed time until failure, are tested. The resulting failure times are $$ \begin{array}{llllllllll} 95 & 16 & 11 & 3 & 42 & 71 & 225 & 64 & 87 & 123 \end{array} $$ Use the test procedure of part (a) to decide whether the data strongly suggests that the true average lifetime is less than the previously claimed value of 75 .

Step by step solution

01

Set Up Hypotheses

We need to establish the null and alternative hypotheses. Our null hypothesis is that the true mean of the exponential distribution, \(\mu\), equals the claimed mean, \(\mu_0 = 75\). Therefore, \(H_0: \mu = 75\). Since we are testing if the actual mean is less than 75, the alternative hypothesis is \(H_a: \mu < 75\).

02

Transform Hypotheses into Lambda

Given \(E(X_i) = \mu = \frac{1}{\lambda}\), \(\mu = \mu_0\) implies \(\lambda = \frac{1}{75}\). The hypotheses thus transform into \(H_0: \lambda = \frac{1}{75}\) and \(H_a: \lambda > \frac{1}{75}\).

03

Express Statistics Using Sum of Samples

We have that \(2\lambda \sum X_i \sim \chi^2(u = 2n)\). The sample size, \(n\), is 10 here, giving \(u = 20\). The test statistic is \(T = 2 \cdot \frac{1}{75} \cdot \sum X_i = \frac{2}{75} \cdot \sum X_i\).

04

Calculate the Sample Sum

Sum up the failure times: \(\sum X_i = 95 + 16 + 11 + 3 + 42 + 71 + 225 + 64 + 87 + 123 = 737\).

05

Compute the Test Statistic

Substitute the calculated sample sum into the test statistic: \(T = \frac{2}{75} \cdot 737\). Calculate \(T = 19.6533\).

06

Determine Rejection Region

Since \(2\lambda \sum X_i\) follows a \(\chi^2(20)\) distribution under the null hypothesis, the rejection region for \(H_0: \mu = 75\) versus \(H_a: \mu < 75\) is \(T < \chi^2_{\alpha, 20}\). Find the critical value for \(\chi^2_{\alpha, 20}\) at significance level \(\alpha\).

07

Compare Test Statistic to Critical Value

Assume a typical significance level (e.g., \(\alpha = 0.05\)). Lookup \(\chi^2_{0.05, 20} = 9.591\) from chi-squared distribution tables. Since 19.6533 is not less than 9.591, we do not reject \(H_0\).

08

Conclusion of the Test

Based on the above steps, and under the assumption \(\alpha=0.05\), we conclude that we lack evidence to claim that the true average lifetime is less than 75.

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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 continuous probability distribution commonly used to model time until an event occurs, such as the failure time of a mechanical component. It is defined by a single parameter, \( \lambda \), which represents the rate of occurrence or the inverse of the mean. For exponential distributions, the expected value (mean) of \(X\) is given by \( \mu = \frac{1}{\lambda} \).
Some key properties of the exponential distribution include:

• It is memoryless, meaning the probability of an event occurring in the future is independent of the past.
• The probability density function (PDF) is \( f(x; \lambda) = \lambda e^{-\lambda x} \), for \( x \ge 0 \).
• Mean and standard deviation are equal: \( \mu = \sigma = \frac{1}{\lambda} \).

Given the sample \( X_1, X_2, \ldots, X_n \), each drawn from an exponential distribution, we use these properties to understand how the sums of these variables behave when applying statistical tests, particularly by leveraging relationships with other distributions like the chi-squared distribution.

Chi-Squared Distribution

A chi-squared distribution is a statistical distribution commonly used for hypothesis testing, especially in the context of variance and goodness-of-fit tests. It is formed by summing the squares of independent standard normal random variables.
Each component in a chi-squared distribution is a squared standard normal variable, and the sum of these forms a chi-squared distributed random variable:

• If there are \(k\) degrees of freedom, then \(X = Z_1^2 + Z_2^2 + \ldots + Z_k^2\) has a chi-squared distribution with \(k\) degrees of freedom.
• In the context of this exercise, it turns out that \(2 \lambda \sum X_i\) follows a chi-squared distribution with \(u = 2n\) degrees of freedom, where \(n\) is the sample size.

This relationship is key in translating the exponential distribution parameters into a form that allows hypothesis testing. By using the chi-squared distribution, we can derive a test statistic and determine critical values for hypothesis spaces.

Test Statistic

The test statistic is a crucial component in hypothesis testing. It measures how far our sample statistic is from the null hypothesis, providing a means to evaluate if an observed effect is statistically significant.
For the exponential distribution, calculating the test statistic involves utilizing the relationship with the chi-squared distribution. By noting that \(2 \lambda \sum X_i \sim \chi^2(2n)\), we derive the test statistic \(T = \frac{2}{\mu_0} \sum X_i\). Here \(\mu_0\) is the mean under the null hypothesis.
To compute the test statistic:

• Sum up all observed values: \(\sum X_i\).
• Calculate \(T = \frac{2}{\mu_0} \sum X_i\).

The test statistic compares the observed sum of the sample to what would be expected under the null hypothesis. Larger deviations from the mean indicate less likelihood of the data coming from the null distribution, prompting us to consider alternative hypotheses.

Rejection Region

The rejection region defines the set of values for which the null hypothesis is rejected. It is established based on a chosen significance level \(\alpha\), which represents the probability of making a Type I error (incorrectly rejecting a true null hypothesis).
To determine the rejection region, we need a critical value from the chi-squared distribution table corresponding to our test statistic's degrees of freedom. For example, at a typical \(\alpha = 0.05\), we use the chi-squared distribution table to find the value \(\chi^2_{0.05, 20}\).
If our test statistic falls into this region:

• For a left-tailed test, we reject the null if \(T < \chi^2_{\alpha, u}\).
• For a right-tailed test, we reject if \(T > \chi^2_{\alpha, u}\).
• For a two-tailed test, we investigate if \(T\) is beyond either tail's critical value.

Determining the appropriate critical value and rejection region ensures that we conclude appropriately about our hypotheses, balancing the risks of Type I and Type II errors.