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Given a random sample of size \(n\) from a normal population with unknown mean and variance, we developed a confidence interval for the population variance \(\sigma^{2}\) in this section. What is the formula for a confidence interval for the population standard deviation \(\sigma ?\)

Short Answer

Expert verified
The confidence interval for \(\sigma\) is \(\left( \sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}} \right)\).

Step by step solution

01

Understanding the Problem

We need to determine the formula for a confidence interval for the population standard deviation \(\sigma\), given a sample from a normal distribution with unknown mean and variance.
02

Confidence Interval for Variance

For a normal population, the confidence interval for the population variance \(\sigma^2\) is given by \[ \left( \frac{(n-1)s^2}{\chi^2_{\alpha/2}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}} \right) \] where \(n\) is the sample size, \(s^2\) is the sample variance, and \(\chi^2_{\alpha/2}\) and \(\chi^2_{1-\alpha/2}\) are the critical values from the chi-squared distribution with \(n-1\) degrees of freedom.
03

Deriving the Confidence Interval for Standard Deviation

To find the confidence interval for the standard deviation \(\sigma\), take the square root of each bound in the confidence interval for the variance. This gives \[ \left( \sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}} \right) \] as the confidence interval for \(\sigma\).

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Key Concepts

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

Population Standard Deviation
When dealing with a set of data, the population standard deviation \( \sigma \) is a measure that describes how much the individual data points in a population differ from the mean of that population. It is a very important concept as it helps to understand the variability or spread of a population.
To calculate \( \sigma \), the formula used is:
  • Find the mean of the population.
  • Subtract the mean from each data point and square the result (these are called squared deviations).
  • Calculate the average of these squared deviations.
  • Finally, take the square root of this average to get the standard deviation.
Understanding the population standard deviation is crucial as it aids in evaluating how representative your sample is of the entire population, especially when establishing confidence intervals for various statistical measures.
Chi-Squared Distribution
The chi-squared distribution is a fundamental concept in statistics, often used when dealing with variances within a normal distribution. It is crucial in testing hypotheses and creating confidence intervals for variance and standard deviation. The notation \( \chi^2 \) is used to denote this distribution, which is characterized by \( k \) degrees of freedom.
Some characteristics of the chi-squared distribution include:
  • It is positively skewed, meaning it has a long tail on the right-hand side.
  • As the number of degrees of freedom increases, the distribution becomes more symmetric and approaches a normal distribution.
This distribution is applied when determining the confidence intervals for the population variance and standard deviation by utilizing the critical chi-squared values \( \chi^2_{\alpha/2} \) and \( \chi^2_{1-\alpha/2} \). These critical values are essential for identifying the bounds of the interval, and can be found in chi-squared distribution tables or computed using statistical software.
Sample Variance
Sample variance, represented as \( s^2 \), serves as an estimator for the population variance. It quantifies the variability within a sample and is a critical component in statistical analyses. To compute the sample variance, use the formula:
  • Calculate the mean of the sample.
  • Subtract the sample mean from each sample data point and square the results.
  • Sum all these squared differences.
  • Divide the sum by \( n-1 \), where \( n \) is the sample size, to account for degrees of freedom.
The sample variance is pivotal in constructing confidence intervals as it enables us to estimate the unknown population variance \( \sigma^2 \). Understanding sample variance lays the foundation for more advanced statistical inference techniques, such as hypothesis testing and constructing confidence intervals.

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Most popular questions from this chapter

The administrators for a hospital wished to estimate the average number of days required for inpatient treatment of patients between the ages of 25 and \(34 .\) A random sample of 500 hospital patients between these ages produced a mean and standard deviation equal to 5.4 and 3.1 days, respectively. Construct a 95\% confidence interval for the mean length of stay for the population of patients from which the sample was drawn.

A random sample of 985 "likely voters" \(-\) those who are judged to be likely to vote in an upcoming election-were polled during a phone-athon conducted by the Republican Party. Of those contacted, 592 indicated that they intended to vote for the Republican running in the election. a. According to this study, the estimate for \(p\), the proportion of all "likely voters" who will vote for the Republican candidate, is \(p=.601 .\) Find a bound for the error of estimation. b. If the "likely voters" are representative of those who will actually vote, do you think that the Republican candidate will be elected? Why? How confident are you in your decision? c. Can you think of reasons that those polled might not be representative of those who actually vote in the election?

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample of size \(n\) from a population with a uniform distribution on the interval \((0, \theta) .\) Let \(Y(n)=\max \left(Y_{1}, Y_{2}, \ldots, Y_{\mathrm{n}}\right)\) and \(U=(1 / \theta) Y_{(n)}\) a. Show that \(U\) has distribution function $$F_{U}(u)=\left\\{\begin{array}{ll} 0, & u<0 \\ u^{n}, & 0 \leq u \leq 1 \\ 1, & u>1 \end{array}\right.$$ b. Because the distribution of \(U\) does not depend on \(\theta, U\) is a pivotal quantity. Find a \(95 \%\) lower confidence bound for \(\theta\)

A precision instrument is guaranteed to read accurately to within 2 units. A sample of four instrument readings on the same object yielded the measurements \(353,351,351,\) and \(355 .\) Find a \(90 \%\) confidence interval for the population variance. What assumptions are necessary? Does the guarantee seem reasonable?

Suppose that \(Y_{1}, Y_{2}, Y_{3}\) denote a random sample from an exponential distribution with density function $$f(y)=\left\\{\begin{array}{ll} \left(\frac{1}{\theta}\right) e^{-y / \theta}, & y>0 \\ 0, & \text { elsewhere } \end{array}\right.$$ Consider the following five estimators of \(\theta\) : $$\hat{\theta}=Y_{1}, \quad \hat{\theta}_{2}=\frac{Y_{1}+Y_{2}}{2}, \quad \hat{\theta}_{3}=\frac{Y_{1}+2 Y_{2}}{3}, \quad \hat{\theta}_{4}=\min \left(Y_{1}, Y_{2}, Y_{3}\right), \quad \hat{\theta}_{5}=\bar{Y}$$ a. Which of these estimators are unbiased? b. Among the unbiased estimators, which has the smallest variance?

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