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Referring to Example \(7.9 .5\) of this section, determine \(c\) so that $$P\left(-c

Short Answer

Expert verified
To get the 95% confidence interval for \(\theta\) given \(T_{2}=t_{2}\) we would calculate \(c\) then construct the interval as everything between \(T_{1}-c\) and \(T_{1}+c\).

Step by step solution

01

Applying the definitions of conditional probability

The given relation can be interpreted based on the definitions of conditional probability \(P\left(-c<T_{1}-\theta<c \mid T_{2}=t_{2}\right)=0.95\). This states the boundary conditions that the difference of \(T_{1}\) and \(\theta\) lies within \(c\) given \(T_{2}=t_{2}\). And this boundary is given to us as 95 % probability.
02

Transforming the inequalities into a more manageable form

Rewrite the probability as \(P\left(T_{1}-c <\theta< T_{1}+c \mid T_{2}=t_{2}\right)=0.95\). This is the standard form of a two-sided confidence interval - the parameter \(\theta\) lies between \(T_{1}-c\) and \(T_{1}+c, given T_{2}=t_{2}\) with 95 % probability.
03

Constructing the confidence interval

This problem doesn't provide functional relationships between \(\theta, T_{1}, \) and \(T_{2}\), it's not possible to provide a specific numeric confidence interval or value of \(c\). In a real world scenario, you would use your estimate of \(T_{1}\) and your calculated value of \(c\), then your 95% confidence interval for \(\theta\) would be \((T_{1}-c, T_{1}+c)\) given \(T_{2}=t_{2}\) .

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Key Concepts

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

Conditional Probability and Its Importance
Conditional probability is a crucial concept in statistics that helps us assess the likelihood of an event occurring, considering that another related event has already occurred. It's expressed as \( P(A | B) \), meaning the probability of event \( A \) given that \( B \) has occurred. This concept helps in understanding many statistical phenomena and is frequently used in various fields
like finance, medicine, and even daily decision-making situations. In this exercise, we use conditional probability to analyze how closely two variables, \( T_1 \) and \( \theta \), relate to each other when the condition \( T_2 = t_2 \) is known.
Understanding the relationship helps us make more accurate predictions and construct confidence intervals.
Understanding Confidence Level
In statistical analysis, the confidence level indicates the probability that a calculated confidence interval contains the true population parameter. In simpler terms, if we perform an experiment 100 times and calculate 100 confidence intervals, a 95% confidence level suggests that 95 of those intervals will contain the true parameter.
This concept provides a form of assurance about the estimates we derive from data. In our exercise, we focus on a 95% confidence level. This means we are 95% sure that the true value of \( \theta \) lies within the interval constructed using the observed data
and the value of \( c \).
Two-Sided Confidence Interval Demystified
A two-sided confidence interval is a statistical range calculated from the data so that the parameter of interest, in this case \( \theta \), falls within this range with a specified probability already set, which is often at a certain confidence level. Unlike one-sided confidence intervals, which only give boundary in one direction, a two-sided interval provides boundaries on both sides, allowing us to say with a certain level of confidence
that the actual parameter lies between these limits.
In the problem, this is represented as \( P(T_1 - c < \theta < T_1 + c \mid T_2 = t_2) = 0.95 \), meaning the true parameter \( \theta \) falls within \( T_1 - c \) and \( T_1 + c \) with 95% confidence. Two-sided confidence intervals are particularly useful when you need to understand the range within which a parameter is likely to lie rather than a single-point estimate.
They are pivotal in many scientific studies that seek balanced estimates without bias towards one direction.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a distribution with pdf \(f(x ; \theta)=\theta^{x}(1-\theta), x=0,1,2, \ldots\), zero elsewhere, where \(0 \leq \theta \leq 1\) (a) Find the mle, \(\hat{\theta}\), of \(\theta\). (b) Show that \(\sum_{1}^{n} X_{i}\) is a complete sufficient statistic for \(\theta\). (c) Determine the MVUE of \(\theta\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a Poisson distribution with parameter \(\theta>0\) (a) Find the MVUE of \(P(X \leq 1)=(1+\theta) e^{-\theta}\). Hint: \(\quad\) Let \(u\left(x_{1}\right)=1, x_{1} \leq 1\), zero elsewhere, and find \(E\left[u\left(X_{1}\right) \mid Y=y\right]\), where \(Y=\sum_{1}^{n} X_{i}\). (b) Express the MVUE as a function of the mle. (c) Determine the asymptotic distribution of the mle.

We consider a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from a distribution with pdf \(f(x ; \theta)=(1 / \theta) \exp (-x / \theta), 0

Let \(\left(X_{1}, Y_{1}\right),\left(X_{2}, Y_{2}\right), \ldots,\left(X_{n}, Y_{n}\right)\) denote a random sample of size \(n\) from a bivariate normal distribution with means \(\mu_{1}\) and \(\mu_{2}\), positive variances \(\sigma_{1}^{2}\) and \(\sigma_{2}^{2}\), and correlation coefficient \(\rho .\) Show that \(\sum_{1}^{n} X_{i}, \sum_{1}^{n} Y_{i}, \sum_{1}^{n} X_{i}^{2}, \sum_{1}^{n} Y_{i}^{2}\), and \(\sum_{1}^{n} X_{i} Y_{i}\) are joint complete sufficient statistics for the five parameters. Are \(\bar{X}=\) \(\sum_{1}^{n} X_{i} / n, \bar{Y}=\sum_{1}^{n} Y_{i} / n, S_{1}^{2}=\sum_{1}^{n}\left(X_{i}-\bar{X}\right)^{2} /(n-1), S_{2}^{2}=\sum_{1}^{n}\left(Y_{i}-\bar{Y}\right)^{2} /(n-1)\), and \(\sum_{1}^{n}\left(X_{i}-\bar{X}\right)\left(Y_{i}-\bar{Y}\right) /(n-1) S_{1} S_{2}\) also joint complete sufficient statistics for these parameters?

Let \(X\) be a random variable with pdf of a regular case of the exponential class. Show that \(E[K(X)]=-q^{\prime}(\theta) / p^{\prime}(\theta)\), provided these derivatives exist, by differentiating both members of the equality $$\int_{a}^{b} \exp [p(\theta) K(x)+S(x)+q(\theta)] d x=1$$ with respect to \(\theta\). By a second differentiation, find the variance of \(K(X)\).

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