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Chapter 8: Problem 85

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 \(v=2\) ). a. Use this fact to obtain a test statistic and rejection region that together specify a level a 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 \(\left.\lambda=1 / \mu_{0} \cdot\right]\) 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 .

Short Answer

Expert verified
The data does not strongly suggest the true average lifetime is less than 75.

Step by step solution

01

Understand the Hypotheses

We are testing the null hypothesis \(H_0: \mu = \mu_0\) versus the alternatives: \(H_1: \mu eq \mu_0\), \(H_1: \mu > \mu_0\), and \(H_1: \mu < \mu_0\). Since \(\mu = 1/\lambda\), \(H_0: \mu = \mu_0\) translates to \(\lambda = 1/\mu_0\).
02

Determine the Test Statistic

Given the fact that \(2 \lambda \sum X_{i}\) follows a chi-squared distribution with \(2n\) degrees of freedom, we can set up the test statistic as \(T = 2 \sum X_i / \mu_0\), which should follow a \(\chi^2\) distribution with \(2n\) degrees of freedom if \(H_0\) is true.
03

Rejection Region

For a level \(\alpha\) test with \(H_1: \mu eq \mu_0\), reject \(H_0\) if the test statistic \(T\) is not in the interval \([\chi^2_{\alpha/2, 2n}, \chi^2_{1-\alpha/2, 2n}]\). For \(H_1: \mu > \mu_0\), reject if \(T < \chi^2_{\alpha, 2n}\). For \(H_1: \mu < \mu_0\), reject if \(T > \chi^2_{1-\alpha, 2n}\).
04

Calculate the Test Statistic for the Given Data

The failure times are \(95, 16, 11, 3, 42, 71, 225, 64, 87, 123\). The sum \(\sum X_i = 737\). Under \(H_0: \mu = 75\), calculate \(\lambda_0 = 1/75\). The test statistic \(T = 2 \cdot 737/75\).
05

Calculate and Compare to Critical Value

Calculate \(T = \frac{2 \times 737}{75} = \frac{1474}{75} \approx 19.6533\). For \(n = 10\), degrees of freedom \(u = 2 \cdot 10 = 20\). For \alpha = 0.05 and \(H_1: \mu < 75\), look up the critical value \(\chi^2_{0.95, 20}\). If \(T > \chi^2_{0.95, 20}\), reject \(H_0\).
06

Conclusion Based on Comparison

With \(T \approx 19.6533\), compare it to \(\chi^2_{0.95, 20} = 31.41\). Since \(19.6533 < 31.41\), we do not reject \(H_0\). Thus, there is not strong evidence to suggest 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.

Chi-Squared Distribution
The chi-squared distribution is a key concept in hypothesis testing. It is used primarily with data obtained from normally distributed samples. The chi-squared distribution, denoted as \( \chi^2 \), is a special case of the gamma distribution. It is particularly useful in testing hypotheses about variances. A chi-squared distribution depends on the degrees of freedom, which usually equals the number of independent values that can vary in an analysis.

When the sum of the squares of k independent standard normal random variables is calculated, the resulting value follows a chi-squared distribution with \( k \) degrees of freedom.

In the context of the given exercise, we use the chi-squared distribution to evaluate the test statistic derived from the exponential distribution. That test statistic, \( T = 2 \lambda \sum X_i \), follows a chi-squared distribution with \( 2n \) degrees of freedom under the null hypothesis \( H_0 \).

Using a chi-squared table, one can look up critical values to determine whether to reject the null hypothesis based on the observed test statistic. This is what makes the chi-squared distribution indispensable in deciding the outcomes of hypothesis tests.
Exponential Distribution
The exponential distribution is a continuous probability distribution often used to model the time until an event occurs, such as failure rates in reliability studies. It is memoryless, meaning the probability of an event occurring in the next unit of time is independent of how much time has already passed.

The exponential distribution has one parameter, \( \lambda \), which is the rate parameter. It describes the rate at which events occur. The mean of an exponential distribution is given by \( 1/\lambda \), which helps in formulating hypotheses like in the exercise: \( H_0: \mu = \mu_0 \) implies \( \lambda = 1/\mu_0 \).

It's utilized in the exercise because the chi-squared test statistic is derived from a data set following an exponential distribution with parameter \( \lambda \). The failure times of components are modeled as being exponentially distributed, which helps in testing their mean lifetime against a claimed value. Knowing the exponential distribution allows us to derive the correct chi-squared statistic for hypothesis testing.
Test Statistic
A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It's used to determine whether to reject the null hypothesis \( H_0 \).

In the exercise, the test statistic \( T = 2 \sum X_i / \mu_0 \) is critical. It is calculated from the sample data and follows a chi-squared distribution under the null hypothesis. For the given exercise, this statistic leverages the properties of the exponential distribution which sums up to present in a form that relates to the chi-squared distribution.

To perform the hypothesis test, one must compute this test statistic and compare it to the critical values of a chi-squared distribution for a certain significance level (denoted by \( \alpha \)). The rejection region for the test is based on whether the test statistic falls into a critical area determined by these values. This helps decide if there's enough evidence to reject \( H_0 \). This process illustrates how the test statistic forms the backbone of hypothesis testing by quantifying deviations from what is stated under the null hypothesis.

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Most popular questions from this chapter

A common characterization of obese individuals is that their body mass index is at least 30 [BMI \(=\) weight/(height \()^{2}\), where height is in meters and weight is in kilograms]. The article "The Impact of Obesity on Illness Absence and Productivity in an Industrial Population of Petrochemical Workers" (Annals of Epidemiology, 2008: 8-14) reported that in a sample of female workers, 262 had BMIs of less than 25,159 had BMIs that were at least 25 but less than 30 , and 120 had BMIs exceeding 30 . Is there compelling evidence for concluding that more than \(20 \%\) of the individuals in the sampled population are obese? a. State and test appropriate hypotheses using the rejection region approach with a significance level of \(.05\). b. Explain in the context of this scenario what constitutes type I and II errors c. What is the probability of not concluding that more than \(20 \%\) of the population is obese when the actual percentage of obese individuals is \(25 \%\) ?

Each of a group of 20 intermediate tennis players is given two rackets, one having nylon strings and the other synthetic gut strings. After several weeks of playing with the two rackets, each player will be asked to state a preference for one of the two types of strings. Let \(p\) denote the proportion of all such players who would prefer gut to nylon, and let \(X\) be the number of players in the sample who prefer gut. Because gut strings are more expensive, consider the null hypothesis that at most \(50 \%\) of all such players prefer gut. We simplify this to \(H_{0}: p=.5\), planning to reject \(H_{0}\) only if sample evidence strongly favors gut strings. a. Which of the rejection regions \(\\{15,16,17,18,19,20\\}\), \(\\{0,1,2,3,4,5\\}\), or \(\\{0,1,2,3,17,18,19,20\\}\) is most appropriate, and why are the other two not appropriate? b. What is the probability of a type I error for the chosen region of part (a)? Does the region specify a level \(.05\) test? Is it the best level .05 test? c. If \(60 \%\) of all enthusiasts prefer gut, calculate the probability of a type II error using the appropriate region from part (a). Repeat if \(80 \%\) of all enthusiasts prefer gut. d. If 13 out of the 20 players prefer gut, should \(H_{0}\) be rejected using a significance level of .10?

The article "Orchard Floor Management Utilizing SoilApplied Coal Dust for Frost Protection" (Agri. and Forest Meteorology, 1988: 71-82) reports the following values for soil heat flux of eight plots covered with coal dust. \(\begin{array}{llllllll}34.7 & 35.4 & 34.7 & 37.7 & 32.5 & 28.0 & 18.4 & 24.9\end{array}\) The mean soil heat flux for plots covered only with grass is 29.0. Assuming that the heat-flux distribution is approximately normal, does the data suggest that the coal dust is effective in increasing the mean heat flux over that for grass? Test the appropriate hypotheses using \(\alpha=.05\).

For a fixed alternative value \(\mu^{\prime}\), show that \(\beta\left(\mu^{\prime}\right) \rightarrow 0\) as \(n \rightarrow \infty\) for either a one-tailed or a two-tailed \(z\) test in the case of a normal population distribution with known \(\sigma\).

The accompanying data on cube compressive strength (MPa) of concrete specimens appeared in the article "Experimental Study of Recycled Rubber-Filled HighStrength Concrete" (Magazine of Concrete Res., 2009: \(549-556)\) \(\begin{array}{rrrrr}112.3 & 97.0 & 92.7 & 86.0 & 102.0 \\ 99.2 & 95.8 & 103.5 & 89.0 & 86.7\end{array}\) a. Is it plausible that the compressive strength for this type of concrete is normally distributed? b. Suppose the concrete will be used for a particular application unless there is strong evidence that true average strength is less than \(100 \mathrm{MPa}\). Should the concrete be used? Carry out a test of appropriate hypotheses using the \(P\)-value method.
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