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Chapter 5: Problem 28

Let \(Y_{1}

Short Answer

Expert verified
In part (a), we have shown that \(P\left(Y_{1}<\mu<Y_{2}\right)=\frac{1}{2}\) and \(E(Y_{2}-Y_{1}) = \sigma\frac{2\sqrt{2}}{\sqrt{\pi}}\). In part (b), the constant \(c\) is approximately -0.6745, and the interval for \(\mu\) centered about the sample mean is longer (approximately \(1.349\sigma\)) than the expected value of the interval defined by the order statistics (\(Y_{2}-Y_{1}\)).

Step by step solution

01

Show \(P\left(Y_{1}

For a random sample from a Normal distribution, each sample observation is equally likely to be the smallest or largest in the sample. Hence, the probability of \(\mu\) (the population mean) falling between \(Y_{1}\) and \(Y_{2}\) can be deduced as \(\frac{1}{2}\). In formal terms, \(P\left(Y_{1}<\mu<Y_{2}\right) = P\left(Y_{1}<Y_{2}\right) = \frac{1}{2}\).
02

Compute Expected Value \(E(Y_{2}-Y_{1})\)

Proceeding to compute the expected length of the random interval \(Y_{2}-Y_{1}\), using E(aX - bY) = aE(X) - bE(Y), where a, b are constants and X, Y are random variables, we get E(Y_{2}-Y_{1}) = E(Y_{2}) - E(Y_{1}). The expected values E(Y_{1}) and E(Y_{2}) can be obtained from the properties of order statistics of normal distribution, where E(Y_{1}) = \(\mu - \sigma\frac{\sqrt{2}}{\sqrt{\pi}}\) and E(Y_{2}) = \(\mu + \sigma\frac{\sqrt{2}}{\sqrt{\pi}}\). Substituting these values we get E(Y_{2}-Y_{1}) = \(\sigma\frac{2\sqrt{2}}{\sqrt{\pi}}\).
03

Solve for constant \(c\)

In part(b), we wish to find a constant \(c\) which results in \(P(\bar{X}-c \sigma<\mu<\bar{X}+c \sigma)=\frac{1}{2}\). This can be interpreted as finding a symmetric \(100(1-0.5)% = 50%\)-confidence interval for \(\mu\) around \(\bar{X}\), which we know has form \(\bar{X} \pm Z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}\), where \(Z_{\frac{\alpha}{2}}\) is a critical value from the standard normal distribution for two-tailed confidence interval of \(100(1-\alpha)%\). Since the interval is 50%, \(\alpha = 0.5\), and \(Z_{\frac{\alpha}{2}} = Z_{0.25}\), which gives \(Z_{\frac{\alpha}{2}} = -0.6745\). Setting this to \(c \sigma\), we get \(c = -0.6745\).
04

Compare lengths of intervals

The length of the interval for \(\mu\) can be calculated as \(2c\sigma\). Using \(c = -0.6745\), we find that the length of this interval is roughly \(1.349\sigma\). Comparing this with the expected length of the interval in part(a) \(E(Y_{2}-Y_{1}) = \sigma\frac{2\sqrt{2}}{\sqrt{\pi}}\) approximately \(1.1283\sigma\), we find that the interval around the sample mean is longer.

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Key Concepts

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

Normal Distribution
The Normal Distribution is a cornerstone of statistics. It's often called the bell curve because of its characteristic shape. The curve is symmetric around its mean, which means the majority of the data are close to the mean, with fewer instances as you move away from the center.
This distribution is defined by two parameters: the mean (\(\mu\) and the variance (\(\sigma^2\)).
  • The mean (\(\mu\) indicates the location of the peak of the curve.
  • The variance (\(\sigma^2\)) describes how spread out the values are around the mean.
Many natural phenomena tend to follow a normal distribution, which is why it plays a significant role in statistical analysis.
Expected Value
The Expected Value is a fundamental concept in probability and statistics that provides the long-run average outcome of a random variable. For independent events, like rolling a dice or drawing a card, it's how we'd anticipate results over several trials.
For a random variable \(X\), with potential outcomes \(x_i\) and corresponding probabilities \(p_i\), the expected value is calculated as:
\[ E(X) = \sum_{i} x_i p_i \]
This allows us to predict the mean or average result if an experiment is repeated many times. Therefore, the expected value doesn't always denote the most likely outcome but rather an average over time.
  • In the exercise, finding \(E(Y_2 - Y_1)\) involves calculating the expected length between the highest and lowest values of the sample.
  • This calculation uses properties of order statistics from the normal distribution.
Confidence Intervals
Confidence Intervals provide a range of values which are believed to contain a parameter of interest, like a population mean, with a certain level of confidence. They are crucial in inferential statistics.
Instead of giving a single estimate, they offer a range (interval) that should include the parameter, with a given probability (confidence level, e.g., 95%).
Some key points about confidence intervals are:
  • The interval is calculated as \( \bar{X} \pm c \cdot \sigma \), where \( \bar{X} \) is the sample mean and \( \sigma \) is the standard deviation.
  • This interval is based on the normal distribution characteristics, which makes it symmetric around the mean.
In the given problem, we use a 50% confidence interval to explore a probability scenario, and we solve for the constant \(c\) to provide this interval's range. Understanding how these intervals are influenced by variance and sample size can help in determining their reliability and accuracy.

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Most popular questions from this chapter

Let \(Y_{1}

. To illustrate Exercise 5.4.22, let \(X_{1}, X_{2}, \ldots, X_{9}\) and \(Y_{1}, Y_{2}, \ldots, Y_{12}\) represent two independent random samples from the respective normal distributions \(N\left(\mu_{1}, \sigma_{1}^{2}\right)\) and \(N\left(\mu_{2}, \sigma_{2}^{2}\right) .\) It is given that \(\sigma_{1}^{2}=3 \sigma_{2}^{2}\), but \(\sigma_{2}^{2}\) is unknown. Define a random variable which has a \(t\) -distribution that can be used to find a 95 percent confidence interval for \(\mu_{1}-\mu_{2}\).

Let \(X\) be a random variable with a continuous cdf \(F(x)\). Assume that \(F(x)\) is strictly increasing on the space of \(X\). Consider the random variable \(Z=F(X)\). Show that \(Z\) has a uniform distribution on the interval \((0,1)\).

. Let \(y_{1}

Proceeding similar to Example 5.8.6, use the Accept-Reject Algorithin to generate an observation from a \(t\) distribution with \(r>1\) degrees of freedom ussing the Cauchy distribution.
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