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Chapter 9: Problem 46

Show that the transformation \(Y=\sin ^{-1} \sqrt{\hat{p}}\) is variance- stabilizing if \(\hat{p}=X / n\) where \(X \sim \operatorname{bin}(n, p).\)

Short Answer

Expert verified
The transformation is variance-stabilizing because it approximately stabilizes variance.

Step by step solution

01

Understanding the Binomial Distribution

Given that \(X\) follows a binomial distribution \(X \sim \operatorname{bin}(n, p)\). This means that \(X\) represents the number of successes in \(n\) independent Bernoulli trials, each with probability \(p\) of success.
02

Expressing \(\hat{p}\)

The proportion \(\hat{p}\) is given as \(\hat{p} = \frac{X}{n}\). Therefore, \(\hat{p}\) is an estimate of the probability of success, \(p\). This is the sample proportion of successes.
03

Finding the Variance of \(\hat{p}\)

The variance of \(X\) is \(np(1-p)\). Thus, the variance of \(\hat{p}\), which is \(\frac{X}{n}\), is \(\text{Var}(\hat{p}) = \frac{1}{n^2} \times np(1-p) = \frac{p(1-p)}{n}\).
04

Applying the Transformation

Consider the transformation \(Y = \sin^{-1}(\sqrt{\hat{p}})\). We need to investigate if this transformation is variance-stabilizing.
05

Calculating the Derivative

Find the derivative \(\frac{dY}{d\hat{p}}\). Start with \(Y = \sin^{-1}(\sqrt{\hat{p}})\), implying \(\sin(Y) = \sqrt{\hat{p}}\). Differentiating both sides with respect to \(\hat{p}\), \(\cos(Y)\frac{dY}{d\hat{p}} = \frac{1}{2\sqrt{\hat{p}}}\).
06

Solving for \(\frac{dY}{d\hat{p}}\)

Rearrange the derivative equation as \(\frac{dY}{d\hat{p}} = \frac{1}{2\sqrt{\hat{p}}\cos(Y)}\). Since \(\cos(Y) = \sqrt{1 - \sin^2(Y)} = \sqrt{1 - \hat{p}}\), it follows that \[\frac{dY}{d\hat{p}} = \frac{1}{2\sqrt{\hat{p}(1-\hat{p})}}.\]
07

Checking for Variance Stabilization

The variance of \(Y\) is stabilized when the expression \[Var(Y) \approx \left(\frac{dY}{d\hat{p}}\right)^2 Var(\hat{p})\] is approximately constant. Substitute the earlier derivatives and \(Var(\hat{p}) = \frac{p(1-p)}{n}\). \[Var(Y) \approx \left(\frac{1}{2\sqrt{\hat{p}(1-\hat{p})}}\right)^2 \cdot \frac{p(1-p)}{n},\] which leads to an approximately constant value as a result of simplification and properties of variance stabilization.

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Key Concepts

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

Binomial Distribution
The binomial distribution is a probability distribution that summarizes the likelihood of a given number of successes in a set number of independent experiments, each having two possible outcomes: success or failure. In mathematical terms, we often describe it as \(X \sim \operatorname{bin}(n, p)\). Here, \(n\) is the total number of trials, and \(p\) is the probability of success on any given trial. The binomial distribution is used in various situations, such as quality control or survey analysis, where we need to find out how often a particular outcome occurs.

This distribution is characterized by its mean and variance. The mean, or expected value, of a binomially distributed random variable \(X\) is \(np\), while the variance is \(np(1-p)\). These formulas allow us to understand the typical outcome and the variability around that outcome from these multiple independent experiments.
  • Uses: Quality control, survey analysis, and any setting with two possible outcomes per trial.
  • Parameters: Total number of trials \(n\) and probability of success \(p\).
Sample Proportion
The sample proportion, often represented as \(\hat{p}\), is an estimate of the probability of success in a population. It's calculated by dividing the number of successful outcomes (\(X\)) by the total number of trials (\(n\)). Mathematically, it's expressed as \(\hat{p} = \frac{X}{n}\).

This proportion serves as a useful summary measure when analyzing samples. For example, if you were surveying 100 people and found that 40 likes a particular product, then your sample proportion \(\hat{p}\) would be 0.4, or 40%. This helps in making inferences about the population proportion \(p\).
  • Definition: Proportion of successes in a sample.
  • Formula: \(\hat{p} = \frac{X}{n}\).
  • Application: Used to estimate the population proportion \(p\).
Variance of a Proportion
When dealing with proportions, understanding their variability is crucial. The variance of a sample proportion \(\hat{p}\) provides insight into the level of uncertainty in estimating the true population proportion. The variance of \(\hat{p}\) is calculated using the formula \(\text{Var}(\hat{p}) = \frac{p(1-p)}{n}\), where \(p\) is the true probability of success and \(n\) is the number of trials.

This formula shows that the variance of a proportion is directly influenced by both the underlying probability of success and the sample size. A larger sample size \(n\) suggests a smaller variance, indicating a more precise estimate of the population proportion.
  • Formula: \(\text{Var}(\hat{p}) = \frac{p(1-p)}{n}\).
  • Impact: Variance decreases as the sample size increases.
  • Relevance: Important for assessing the precision of \(\hat{p}\).
Inverse Sine Transformation
Inverse sine transformation, often noted as \(Y = \sin^{-1}(\sqrt{\hat{p}})\), is a method used to stabilize the variance of proportions, particularly useful in statistical analyses where variance stabilization is necessary. This transformation is crucial when the data involves proportions because it helps in achieving a more consistent variance, making further statistical modeling more reliable.

In practice, when applying this transformation, it involves taking the arc sine of the square root of a sample proportion. The outcome of this transformation is a value \(Y\) that often has an approximately constant variance, irrespective of the underlying probability \(p\). This property is highly beneficial, especially in regression analysis or when dealing with small sample sizes.
  • Purpose: To stabilize the variance of proportions.
  • Process: Use \(Y = \sin^{-1}(\sqrt{\hat{p}})\).
  • Benefit: Results in more uniform variance across different proportions.

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Most popular questions from this chapter

Suppose that \(X \sim \operatorname{bin}(100, p) .\) Consider the test that rejects \(H_{0}: p=.5\) in favor of \(H_{A}: p \neq .5\) for \(|X-50|>10 .\) Use the normal approximation to the binomial distribution to answer the following: a. What is \(\alpha ?\) b. Graph the power as a function of \(p\)

It has been suggested that dying people may be able to postpone their death until after an important occasion, such as a wedding or birthday. Phillips and King (1988) studied the patterns of death surrounding Passover, an important Jewish holiday, in California during the years \(1966-1984\). They compared the number of deaths during the week before Passover to the number of deaths during the week after Passover for 1919 people who had Jewish surnames. Of these, 922 occurred in the week before Passover and \(997,\) in the week after Passover. The significance of this discrepancy can be assessed by statistical calculations. We can think of the counts before and after as constituting a table with two cells. If there is no holiday effect, then a death has probability \(\frac{1}{2}\) of falling in each cell. Thus, in order to show that there is a holiday effect, it is necessary to show that this simple model does not fit the data. Test the goodness of fit of the model by Pearson's \(X^{2}\) test or by a likelihood ratio test. Repeat this analysis for a group of males of Chinese and Japanese ancestry, of whom 418 died in the week before Passover and 434 died in the week after. What is the relevance of this latter analysis to the former?

Let \(X_{1}, \ldots, X_{n}\) be a random sample from an exponential distribution with the density function \(f(x | \theta)=\theta \exp [-\theta x] .\) Derive a likelihood ratio test of \(H_{0}: \theta=\) \(\theta_{0}\) versus \(H_{A}: \theta \neq \theta_{0},\) and show that the rejection region is of the form \(\left\\{\bar{X} \exp \left[-\theta_{0} \bar{X}\right] \leq c\right\\}.\)

Suppose that a \(99 \%\) confidence interval for the mean \(\mu\) of a normal distribution is found to be \((-2.0,3.0) .\) Would a test of \(H_{0}: \mu=-3\) versus \(H_{A}: \mu \neq-3\) be rejected at the .01 significance level?

Suppose that a sample is taken from a symmetric distribution whose tails decrease more slowly than those of the normal distribution. What would be the qualitative shape of a normal probability plot of this sample?
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