91Ó°ÊÓ

Let \(0 \leq \gamma \leq \alpha\). Then a \(100(1-\alpha)\) \% CI for \(\mu\) when \(n\) is large is $$ \left(\bar{x}-z_{\gamma} \cdot \frac{s}{\sqrt{n}}, \bar{x}+z_{\alpha-\gamma} \cdot \frac{s}{\sqrt{n}}\right) $$ The choice \(\gamma=\alpha / 2\) yields the usual interval derived in Section 7.2; if \(\gamma \neq \alpha / 2\), this interval is not symmetric about \(\bar{x}\). The width of this interval is \(w=s\left(z_{\gamma}+z_{\alpha-\gamma}\right) / \sqrt{n}\). Show that \(w\) is minimized for the choice \(\gamma=\alpha / 2\), so that the symmetric interval is the shortest. [Hints: (a) By definition of \(z_{a}, \Phi\left(z_{a}\right)=1-\alpha\), so that \(z_{a}=\Phi^{-1}(1-\alpha)\); (b) the relationship between the derivative of a function \(y=f(x)\) and the inverse function \(x=f^{-1}(y)\) is \((d / d y)\) \(\left.f^{-1}(y)=1 / f^{\prime}(x) .\right]\)

Step by step solution

01

Define the Interval Width Function

The width of the confidence interval is given by the expression: \[ w = \frac{s}{\sqrt{n}} \left( z_{\gamma} + z_{\alpha - \gamma} \right). \] We need to show that this function is minimized when \( \gamma = \frac{\alpha}{2} \).

02

Understand z-values Behavior

Recall that \(z_a\) represents the critical value where the cumulative distribution \(\Phi(z_a) = 1-a\). For any \(\gamma\), \[ z_{\gamma} = \Phi^{-1}(1 - \gamma) \] and \[ z_{\alpha - \gamma} = \Phi^{-1}(1 - (\alpha - \gamma)). \]

03

Find the Derivative of the Width Function

The function for the width can be rewritten as the composition of inverse functions based on \(\Phi\):\[ w(\gamma) = \frac{s}{\sqrt{n}}(\Phi^{-1}(1 - \gamma) + \Phi^{-1}(1 - (\alpha - \gamma))). \] To find the minimum, we differentiate this function with respect to \( \gamma \).

04

Use Chain Rule for Derivatives

Using the chain rule, the derivative of the width function \( w(\gamma) \) is:\[ \frac{d w}{d \gamma} = \frac{s}{\sqrt{n}} \left(-\frac{1}{\Phi'(z_{\gamma})} + \frac{1}{\Phi'(z_{\alpha - \gamma})} \right), \]where \( \Phi' \) represents the derivative of the normal CDF, or the PDF of the standard normal distribution.

05

Set Derivative to Zero for Minimization

We want to find where \( \frac{d w}{d \gamma} = 0 \):\[ \frac{1}{\Phi'(z_{\gamma})} = \frac{1}{\Phi'(z_{\alpha - \gamma})}. \]This implies \( \Phi'(z_{\gamma}) = \Phi'(z_{\alpha - \gamma}) \), meaning \( z_{\gamma} = z_{\alpha - \gamma} \).

06

Determine Symmetry Condition

If \( z_{\gamma} = z_{\alpha - \gamma} \), then:\[ 1 - \gamma = \alpha - \gamma \Rightarrow \gamma = \frac{\alpha}{2}. \]This calculation shows that the width is minimized for symmetric intervals where \( \gamma = \frac{\alpha}{2} \).

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Key Concepts

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

Symmetric Interval

A symmetric interval, in the context of confidence intervals, is one that is evenly distributed around the sample mean, \( \bar{x} \). This means that the lower and upper bounds of the interval are equidistant from \( \bar{x} \).
The formula provided gives a confidence interval of the form \( (\bar{x}-z_{\gamma} \cdot \frac{s}{\sqrt{n}}, \bar{x}+z_{\alpha-\gamma} \cdot \frac{s}{\sqrt{n}}) \).

• When \( \gamma = \frac{\alpha}{2} \), the interval becomes symmetric.
• This configuration is commonly used because it balances the probability error between the two tails of a normal distribution curve, ensuring that the interval captures the true mean accurately.

A symmetric interval is efficient because it takes advantage of the properties of the standard normal distribution to ensure that both sides of the mean have the same confidence level.

Interval Width Minimization

Minimizing the width of a confidence interval is crucial for obtaining a precise estimate of the true population parameter. The narrower the interval, the more precise the estimate. For the interval \( w = \frac{s}{\sqrt{n}}(z_{\gamma} + z_{\alpha - \gamma}) \), the goal is to find the value of \( \gamma \) that makes \( w \) as small as possible.
The key to minimizing this interval width lies in setting \( \gamma \) to \( \frac{\alpha}{2} \). This selection balances the critical values, making \( z_{\gamma} = z_{\alpha - \gamma} \), based on the properties of the normal distribution. This equality ensures that

• the cumulative probabilities of both tails are equal, and
• ensures that the overall interval around \( \bar{x} \) is shortest while still maintaining the desired confidence level.

In simpler terms, this choice of \( \gamma \) provides the most statistically efficient interval length, reducing possible uncertainty.

Standard Normal Distribution

The standard normal distribution is a fundamental concept in statistics, representing a normal distribution with a mean of 0 and a standard deviation of 1. It is crucial for determining critical values and confidence intervals.
Within this distribution:

• The cumulative distribution function (CDF), denoted as \( \Phi \), gives the probability that a standard normal random variable is less than a given value.
• The inverse function, \( \Phi^{-1} \), is used to find the critical value that corresponds to a specific probability \( 1-a \).

For example, if we need an interval that captures 95% of the distribution, we find critical values such that the area under the curve between them is 0.95. The function \( \Phi^{-1} \) is essential in finding these values, enabling us to construct confidence intervals correctly.

Critical Value

A critical value is a key component in constructing confidence intervals. It is a point on the distribution that relates to a specified probability level. In the standard normal distribution, these values are denoted as \( z \), such as \( z_{\gamma} \) and \( z_{\alpha - \gamma} \).

• The critical value \( z_{a} \) corresponds to the point where \( \Phi(z_{a}) = 1-a \).
• This means that a proportion \( a \) of the data lies outside \( z_{a} \) in the tails of the distribution.

These values are vital for calculating the bounds of confidence intervals.
By selecting appropriate \( \gamma \) and critical values, it is possible to fine-tune the confidence interval to ensure it meets a specific confidence level.
The specific choice of \( \gamma = \frac{\alpha}{2} \) aligns the interval symmetrically around the mean, ensuring efficiency.