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 \(2 \lambda \Sigma X_{i}\) has a chi-squared distribution with \(v=2 n\) (by first showing that \(2 \lambda X_{i}\) has a chi-squared distribution with \(v=2\) ). 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

Understand the Hypotheses

For testing \(H_0: \mu = \mu_0\) against alternatives, note that \(\mu = \frac{1}{\lambda}\). Thus, \(\mu = \mu_0\) implies \(\lambda = \frac{1}{\mu_0}\). For part (b), we test \(H_0: \mu = 75\). The null hypothesis assumes that the mean lifetime is 75, which corresponds to \(\lambda = \frac{1}{75}\).

02

Calculate Test Statistic

Given \(2 \lambda \sum X_i\) has a chi-squared distribution with \(v = 2n\), under the null hypothesis, we have \(\lambda = \frac{1}{75}\). Thus, the test statistic is \(2 \times \frac{1}{75} \times \sum X_i\), where \(\sum X_i = 737\) (sum of the ten failure times).

03

Chi-Squared Distribution

Calculate the test statistic: \(2 \times \frac{1}{75} \times 737 = \frac{1474}{75} \approx 19.653\). This is compared to the chi-squared distribution with \(2 \times 10 = 20\) degrees of freedom for the test.

04

Determine Rejection Region

Depending on the alternative hypothesis, different critical values from chi-squared distribution tables at a significance level \(\alpha\) determine the rejection region. For \(H_a: \mu < 75\), reject \(H_0\) if the test statistic is lower than the critical value from the chi-squared table.

05

Apply the Decision Rule

For a one-tailed test with \(H_a: \mu < 75\), we find from chi-squared tables: for \(\alpha = 0.05\) and \(20\) degrees of freedom, the critical value is approximately \(10.851\). Since \(19.653 > 10.851\), we do not reject \(H_0\).

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Key Concepts

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

Chi-squared Distribution

The Chi-squared distribution is an important concept in statistics, especially when dealing with hypothesis testing and variance analysis. It is a type of probability distribution that arises when a squared sum of independent standard normal random variables is calculated.
In the context of exponential distributions, the Chi-squared distribution is particularly useful. When you have a random sample from an exponential distribution with parameter \(\lambda\), it can be shown that \(2 \lambda \Sigma X_{i}\) follows a Chi-squared distribution. Specifically, if \(X_i\) are your sample observations, then the sum of these observations, when adjusted by \(2 \lambda\), can be tested using a Chi-squared test.

• The degrees of freedom (\(v\)) in a Chi-squared distribution usually depends on the number of observations. For example, \(v = 2n\) when dealing with samples from an exponential distribution where \(n\) is the sample size.
• Chi-squared tests are particularly valuable for determining how much a set of observed values deviates from what would be expected under a specific hypothesis.

Hypothesis Testing

Hypothesis testing is a core method in inferential statistics that allows you to make decisions about a population parameter based on sample data. The goal is to test an assumption (the null hypothesis \(H_0\)) against an alternative hypothesis \(H_a\).
In this specific exercise, we are testing whether the mean lifetime \(\mu\) of components is equal to a specific claimed value, 75, using observed failure times.

• The null hypothesis \(H_0\) states that the mean \(\mu\) of the distribution is 75, implying that \(\lambda = \frac{1}{75}\).
• The alternative hypothesis \(H_a\) proposes that the mean lifetime \(\mu\) is less than 75.

Using failure time data, we test if the process mean diverges significantly from 75, using steps outlined in the hypothesis testing procedure.

Test Statistic

A test statistic is a number derived from the sample data that allows you to decide whether to reject the null hypothesis. It involves calculating a value based on your data and comparing it against a theoretical distribution.
For our problem, we compute the test statistic using the formula derived from the Chi-squared distribution: \[ 2 \times \frac{1}{75} \times \sum X_i \] where \(\sum X_i = 737\) is the total sum of the observed failure times of the components. This results in a test statistic of approximately 19.653.

• The test statistic value will be compared against a critical value from a Chi-squared table.
• The decision to reject or fail to reject the null hypothesis depends on this comparison.

Significance Level

The significance level, denoted as \(\alpha\), is a critical concept in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is true (Type I error). Common significance levels are 0.05, 0.01, and 0.10. In our exercise, we use \(\alpha = 0.05\).
A significance level helps determine the rejection region by providing a threshold:

• If the test statistic falls into this rejection region, \(H_0\) is rejected in favor of \(H_a\).
• For \(\alpha = 0.05\) and with 20 degrees of freedom, the critical value from the Chi-squared distribution is approximately 10.851.

In our scenario, since the test statistic (19.653) is not less than the critical value, we do not reject \(H_0\), indicating the observed data does not significantly suggest the mean lifetime is less than the claimed 75.