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Chapter 9: Problem 11

Construct the appropriate confidence interval. A simple random sample of size \(n=12\) is drawn from a population that is normally distributed. The sample variance is found to be \(s^{2}=23.7\). Construct a \(90 \%\) confidence interval for the population variance.

Short Answer

Expert verified
The 90% confidence interval for the population variance is (13.26, 42.01).

Step by step solution

01

Identify the key values

The sample size is given as =12, the sample variance is given as \(s^2=23.7\), and the confidence level is 90%.
02

Determine degrees of freedom

Degrees of freedom (df) can be calculated as - 1. In this case, =12, so df = 12 - 1 = 11.
03

Find the critical chi-square values

For a 90% confidence interval and 11 degrees of freedom, find the chi-square values \(Chi_{1-alpha/2}^2\)\right) and \(Chi_{alpha/2}^2\).\right). Use a chi-square table or calculator to find the critical values: \(Chi_{0.05,11}^2 = 19.675\)\right) and \(Chi_{0.95,11}^2 = 6.204\).\right).
04

Calculate the confidence interval

Use the formula for the confidence interval of the population variance: [ \left(\frac{(n-1)s^2}{Chi_{1-alpha/2}^2}\right, \frac{(n-1)s^2}{Chi_{alpha/2}^2}\right) \right]. Plugging in the values: =12, s^2=23.7, \(Chi_{1-alpha/2}^2 = 19.675\), and \(Chi_{alpha/2}^2 = 6.204\), we get the interval bounds: [\left(\frac{11 \times 23.7}{19.675}\right, \frac{11 \times 23.7}{6.204}\right)\right] = (13.26, 42.01).

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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 crucial concept when working with confidence intervals for population variance. This distribution is used because the variance of a sample follows a chi-square distribution when the underlying population is normally distributed. The chi-square distribution has a shape that depends on the degrees of freedom (df), which we will explore further. One key feature is that it is skewed to the right, and as the degrees of freedom increase, the distribution becomes more symmetrical. Recognizing and using the chi-square distribution helps in finding the critical values necessary for calculating confidence intervals properly.
sample variance
Sample variance, denoted as \(s^2\), measures the spread of sample data points around their mean. It's calculated as the sum of squared deviations from the sample mean, divided by the number of observations minus one (the degrees of freedom). In formulas, it can be written as follows:
  • Step 1: Calculate the mean (\bar{x}) of the sample data.
  • Step 2: Compute the squared differences from the mean for each data point.
  • Step 3: Sum all the squared differences.
  • Step 4: Divide this sum by the number of data points minus one.
The sample variance is an unbiased estimate of the population variance when the population is normally distributed. In our example exercise, the sample variance provided is 23.7, which we use in the formula to construct the confidence interval for population variance.
degrees of freedom
Degrees of freedom (df) is an essential concept in statistics that corresponds to the number of independent values or quantities that can vary in the analysis. In the context of calculating sample variance and confidence intervals, it is calculated as the sample size \(n\) minus one, i.e., \(df = n - 1\).
  • For example, if the sample size \(n = 12\), then \(df = 12 - 1 = 11\).
Degrees of freedom play a vital role in determining the shape and spread of the chi-square distribution. Understanding how to calculate and apply df is crucial for accurate statistical analysis, as seen when we retrieve the critical chi-square values from the chi-square table or calculator.
critical values
Critical values are points on the scale of the test statistic beyond which we reject the null hypothesis, and they play an important role in confidence intervals. For a chi-square distribution, the critical values come from tables based on the desired confidence level and degrees of freedom (df). For a 90% confidence interval and df = 11:
  • Lower critical value (\(\chi^2_{0.05,11}\)): 19.675
  • Upper critical value (\(\chi^2_{0.95,11}\)): 6.204
These values are then used to calculate the bounds of the confidence interval using the formula:
\[ \left(\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}, \frac{(n-1)s^2}{\chi^2_{\alpha/2}} \right)\]This ensures we capture the population variance within the calculated interval with the specified level of confidence.

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Most popular questions from this chapter

A USA Today/Gallup poll asked 1006 adult Americans how much it would bother them to stay in a room on the 13 th floor of a hotel. Interestingly, \(13 \%\) said it would bother them. The margin of error was 3 percentage points with \(95 \%\) confidence. Which of the following represents a reasonable interpretation of the survey results? For those not reasonable, explain the flaw. (a) We are \(95 \%\) confident that the proportion of adult Americans who would be bothered to stay in a room on the 13th floor is between 0.10 and 0.16 . (b) We are between \(92 \%\) and \(98 \%\) confident that \(13 \%\) of adult Americans would be bothered to stay in a room on the 13th floor. (c) In \(95 \%\) of samples of adult Americans, the proportion who would be bothered to stay in a room on the 13 th floor is between 0.10 and 0.16 . (d) We are \(95 \%\) confident that \(13 \%\) of adult Americans would be bothered to stay in a room on the 13 th floor.

Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 20 students, she finds 2 who eat cauliflower. Obtain and interpret a \(95 \%\) confidence interval for the proportion of students who eat cauliflower on Jane's campus using Agresti and Coull's method.

Construct a confidence interval of the population proportion at the given level of confidence. \(x=540, n=900,96 \%\) confidence

Explain what "95\% confidence" means in a \(95 \%\) confidence interval.

Blood Alcohol Concentration A random sample of 51 fatal crashes in 2013 in which the driver had a positive blood alcohol concentration (BAC) from the National Highway Traffic Safety Administration results in a mean BAC of 0.167 gram per deciliter \((\mathrm{g} / \mathrm{dL})\) with a standard deviation of \(0.010 \mathrm{~g} / \mathrm{dL}\) (a) A histogram of blood alcohol concentrations in fatal accidents shows that BACs are highly skewed right. Explain why a large sample size is needed to construct a confidence interval for the mean BAC of fatal crashes with a positive \(\mathrm{BAC}\) (b) In \(2013,\) there were approximately 25,000 fatal crashes in which the driver had a positive BAC. Explain why this, along with the fact that the data were obtained using a simple random sample, satisfies the requirements for constructing a confidence interval. (c) Determine and interpret a \(90 \%\) confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC. (d) All 50 states and the District of Columbia use a BAC of \(0.08 \mathrm{~g} / \mathrm{dL}\) as the legal intoxication level. Is it possible that the mean BAC of all drivers involved in fatal accidents who are found to have positive BAC values is less than the legal intoxication level? Explain.
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