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Chapter 9: Problem 94

A service station has six gas pumps. When no vehicles are at the station, let \(p_{i}\) denote the probability that the next vehicle will select pump \(i\) \((i=1,2, \ldots, 6) .\) Based on a sample of size \(n\), we wish to test \(H_{0}: p_{1}=\ldots=p_{6}\) versus the alternative \(H_{\mathrm{a}}: p_{1}=p_{3}=p_{5}, p_{2}=p_{4}=p_{6}\) (note that \(H_{\mathrm{a}}\) is not a simple hypothesis). Let \(X\) be the number of customers in the sample that select an even-numbered pump. a. Show that the likelihood ratio test rejects \(H_{0}\) if either \(X \geq c\) or \(X \leq n-c\). [Hint: When \(H_{\mathrm{a}}\) is true, let \(\theta\) denote the common value of \(p_{2}, p_{4}\), and \(\left.p_{6} \cdot\right]\) b. Let \(n=10\) and \(c=9\). Determine the power of the test both when \(H_{0}\) is true and also when \(p_{2}=p_{4}=p_{6}=\frac{1}{10}, p_{1}=p_{3}=p_{5}=\frac{7}{30} .\)

Short Answer

Expert verified
The test rejects \(H_0\) if \(X \geq 9\) or \(X \leq 1\). Compute powers for both hypotheses using binomial probabilities.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis is that each pump is equally likely to be chosen: \[ H_{0}: p_{1} = p_{2} = p_{3} = p_{4} = p_{5} = p_{6} = \frac{1}{6} \]Under the alternative hypothesis, the probabilities are:\[ H_{a}: p_{1} = p_{3} = p_{5}, \; p_{2} = p_{4} = p_{6} \]
02

Define Likelihood Functions

Under the null hypothesis, probabilities for each pump are equal, so:\[ L(H_0) = \left(\frac{1}{6}\right)^{n} \]Under the alternative hypothesis, let \( \theta \) be the common value of \( p_{2}, p_{4}, p_{6} \) and thus \( 1 - \theta \) is the common value of \( p_{1}, p_{3}, p_{5} \). Therefore, the likelihood is:\[ L(H_a) = (1-\theta)^{k} \theta^{n-k} \]
03

Compute the Likelihood Ratio

The likelihood ratio test statistic is:\[ \Lambda = \frac{L(H_0)}{L(H_a)} \ = \frac{(1/6)^{n}}{(1-\theta)^{k} \theta^{n-k}} \]To reject \( H_0 \), we need to compute the critical region for \( X \).
04

Analyze Critical Region

In order to reject \( H_0 \), set \( X \geq c \) and \( X \leq n-c \). This form ensures capturing large deviations in \( X \) which suggests higher likelihood under \( H_a \).
05

Calculate Power of Test for H鈧� True

When \( H_0 \) is true, each pump is equally likely, thus \[ P(X \geq 9 \; \text{or} \; X \leq 1) = P(X = 9) + P(X = 10) + P(X = 0) + P(X = 1) \]Use the binomial distribution: \[ X \sim \text{Binomial}(n=10, p=0.5) \] Calculate \( P(X=k) \) for \( k = 0, 1, 9, 10 \) using:\[ P(X=k) = \binom{10}{k}(0.5)^{k}(0.5)^{10-k} \]
06

Calculate Power of Test for H鈧� True

Under \( H_a : p_{2} = p_{4} = p_{6} = 0.1 \), calculate:\[ P(X \geq 9 \; \text{or} \; X \leq 1) \] Here, \( X \sim \text{Binomial}(n=10, p=0.3) \).Calculate \( P(X=k) \) for \( k=0, 1, 9, 10 \) similarly. Substitute \( p=0.3 \) in:\[ P(X=k) = \binom{10}{k}(0.3)^{k}(0.7)^{10-k} \]
07

Conclusion of Powers

Complete the probabilities with results from Steps 5 and 6 for both scenarios:- Power for \( H_0 \), sum probabilities as calculated.- Power for \( H_a \), sum respective probabilities as calculated in Step 6.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental aspect of statistical analysis. It involves deciding whether to accept or reject a specific hypothesis based on sample data. In this case, we evaluate whether each gas pump at a service station is equally likely to be selected. The null hypothesis, denoted as \(H_0\), asserts that all pumps have equal selection probability. Meanwhile, the alternative hypothesis, \(H_a\), proposes that certain pumps have different probabilities than others. This alternative isn't simple, meaning it incorporates multiple parameters or conditions.
This type of testing is crucial because it helps to make informed decisions based on statistical evidence rather than relying on assumptions. Hypothesis testing usually involves the following steps:
  • Formulating the null and alternative hypotheses.
  • Determining the test statistic that needs to be calculated.
  • Establishing the critical region or cutoff points to decide whether to reject \(H_0\).
  • Making a conclusion based on the comparison of test statistics to the critical region.
In this exercise, we use the likelihood ratio test, a common method for comparing hypotheses. This test is particularly powerful when dealing with complex hypotheses, like those with multiple conditions or parameters.
Binomial Distribution
The binomial distribution is an essential type of probability distribution in statistics. It applies to situations where there are exactly two possible outcomes for a fixed number of trials, such as success or failure, heads or tails. In the context of the gas station problem, selecting an even-numbered pump or an odd-numbered one can be considered the two outcomes.
When we say a random variable \(X\) follows a binomial distribution, we refer to it as:
\[ X \sim \text{Binomial}(n, p) \]
where:
  • \(n\) is the number of trials (e.g., the number of customers, 10 in this exercise).
  • \(p\) is the probability of success on a given trial (e.g., selecting an even-numbered pump).
In this scenario, we are particularly interested in the number of customers selecting even-numbered pumps. The binomial distribution allows us to calculate the probability of a specific number of successes given \(n\) trials and a probability \(p\). For hypothesis testing, especially in step 5 and 6 of the solution, knowing the distribution helps us compute the likelihood of observing given results under different hypotheses.
Statistical Power
Statistical power is an important concept in hypothesis testing that refers to the likelihood of correctly rejecting a false null hypothesis. In other words, power helps us detect a true effect when there is one. A test with high power reduces the risk of a Type II error, which occurs when we fail to reject a false null hypothesis.
Power is influenced by several factors, including:
  • The sample size \(n\) 鈥� larger samples typically provide more information, increasing power.
  • The significance level (\(\alpha\)) 鈥� choosing a higher level will result in a higher power.
  • The true effect size 鈥� larger differences between \(H_0\) and \(H_a\) are easier to detect.
In this exercise, the power of the test is calculated under two conditions: when \(H_0\) is true and when the alternative hypothesis \(H_a\) expresses different probabilities. This involves assessing the probabilities of observing extreme values of \(X\) (either very low or very high), which leads us to either accept or reject \(H_0\). Calculating these probabilities gives us a way to estimate the power.
Probability Distribution
Probability distribution is a concept that describes how probabilities are assigned over different possible outcomes of a random variable. In statistics, understanding the probability distribution of your data is crucial for accurate analysis and interpretation.
For example, in this gas pump selection problem, each observed outcome (e.g., a customer selecting a gas pump) can be modeled as a random variable. This random variable can take on different values, and the probability distribution tells us how likely each outcome is.
There are different types of probability distributions, and the binomial distribution we explored earlier is just one of them. Each distribution type is defined by specific parameters. In the binomial distribution, parameters include the number of trials (\(n\)) and the probability of success (\(p\)).
Knowing the probability distribution of a variable helps in making predictions and in understanding the underlying data structure. It provides a mathematical framework to interpret the random behavior within the sample data, aiding in hypothesis testing and decision-making.
In the context of this exercise, the binomial distribution helps to model the probability of different numbers of customers choosing even-numbered pumps, which is integral to conducting the likelihood ratio test effectively.

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Most popular questions from this chapter

A plan for an executive traveler's club has been developed by an airline on the premise that \(5 \%\) of its current customers would qualify for membership. A random sample of 500 customers yielded 40 who would qualify. a. Using this data, test at level \(.01\) the null hypothesis that the company's premise is correct against the alternative that it is not correct. b. What is the probability that when the test of part (a) is used, the company's premise will be judged correct when in fact \(10 \%\) of all current customers qualify?

A regular type of laminate is currently being used by a manufacturer of circuit boards. A special laminate has been developed to reduce warpage. The regular laminate will be used on one sample of specimens and the special laminate on another sample, and the amount of warpage will then be determined for each specimen. The manufacturer will then switch to the special laminate only if it can be demonstrated that the true average amount of warpage for that laminate is less than for the regular laminate. State the relevant hypotheses, and describe the type I and type II errors in the context of this situation.

The sample average unrestrained compressive strength for 45 specimens of a particular type of brick was computed to be \(3107 \mathrm{psi}\), and the sample standard deviation was 188. The distribution of unrestrained compressive strength may be somewhat skewed. Does the data strongly indicate that the true average unrestrained compressive strength is less than the design value of 3200 ? Test using \(\alpha=.001\).

One method for straightening wire before coiling it to make a spring is called "roller straightening." The article "The Effect of Roller and Spinner Wire Straightening on Coiling Performance and Wire Properties" (Springs, 1987: 27-28) reports on the tensile properties of wire. Suppose a sample of 16 wires is selected and each is tested to determine tensile strength \(\left(\mathrm{N} / \mathrm{mm}^{2}\right)\). The resulting sample mean and standard deviation are 2160 and 30 , respectively. a. The mean tensile strength for springs made using spinner straightening is \(2150 \mathrm{~N} / \mathrm{mm}^{2}\). What hypotheses should be tested to determine whether the mean tensile strength for the roller method exceeds 2150 ? b. Assuming that the tensile strength distribution is approximately normal, what test statistic would you use to test the hypotheses in part (a)? c. What is the value of the test statistic for this data? d. What is the \(P\)-value for the value of the test statistic computed in part (c)?

A new method for measuring phosphorus levels in soil is described in the article "A Rapid Method to Determine Total Phosphorus in Soils" (Soil Sci. Amer. \(J ., 1988: 1301-1304)\). Suppose a sample of 11 soil specimens, each with a true phosphorus content of \(548 \mathrm{mg} / \mathrm{kg}\), is analyzed using the new method. The resulting sample mean and standard deviation for phosphorus level are 587 and 10 , respectively. a. Is there evidence that the mean phosphorus level reported by the new method differs significantly from the true value of \(548 \mathrm{mg} / \mathrm{kg}\) ? Use \(\alpha=.05\). b. What assumptions must you make for the test in part (a) to be appropriate?
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