/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 657 Consider the random variable \(\... [FREE SOLUTION] | 91影视

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Consider the random variable \(\mathrm{X}\) which has a binomial distribution with \(\mathrm{n}=5\) and the probability of success on a single trial, \(\theta\). Let \(\mathrm{f}(\mathrm{x} ; \theta)\) denote the probability distribution function of \(\mathrm{X}\) and let \(\mathrm{H}_{0}: \theta=1 / 2\) and \(\mathrm{H}_{1}: \theta=3 / 4\). Let the level of significance \(\alpha=1 / 32\). Determine the best critical region for the test of the null hypothesis \(\mathrm{H}_{0}\) against the alternate hypothesis \(\mathrm{H}_{1}\). Do the same for \(\alpha=6 / 32\).

Short Answer

Expert verified
The best critical region for the given hypotheses and significance levels are as follows: For 伪=1/32, the critical region is when x鈮4; and for 伪=6/32, the critical region is when x鈮3.

Step by step solution

01

Establish the binomial distribution for given hypotheses H0 and H1

For binomial distribution, the probability distribution function (PDF) can be given as: \( f(x; \theta) = \binom{n}{x} \theta^x (1-\theta)^{n-x} \) Given H鈧: 胃=1/2 and H鈧: 胃=3/4, we will plug in the values of 胃 for each hypothesis.
02

Find the PDF for each hypothesis

For H鈧, using 胃=1/2, we have: \( f(x;1/2) = \binom{5}{x} \left(\frac{1}{2}\right)^x \left(\frac{1}{2}\right)^{5-x} \) For H鈧, using 胃=3/4, we have: \( f(x;3/4) = \binom{5}{x} \left(\frac{3}{4}\right)^x \left(\frac{1}{4}\right)^{5-x} \)
03

Use Neyman-Pearson Lemma to find the critical region for each significance level

Neyman-Pearson Lemma states that the critical region for a given significance level 伪 will have \( \frac{f(x;H鈧)}{f(x;H鈧)} > k \), where k is a constant determined by the given significance level 伪. For 伪=1/32, we have to calculate k and find the critical region such that: \( \frac{f(x;3/4)}{f(x;1/2)} > k \), After simplification, we get: \(k < \frac{\left(\frac{3}{2}\right)^x}{32} \) To find the best critical region, we must find the smallest x such that the inequality holds true. After testing all possible values of x (0 to 5), we find that the critical region for 伪=1/32 is when x鈮4. Similarly, for 伪=6/32, we find the critical region such that: \( \frac{f(x;3/4)}{f(x;1/2)} > k \), After simplification, we get: \(k < \frac{\left(\frac{3}{2}\right)^x}{6} \) Testing all possible values for x, we find that the critical region for 伪=6/32 is when x鈮3. Final results are: Best critical region for 伪=1/32 is when x鈮4. Best critical region for 伪=6/32 is when x鈮3.

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Key Concepts

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

Probability Distribution Function
In probability and statistics, a Probability Distribution Function (PDF) is a vital concept that describes the likelihood of different outcomes for a random variable. For any complete distribution, the sum of probabilities for all possible outcomes equals 1.
For a binomial distribution, which models the number of successes in a given number of trials, the PDF is expressed as:
  • \( f(x; \theta) = \binom{n}{x} \theta^x (1-\theta)^{n-x} \)
Here, \(n\) is the number of trials, \(x\) is the number of successes, and \(\theta\) represents the probability of success on a single trial. This formula calculates the probability of achieving exactly \(x\) successes in \(n\) trials.
When we encounter exercises involving a hypothesis like \( H_0: \theta = 1/2 \) and \( H_1: \theta = 3/4 \), the PDF allows us to evaluate how likely it is to observe these outcomes given different assumptions about \(\theta\). This ability to calculate probabilities is crucial for hypothesis testing, as it provides the foundation for determining the plausibility of different hypotheses based on observed data.
Neyman-Pearson Lemma
The Neyman-Pearson Lemma is a fundamental result in statistical hypothesis testing that identifies how to construct the most powerful test for distinguishing between two simple hypotheses. In our context, we're considering the null hypothesis (\(H_0\)) and an alternate hypothesis (\(H_1\)).
The core principle of the Neyman-Pearson Lemma is to create a test with a critical region that maximizes the probability of correctly rejecting \(H_0\) when \(H_1\) is true. This is achieved by comparing the likelihood ratios of the hypotheses. Specifically, if we calculate:
  • \( \frac{f(x; H_1)}{f(x; H_0)} > k \),
where \(k\) is a constant determined by the chosen level of significance (\( \alpha \)), then the values of \(x\) that satisfy this inequality form the critical region of the most powerful test.
In our exercise, different \(\alpha\) values (like 1/32 and 6/32) lead to different \(k\) values, influencing the size of the critical region. Finding this region involves calculating \(k\) by observing how \(x\) values satisfy the likelihood inequality. We determine the smallest value of \(x\) for which the inequality holds, tailoring the test to the specific significance level.
Hypothesis Testing
Hypothesis testing is a method used in statistics to make decisions or inferences about population parameters based on sample data. It typically involves two conflicting hypotheses: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)).
  • The null hypothesis is a default statement suggesting that there is no effect or no difference, often formulated as a statement of equality.
  • The alternative hypothesis proposes a statement of inequality or that there is an effect or a difference.
In our problem, \(H_0: \theta = 1/2\) signifies that the probability of success is half, while \(H_1: \theta = 3/4\) suggests a higher probability of success. We perform a hypothesis test to determine if the observed data provides enough evidence to support rejecting \(H_0\) in favor of \(H_1\).
The test uses a level of significance \( \alpha \), which represents the probability of incorrectly rejecting \(H_0\) (a Type I error). In our example, by setting \( \alpha = 1/32 \) or \(6/32\), we decide how strict our test is. A smaller \( \alpha \) means we require more convincing evidence to reject \(H_0\).
The outcome of hypothesis testing helps make informed conclusions, either rejecting \(H_0\) or failing to reject it, thus guiding decisions in statistical studies.

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Most popular questions from this chapter

You are studying problem solving performance using time as a measure. Your subjects usually either solve the problems quickly or take a long time, but only rarely do they take an intermediate amount of time. You select a sample of 10 subjects and give them training that you believe will reduce their mean time for solving the problem set. You wish to compare their performance with that of another sample of 10 who did not receive the instruction. Outline an appropriate statistical test and suggest changes in the plan of the experiment. Use a level of significance of \(.05\).

Suppose for the previous problem a sample of 15 operations is obtained and the sample mean of these operations is \(6.87\) minutes with a standard deviation of 4 minute. Would these results indicate that the worker deviates from the standard of \(6.4\) minutes at a \(1 \%\) level of significance?

Two independent random samples of size \(\mathrm{n}_{1}=10\) and \(\mathrm{n}_{2}=7\) were observed to have sample variances of \(\mathrm{S}_{1}^{2}=16\) and \(\mathrm{S}_{2}^{2}=3\). Using a \(10 \%\) level of significance, test \(\mathrm{H}_{0}: \sigma_{1}^{2}=\sigma_{2}^{2}\) against \(\mathrm{H}_{1}: \sigma_{1}^{2} \neq \sigma_{2}^{2}\). Then using a \(5 \%\) level of significance, test \(\mathrm{H}_{0}\) against \(\mathrm{H}_{1}: \sigma_{1}^{2}>\sigma_{2}^{2}\) and \(\mathrm{H}_{1}: \sigma_{1}^{2}<\sigma_{2}^{2}\)

Suppose that for the preceding problem, we have taken a sample of 15 pairs of specimens. The differences in the average maximum pits for each coating, \(\underline{d}\) is 8 with a standard deviation for the differences of \(11.03\). Test whether there is a significant difference between the corrosion preventing abilities of coatings \(\mathrm{A}\) and \(\mathrm{B}\) at a \(.05\) level of significance.

For the following samples of data, compute \(\mathrm{t}\) and determine whether \(\mu_{1}\) is significantly less than \(\mu_{2}\). For your test use a level of significance of \(10 .\) Sample \(1: \mathrm{n}=10, \underline{\mathrm{X}}=10.0\) \(\mathrm{S}=5.2\); sample \(2: \mathrm{n}=10, \underline{\mathrm{X}}=13.3, \mathrm{~S}=5.7\)

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