/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 If a \(90 \%\) confidence interv... [FREE SOLUTION] | 91Ó°ÊÓ

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If a \(90 \%\) confidence interval for \(\sigma^{2}\) is reported to be \((51.47,261.90)\), what is the value of the sample standard deviation?

Short Answer

Expert verified
The value of the sample standard deviation is the square root of the square root of the product of 51.47 and 261.90.

Step by step solution

01

Confirm the given values

First, we acknowledge the given 90% confidence interval for the population variance \(\sigma^{2}\), which spans from 51.47 to 261.90.
02

Calculate sample variance

The sample variance \(s^{2}\) can be estimated as the square root of the product of the lower and upper bounds of the confidence interval for \(\sigma^{2}\). Hence, \(s^{2}\) is the square root of \(51.47 * 261.90\).
03

Find sample standard deviation

The sample standard deviation \(s\) is the square root of the sample variance \(s^{2}\). Therefore, \(s\) is the square root of the already calculated \(s^{2}\).

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Key Concepts

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

Population Variance
When we talk about population variance, we're referring to a measure that expresses the diversity or spread of all possible outcomes in a population. One way to think about it is by considering it as the average of the squared differences from the mean of the population. Population variance, symbolized by \(\sigma^{2}\), gives us insight into how similar or different each data point is relative to the population mean.In the context of confidence intervals, such as the one mentioned \((51.47, 261.90)\), this interval gives us a range within which the true population variance \(\sigma^{2}\)is expected to fall a specific percentage of the time (e.g., 90%). These intervals are a powerful tool because they help in making statistical inferences about the populationbased on the sample data.
Sample Standard Deviation
The sample standard deviation, denoted by \(s\), is a measure of how much the observations in a sample differ from the sample mean.It is essentially an estimate of the population standard deviation based on sample data. To calculate it, we first need the sample variance \(s^{2}\) and then take its square root.Here’s a quick breakdown:
  • Start from the sample variance, \(s^{2}\).
  • The sample standard deviation \(s\) is simply the square root of this sample variance.
This measure is crucial because it helps us understand how much actual variation the sample data exhibits. It's particularly useful when making predictions or assessing the confidence in data results.
Sample Variance
The sample variance \(s^{2}\) serves as a concrete method to approximate the population variance from a sample.This value tells us how much the values in the sample differ from their mean. It is calculated by taking the differences of each data point from the sample mean, squaring these differences, and then averaging them.To connect with the exercise, once we've identified the confidence interval of the population variance \(\sigma^{2}\),we use the method of estimating \(s^{2}\) by leveraging these bounds. Specifically, in the original exercise, \(s^{2}\) is estimated as the square root of the product of the confidence interval boundaries \(51.47\) and \(261.90\). Once the sample variance is known, it can be used in further calculations, such as determining the sample standard deviation.Understanding sample variance is key to grasping the variability and reliability of the sample as a representative of the broader population.

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Most popular questions from this chapter

Find the value \(y\) that satisfies each of the following equations: (a) \(P\left(\chi_{9}^{2} \geq y\right)=0.99\) (b) \(P\left(\chi_{15}^{2} \leq y\right)=0.05\) (c) \(P\left(9.542 \leq x_{22}^{2} \leq y\right)=0.09\) (d) \(P\left(y \leq \chi_{31}^{2} \leq 48.232\right)=0.95\)

Explain why the distribution of \(t\) ratios calculated from small samples drawn from the exponential pdf, \(f_{Y}(y)=e^{-y}, y \geq 0\), will be skewed to the left [recall Figure 7.4.6(b)]. (Hint: What does the shape of \(f_{Y}(y)\) imply about the possibility of each \(y_{i}\) being close to 0 ? If the entire sample did consist of \(y_{i}\) 's close to 0 , what value would the \(t\) ratio have?)

What "confidence" is associated with each of the following random intervals? Assume that the \(Y_{i}\) 's are normally distributed. (a) \(\left[\bar{Y}-2.0930\left(\frac{S}{\sqrt{20}}\right), \bar{Y}+2.0930\left(\frac{S}{\sqrt{20}}\right)\right]\) (b) \(\left[\bar{Y}-1.345\left(\frac{S}{\sqrt{15}}\right), \bar{Y}+1.345\left(\frac{S}{\sqrt{15}}\right)\right]\) (c) \(\left[\bar{Y}-1.7056\left(\frac{S}{\sqrt{27}}\right), \bar{Y}+2.7787\left(\frac{S}{\sqrt{27}}\right)\right]\) (d) \(\left[-\infty, \bar{Y}+1.7247\left(\frac{S}{\sqrt{21}}\right)\right]\)

In Case Study 7.5.1, the \(95 \%\) confidence interval was constructed for \(\sigma\) rather than for \(\sigma^{2}\). In practice, is an experimenter more likely to focus on the standard deviation or on the variance, or do you think that both formulas in Theorem \(7.5 .1\) are likely to be used equally often? Explain.

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be a random sample of size \(n\) from the pdf $$ f_{Y}(y)=\left(\frac{1}{\theta}\right) e^{-y / \theta}, \quad y>0 ; \quad \theta>0 $$ (a) Use moment-generating functions to show that the ratio \(2 n \bar{Y} / \theta\) has a chi square distribution with \(2 n\) df. (b) Use the result in part (a) to derive a \(100(1-\alpha) \%\) confidence interval for \(\theta\).

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