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Chapter 8: Problem 33

Suppose that \(X_{1}, X_{2}, \ldots, X_{n}\) are i.i.d. \(N\left(\mu, \sigma^{2}\right),\) where \(\mu\) and \(\sigma\) are unknown. How should the constant \(c\) be chosen so that the interval \((-\infty, \bar{X}+c)\) is a \(95 \%\) confidence interval for \(\mu ;\) that is, \(c\) should be chosen so that \(P(-\infty<\mu \leq\) \(\bar{X}+c)=.95\)

Short Answer

Expert verified
Choose \(c = 1.645 \frac{\sigma}{\sqrt{n}}\).

Step by step solution

01

Understand the Problem

We need to find a constant \(c\) such that the interval \((-\infty, \overline{X} + c)\) is a 95% confidence interval for the mean \(\mu\) of a normal distribution. This means that 95% of the time, the true mean \(\mu\) should be less than or equal to \(\overline{X} + c\).
02

Recognize the Standard Normal Distribution

Since \(X_1, X_2, \ldots, X_n\) are i.i.d. normal variables, any linear combination of these variables, such as the sample mean \(\overline{X}\), is also normally distributed with mean \(\mu\) and variance \(\frac{\sigma^2}{n}\). We convert this into a standard normal variable \(Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}\).
03

Apply the Concept of Confidence Interval

To make \(( -\infty, \overline{X} + c)\) a 95% confidence interval, we need to equate this probability to being less than the point where 95% of the standard normal distribution falls. This can be expressed as \( P(-\infty < \mu \leq \overline{X} + c) = 0.95 \).
04

Find the Critical Value

From the standard normal distribution table, the z-score that cuts off 5% in the upper tail is approximately 1.645. Therefore, we set \( \overline{X} + c = \overline{X} + 1.645 \frac{\sigma}{\sqrt{n}} \).
05

Solve for c

Thus, solve for \(c\) to align with the z-score: \(c = 1.645 \frac{\sigma}{\sqrt{n}}\). This ensures that \(\mu\) falls within the interval \((-\infty, \overline{X} + c)\) with a probability of 95%.

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Key Concepts

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

Normal Distribution
A normal distribution, often referred to as a bell curve due to its shape, is a type of continuous probability distribution for a real-valued random variable. It plays a critical role in statistics because many patterns in nature and data science fit a normal distribution. One key characteristic is that it's symmetrical; the mean, median, and mode all coincide at the highest point of the curve.
This symmetry implies that most values cluster around a central region, with occurrences tapering off as you move away from the mean. Many statistical tests and estimates, including confidence intervals, assume that the data follows a normal distribution.
  • The shape is entirely determined by two parameters: the mean (\( \mu \)), which indicates where the bell is centered, and the standard deviation (\( \sigma \)), which determines the width of the bell.
  • In the context of an exercise like this, understanding that a sample mean is also normally distributed is crucial as it justifies using the normal distribution for interval estimation.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how much the values in the data set deviate from the mean on average. A small standard deviation means that the values cluster close to the mean, while a large standard deviation indicates that the values are spread out over a wider range.
In statistics, standard deviation is especially important as it influences how we interpret data. For instance, in evaluating the reliability of the sample mean from a set of observations, a smaller standard deviation suggests more confidence in the sample mean as a representation of the population mean.
  • Formulaically, standard deviation is denoted as \( \sigma \) (for a population) or \( s \) (for a sample) and calculated as the square root of the variance.
  • In the context of the confidence interval exercise, the standard deviation paired with sample size helps estimate the margin of error around the sample mean.
Z-Score
A z-score measures how many standard deviations an element is from the mean. In a normal distribution, z-scores allow us to determine the probability of a value occurring under the bell curve. Essentially, it helps standardize different data points to be compared on the same scale.
Z-scores are especially useful for confidence intervals, as they determine the extent to which the sample mean lays within a specific range. For instance, knowing that a z-score of 1.645 indicates the 95% quantile in a standard normal distribution helps in calculating the appropriate margin in our confidence interval.
  • The z-score is calculated as \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value in question, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
  • In the exercise, we determine \( c = 1.645 \frac{\sigma}{\sqrt{n}} \), connecting directly to the z-score needed to establish the confidence interval.
Sample Mean
The sample mean, denoted as \( \bar{X} \), is the average of all sample data points and often acts as an estimate of the population mean \( \mu \). It provides a measure of the center of the data set.
Understanding the sample mean is critical for creating confidence intervals because it acts as the best estimate for the unknown population mean. Any statistical deductions or confidence intervals drawn generally revolve around this estimate.
  • The formula for the sample mean is \( \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i \).
  • In the problem at hand, the sample mean \( \bar{X} \) forms one end of the interval, with the term \( + c \) denoting how far to extend this interval to achieve the desired confidence level.

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Most popular questions from this chapter

In Example A of Section \(8.4,\) we used knowledge of the exact form of the sampling distribution of \(\hat{\lambda}\) to estimate its standard error by $$s_{\hat{\lambda}}=\sqrt{\frac{\hat{\lambda}}{n}}$$ This was arrived at by realizing that \(\sum X_{i}\) follows a Poisson distribution with parameter \(n \lambda_{0} .\) Now suppose we hadn't realized this but had used the bootstrap, letting the computer do our work for us by generating \(B\) samples of size \(n=23\) of Poisson random variables with parameter \(\lambda=24.9,\) forming the mle of \(\lambda\) from each sample, and then finally computing the standard deviation of the resulting collection of estimates and taking this as an estimate of the standard error of \(\hat{\lambda}\) Argue that as \(B \rightarrow \infty,\) the standard error estimated in this way will tend to \(s_{\hat{\lambda}}\).

Suppose that \(X_{1}, X_{2}, \ldots, X_{n}\) are i.i.d. \(N\left(\mu_{0}, \sigma_{0}^{2}\right)\) and \(\mu\) and \(\sigma^{2}\) are estimated by the method of maximum likelihood, with resulting estimates \(\hat{\mu}\) and \(\hat{\sigma}^{2} .\) Suppose the bootstrap is used to estimate the sampling distribution of \(\hat{\mu}\) a. Explain why the bootstrap estimate of the distribution of \(\hat{\mu}\) is \(N\left(\hat{\mu}, \frac{\hat{\sigma}^{2}}{n}\right)\). b. Explain why the bootstrap estimate of the distribution of \(\hat{\mu}-\mu_{0}\) is \(N\left(0, \frac{\hat{\sigma}^{2}}{n}\right)\). c. According to the result of the previous part, what is the form of the bootstrap confidence interval for \(\mu,\) and how does it compare to the exact confidence interval based on the \(t\) distribution?

The upper quartile of a distribution with cumulative distribution \(F\) is that point \(q_{.25}\) such that \(F\left(q_{.25}\right)=.75 .\) For a gamma distribution, the upper quartile depends on \(\alpha\) and \(\lambda,\) so denote it as \(q(\alpha, \lambda) .\) If a gamma distribution is fit to data as in Example \(\mathrm{C}\) of Section 8.5 and the parameters \(\alpha\) and \(\lambda\) are estimated by \(\hat{\alpha}\) and \(\hat{\lambda}\) the upper quartile could then be estimated by \(\hat{q}=q(\hat{\alpha}, \hat{\lambda}) .\) Explain how to use the bootstrap to estimate the standard error of \(\hat{q}\).

Show that the gamma distribution is a conjugate prior for the exponential distribution. Suppose that the waiting time in a queue is modeled as an exponential random variable with unknown parameter \(\lambda,\) and that the average time to serve a random sample of 20 customers is 5.1 minutes. A gamma distribution is used as a prior. Consider two cases: (1) the mean of the gamma is 0.5 and the standard deviation is \(1,\) and (2) the mean is 10 and the standard deviation is \(20 .\) Plot the two posterior distributions and compare them. Find the two posterior means and compare them. Explain the differences.

Suppose that \(X_{1}, X_{2}, \ldots, X_{n}\) are i.i.d. random variables on the interval [0,1] with the density function $$f(x | \alpha)=\frac{\Gamma(2 \alpha)}{\Gamma(\alpha)^{2}}[x(1-x)]^{\alpha-1}$$ where \(\alpha>0\) is a parameter to be estimated from the sample. It can be shown that $$\begin{aligned} E(X) &=\frac{1}{2} \\ \operatorname{Var}(X) &=\frac{1}{4(2 \alpha+1)} \end{aligned}$$ a. How does the shape of the density depend on \(\alpha ?\) b. How can the method of moments be used to estimate \(\alpha ?\) c. What equation does the mle of \(\alpha\) satisfy? d. What is the asymptotic variance of the mle? e. Find a sufficient statistic for \(\alpha .\)
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