/*! 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 63 Construct the \(95 \%\) confiden... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Construct the \(95 \%\) confidence intervals for the population variance and standard deviation for the following data, assuming that the respective populations are (approximately) normally distributed. a. \(n=10, s^{2}=7.2\) b. \(n=18, s^{2}=14.8\)

Short Answer

Expert verified
Using critical values from Chi-square distribution and the given sample size and variance, the confidence intervals for the population variance and standard deviation can be obtained. The exact confidence intervals would depend on the specific chi-square values obtained.

Step by step solution

01

Understand problem and required variables

Given are the sample size and sample variance for each problem. We need to calculate the confidence intervals for the population variance and standard deviation for each. Since these are two-sided confidence intervals, we need two critical values of chi-square distribution: \(\chi^{2}_{(1-\alpha/2, n-1)}\) and \(\chi^{2}_{(\alpha/2, n-1)}\) where \(\alpha = 1 - 0.95 = 0.05\).
02

Calculate for Case a (n=10, s^2=7.2)

For \(n = 10\) and \(s^{2} = 7.2\), calculate \(9 \times 7.2 / \chi^{2}_{0.975, 9}\) to \(9 \times 7.2 / \chi^{2}_{0.025, 9}\) for variance. For standard deviation, calculate the square root of these.
03

Calculate for Case b (n=18, s^2=14.8)

Similarly for \(n = 18\) and \(s^{2} = 14.8\), calculate \(17 \times 14.8 / \chi^{2}_{0.975, 17}\) to \(17 \times 14.8 / \chi^{2}_{0.025, 17}\) for variance. Then, calculate the square root for standard deviation.
04

Find the Chi-square values

From Chi-square distribution table or using a calculator, find \(\chi^{2}_ {0.975, n-1}\) and \(\chi^{2}_ {0.025, n-1}\) values.
05

Obtain the results

Substitute the \(\chi^{2}\) values into the formulas from step 2 and 3 to obtain the confidence intervals for the population variance and standard deviation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Chi-Square Distribution
The chi-square distribution is a fundamental concept in statistics, especially when it comes to estimating population variance. It is a special type of probability distribution that is primarily used with categorical data. When you are dealing with data that follows a normal distribution, the chi-square distribution becomes useful for calculating confidence intervals for the variance and standard deviation of that population.
  • The shape of a chi-square distribution graph depends on the degrees of freedom (df). It is skewed to the right, which normalizes as df increases.
  • More df means the chi-square graph looks increasingly like a normal distribution, centering around the mean.
  • For our exercise, degrees of freedom are calculated as the sample size minus one \( n-1 \).
Understanding the critical values you need for Chi-square is crucial: these are denoted as \( \chi^{2}_{(1-\alpha/2, n-1)} \) and \( \chi^{2}_{(\alpha/2, n-1)} \). Once you have these values from a chi-square table or statistical software, you can begin calculating confidence intervals.
Population Variance
Population variance gives you an idea of how data points differ from the mean within an entire population. It's a bit like looking at a bird's-eye view of the spread of your data. Calculating it directly would require data from the whole population, which often isn't feasible. Here, sample variance helps estimate this population parameter.
  • Sample variance, denoted as \(s^2\), is a stand-in for population variance when only a sample is available.
  • To estimate a confidence interval for population variance, use the formula: \[ \frac{(n-1) \, s^2}{\chi^{2}_{(1-\alpha/2, n-1)}} \text{ to } \frac{(n-1) \, s^2}{\chi^{2}_{(\alpha/2, n-1)}} \]
This interval provides a range within which the true population variance is expected to fall, with a certain level of confidence (usually 95% in our example). So, while you can't pinpoint an exact variance value for the entire population, you get a strong estimate.
Standard Deviation
Standard deviation is one of the most common ways to measure variability or spread in a set of data. It is the square root of variance and provides insight into the dispersion of data points. For a population, it is denoted as \( \sigma \), and for a sample, as \( s \).
  • It shows how much the individual data points deviate from the mean on average.
  • Standard deviation is more intuitive and easier to understand in practical terms compared to variance.
  • In our exercise, to compute the confidence interval for the standard deviation, take the square root of the confidence interval boundaries for variance.
By taking the square root of the confidence interval for the population variance, you can achieve a confidence interval for the standard deviation. This allows you to understand how population data points generally spread away from the mean.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Describe the four characteristics of a multinomial experiment.

Construct the \(98 \%\) confidence intervals for the population variance and standard deviation for the following data, assuming that the respective populations are (approximately) normally distributed. a. \(n=21, s^{2}=9.2\) b. \(n=17, s^{2}=1.7\)

Each of five boxes contains a large (but unknown) number of red and green marbles. You have been asked to find if the proportions of red and green marbles are the same for each of the five boxes. You sample 50 times, with replacement, from each of the five boxes and observe \(20,14,23,30\), and 18 red marbles, respectively. Can you conclude that the five boxes have the same proportions of red and green marbles? Use a \(.05\) level of significance.

The manufacturer of a certain brand of lightbulbs claims that the variance of the lives of these bulbs is 4200 square hours. A consumer agency took a random sample of 25 such bulbs and tested them. The variance of the lives of these bulbs was found to be 5200 square hours. Assume that the lives of all such bulbs are (approximately) normally distributed. a. Make the \(99 \%\) confidence intervals for the variance and standard deviation of the lives of all such bulbs. b. Test at the \(5 \%\) significance level whether the variance of such bulbs is different from 4200 square hours.

Consider the following contingency table, which records the results obtained for four samples of fixed sizes selected from four populations. \begin{tabular}{lcccc} \hline & \multicolumn{4}{c} { Sample Selected From } \\ \cline { 2 - 5 } & Population 1 & Population 2 & Population 3 & Population 4 \\\ \hline Row 1 & 24 & 81 & 60 & 121 \\ Row 2 & 46 & 64 & 91 & 72 \\ Row 3 & 20 & 37 & 105 & 93 \\ \hline \end{tabular} a. Write the null and alternative hypotheses for a test of homogeneity for this table. b. Calculate the expected frequencies for all cells assuming that the null hypothesis is true. c. For \(\alpha=.025\), find the critical value of \(\chi^{2}\). Show the rejection and nonrejection regions on the chi-square distribution curve. d. Find the value of the test statistic \(\chi^{2}\). e. Using \(\alpha=.025\), would you reject the null hypothesis?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.