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Consider the hypotheses (10.4.4). Suppose we select the score function \(\varphi(u)\) and the corresponding test based on \(W_{\varphi} .\) Suppose we want to determine the sample size \(n=n_{1}+n_{2}\) for this test of significance level \(\alpha\) to detect the alternative \(\Delta^{*}\) with approximate power \(\gamma^{*}\). Assuming that the sample sizes \(n_{1}\) and \(n_{2}\) are the same, show that $$ n \approx\left(\frac{\left(z_{\alpha}-z_{\gamma^{*}}\right) 2 \tau_{\varphi}}{\Delta^{*}}\right)^{2} $$

Short Answer

Expert verified
The required sample size \(n\) for the test of significance with level \(\alpha\), power \(\gamma^{*}\), and alternative \(\Delta^{*}\) is approximately \(n \approx\left(\frac{\left(z_{\alpha}-z_{\gamma^{*}}\right) 2 \tau_{\varphi}}{\Delta^{*}}\right)^{2}\), assuming the sample sizes \(n_{1}\) and \(n_{2}\) are the same.

Step by step solution

01

Understanding the Z-scores

The z-scores \(z_{\alpha}\) and \(z_{\gamma^{*}}\) come from the standard normal distribution. The score \(z_{\alpha}\) is the z-value such that the probability to the right of \(z_{\alpha}\) under the standard normal curve is \(\alpha\). Similar, \(z_{\gamma^{*}}\) is the z-score such that the power of the test is \(\gamma^{*}\), meaning the probability to the right of \(z_{\gamma^{*}}\) under the standard normal curve is \(1-\gamma^{*}\).
02

Express \(n\) in terms of other variables

If we rearrange the equation \(n \approx\left(\frac{\left(z_{\alpha}-z_{\gamma^{*}}\right) 2 \tau_{\varphi}}{\Delta^{*}}\right)^{2}\), we get \(n \approx \left(\frac{(z_{\alpha} - z_{\gamma^{*}})2\tau_{\varphi}}{\Delta^{*}}\right)^2\), where \(2\tau_{\varphi}\) is the standard deviation of the test statistic \(W_{\varphi}\) when the null hypothesis is true, and \(\Delta^{*}\) represents the alternative hypothesis.
03

Interpret the result

The expression for \(n\) tells us that the sample size needed for the test increases if either the level of significance \(\alpha\) decreases or the power \(\gamma^{*}\) of the test increases. It also increases as the standard deviation \(\tau_{\varphi}\) of the test statistic increases, and decreases as the magnitude of the difference represented by the alternative hypothesis \(\Delta^{*}\) increases. Essentially, as the test becomes more stringent or if there's greater variability, more data is needed. Conversely, if the magnitude of the difference we are attempting to detect increases, less data is needed.

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Key Concepts

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

Hypothesis Testing
In the realm of statistics, hypothesis testing is a critical method used to evaluate various claims about a population based on sample data. At its core, hypothesis testing involves comparing two hypotheses: the null hypothesis (ull) which is a statement of no effect or no difference, and the alternative hypothesis (ull) which is what the researcher aims to support.

For instance, if we want to test the new teaching method's effectiveness, our null hypothesis might claim that the new method does not improve test scores, while the alternative hypothesis might state that it does improve scores. A significance level (ull) is chosen to determine the threshold for rejecting the null hypothesis. Common practice is to set ull at 0.05, indicating that there is a 5% chance that we reject the null hypothesis when it is actually true (a type I error).

If evidence from the sample is strong enough to reject the null hypothesis, we conclude that our alternative hypothesis may be true. But the conclusion is never definitive; it's about probability and the avoidance of incorrect conclusions to a reasonable degree based on the data and the standard of evidence we've set (significance level).
Z-Score
A z-score is a numerical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. It is a way to standardize scores on the same scale by subtracting the mean and dividing by the standard deviation.

In hypothesis testing, the z-score helps determine how unusual or typical a data point is relative to the data set. When the z-score is far from 0, it indicates that the data point is far from the mean. Z-scores are crucial in calculating the probability of a score occurring within a standard normal distribution, and consequently in deciding whether to reject the null hypothesis.
Statistical Power
Statistical power is the likelihood that a test will correctly reject a false null hypothesis; it measures a test's ability to detect an effect if there is one. A higher power means a lower chance of making a type II error, which occurs when the test fails to reject a false null hypothesis.

Several factors influence statistical power, such as sample size, effect size, and significance level. Increasing the sample size or the effect size (the difference between the null and alternative hypotheses) typically increases power. Conversely, a lower significance level (ull) can decrease power, as it makes it harder to reject the null hypothesis.
Standard Normal Distribution
The standard normal distribution, also known as the z-distribution, is a special case of the normal distribution. It is a symmetric, bell-shaped distribution that has a mean of 0 and a standard deviation of 1. This standardization allows for the creation of z-tables, which show the probability of a z-score occurring within the distribution.

Understanding the standard normal distribution is crucial in hypothesis testing, especially when working with z-scores. It allows researchers to convert raw data into a standardized form where probabilities and critical values can be easily looked up and compared to determine the statistical significance of their results.
Alternative Hypothesis
The alternative hypothesis (ull) represents what a statistical test is set up to investigate. It is a statement that indicates the presence of an effect or difference that the researcher wants to support. Unlike the null hypothesis, which is often a statement of 'no difference' or 'no effect,' the alternative hypothesis is one of 'difference' or 'effect'.

In the context of sample size determination, the alternative hypothesis is symbolized as ull, which represents the minimum effect size we want to detect. It is central to determining the necessary sample size because the more substantial the effect we are interested in, the fewer subjects we may need to detect it. This is because larger effects are easier to distinguish from random noise or variation in the data.

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Most popular questions from this chapter

Let \(\theta\) denote the median of a random variable \(X\). Consider testing $$ H_{0}: \theta=0 \text { versus } H_{1}: \theta>0 . $$ Suppose we have a sample of size \(n=25\). (a) Let \(S(0)\) denote the sign test statistic. Determine the level of the test: reject \(H_{0}\) if \(S(0) \geq 16\) (b) Determine the power of the test in part (a) if \(X\) has \(N(0.5,1)\) distribution. (c) Assuming \(X\) has finite mean \(\mu=\theta\), consider the asymptotic test of rejecting \(H_{0}\) if \(\bar{X} /(\sigma / \sqrt{n}) \geq k\). Assuming that \(\sigma=1\), determine \(k\) so the asymptotic test has the same level as the test in part (a). Then determine the power of this test for the situation in part (b).

By considering the asymptotic power lemma, Theorem \(10.4 .2\), show that the equal sample size situation \(n_{1}=n_{2}\) is the most powerful design among designs with \(n_{1}+n_{2}=n, n\) fixed, when level and alternatives are also fixed. Hint: Show that this problem is equivalent to maximizing the function $$ g\left(n_{1}\right)=\frac{n_{1}\left(n-n_{1}\right)}{n^{2}} $$ and then obtain the result.

Often influence functions are derived by differentiating implicitly the defining equation for the functional at the contaminated cdf \(F_{x, e}(t),(10.9 .13) .\) Consider the mean functional with the defining equation (10.9.10). Using the linearity of the differential, first show that the defining equation at the cdf \(F_{x, \epsilon}(t)\) can be expressed as $$ \begin{aligned} 0=\int_{-\infty}^{\infty}\left[t-T\left(F_{x, \epsilon}\right)\right] d F_{x, \epsilon}(t)=&(1-\epsilon) \int_{-\infty}^{\infty}\left[t-T\left(F_{x, \epsilon}\right)\right] f_{X}(t) d t \\ &+\epsilon \int_{-\infty}^{\infty}\left[t-T\left(F_{x, \epsilon}\right)\right] d \Delta(t) \end{aligned} $$ Recall that we want \(\partial T\left(F_{x, \epsilon}\right) / \partial \epsilon .\) Obtain this by implicitly differentiating the above equation with respect to \(\epsilon\).

Consider the location model as defined in expression (10.9.1). Let $$ \widehat{\theta}=\operatorname{Argmin}_{\theta}\|\mathbf{X}-\theta \mathbf{1}\|_{\mathrm{LS}}^{2} $$ where \(\|\cdot\|_{\mathrm{LS}}^{2}\) is the square of the Euclidean norm. Show that \(\widehat{\theta}=\bar{x}\).

Show that Kendall's \(\tau\) satisfies the inequality \(-1 \leq \tau \leq 1\).

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