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Chapter 9: Problem 12

Construct the appropriate confidence interval. A simple random sample of size \(n=25\) is drawn from a population that is normally distributed. The sample variance is found to be \(s^{2}=3.97\). Construct a \(95 \%\) confidence interval for the population standard deviation.

Short Answer

Expert verified
The 95% confidence interval for the population standard deviation is (1.56, 2.77).

Step by step solution

01

- Identify the Sample Statistics

The sample size is given as \(n = 25\). The sample variance is given as \(s^2 = 3.97\).
02

- Determine the Confidence Level and Degrees of Freedom

The confidence level is \(95\text{\b backslash%}\), so the significance level is \(\alpha = 0.05\). The degrees of freedom (df) are calculated as \(n - 1 = 25 - 1 = 24\).
03

- Find the Chi-Square Critical Values

Use a chi-square distribution table to find the critical values for \(\chi^2_{0.025\text{\b backslash, 24}}\) and \(\chi^2_{0.975\text{\b backslash, 24}}\). These values are approximately \(\chi^2_{0.025\text{\b backslash, 24}} = 39.364\) and \(\chi^2_{0.975\text{\b backslash, 24}} = 12.401\).
04

- Calculate the Confidence Interval for the Population Variance

The confidence interval for the population variance \(\sigma^2\) is calculated using the formula: \[\left( \frac{(n-1)s^2}{\chi^2_{\alpha/2\text{\b backslash, df}}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2\text{\b backslash, df}}} \right)\]Substituting the values: \[\left( \frac{24 \times 3.97}{39.364}, \frac{24 \times 3.97}{12.401} \right) = (2.42, 7.68)\]
05

- Convert the Variance Interval to Standard Deviation Interval

Since the population standard deviation \(\sigma\) is the square root of the population variance \(\sigma^2\), take the square root of the endpoints of the interval to get the confidence interval for \(\sigma\):\[(\sqrt{2.42}, \sqrt{7.68}) = (1.56, 2.77)\]

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Key Concepts

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

Sample Variance
When it comes to statistics, the sample variance is a crucial concept. It measures how spread out the numbers in a sample are around the mean. Mathematically, it is represented as \( s^2 \), where \( s \) is the sample standard deviation.
To calculate the variance, you take each number in the sample, subtract the sample mean, square the result, and then take the average of those squared differences.
This tells you how much the sample numbers differ from the mean value.
Understanding sample variance is important because it forms the basis for calculating the confidence interval for the population standard deviation.
Chi-Square Distribution
The chi-square distribution is a special kind of distribution that is useful when dealing with variance. It is skewed to the right and is used in various hypothesis testing scenarios.
Specifically, it is used when constructing confidence intervals for variances and standard deviations.
The shape of the chi-square distribution depends on the degrees of freedom (df). In our case, with a sample size of 25, we calculated df as n - 1, which is 24.
Critical values from the chi-square distribution table help us find the endpoints of the confidence interval for the population variance.
In this exercise, the chi-square critical values we used were approximately 39.364 and 12.401.
Degrees of Freedom
Degrees of freedom (df) is a concept that helps us understand the number of independent values that can vary in an analysis without breaking any constraints.
In simpler terms, it tells us the number of values that have the freedom to vary. For calculations involving sample variance, df is usually n - 1, where n is the sample size.
This adjustment is made because one value is sacrificed to calculate the mean of the sample.
In our exercise, with a sample size of 25, we used df = 24. This degrees of freedom value is used to look up critical values in the chi-square distribution table.
These critical values are essential for determining the range of our confidence interval.
Confidence Level
The confidence level is the percentage that quantifies how sure we are that our interval estimate contains the true population parameter.
A high confidence level, like 95%, means we are very certain our interval includes the real value.
In this exercise, a 95% confidence level implies that if we were to take 100 different samples and create a confidence interval from each sample, about 95 of those intervals would contain the true population standard deviation.
The confidence level helps us find statistical significance and is represented by \( 1 - \alpha \), where \( \alpha \) is the significance level.
For a 95% confidence level, \( \alpha \) is 0.05. This level determines the critical chi-square values we need to use.

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Most popular questions from this chapter

In March 2014, Harris Interactive conducted a poll of a random sample of 2234 adult Americans 18 years of age or older and asked, "Which is more annoying to you, tailgaters or slow drivers who stay in the passing lane?" Among those surveyed, 1184 were more annoyed by tailgaters. (a) Explain why the variable of interest is qualitative with two possible outcomes. What are the two outcomes? (b) Verify the requirements for constructing a \(90 \%\) confidence interval for the population proportion of all adult Americans who are more annoyed by tailgaters than slow drivers in the passing lane. (c) Construct a \(90 \%\) confidence interval for the population proportion of all adult Americans who are more annoyed by tailgaters than slow drivers in the passing lane.

Suppose you have two populations: Population \(\mathrm{A}-\) All students at Illinois State University \((N=21,000)\) and Population \(\mathrm{B}-\) All residents of the city of Homer Glen, IL \((N=21,000)\). You want to estimate the mean age of each population using two separate samples each of size \(n=75\). If you construct a \(95 \%\) confidence interval for each population mean, will the margin of error for population A be larger, the same, or smaller than the margin of error for population \(\mathrm{B}\) ? Justify your reasoning.

True or False: The value of \(t_{0.10}\) with 5 degrees of freedom is greater than the value of \(t_{0.10}\) with 10 degrees of freedom.

The American Society for Microbiology (ASM) and the Soap and Detergent Association (SDA) jointly commissioned two separate studies, both of which were conducted by Harris Interactive. In one of the studies, 1001 adults were interviewed by telephone and asked about their handwashing habits. In the telephone interviews, 921 of the adults said they always wash their hands in public restrooms. In the other study, the hand-washing behavior of 6076 adults was inconspicuously observed within public restrooms in four U.S. cities and 4679 of the 6076 adults were observed washing their hands. (a) In the telephone survey, what is the variable of interest? Is it qualitative or quantitative? (b) What is the sample in the telephone survey? What is the population to which this study applies? (c) Verify that the requirements for constructing a confidence interval for the population proportion of adults who say they always wash their hands in public restrooms are satisfied. (d) Using the results from the telephone interviews, construct a \(95 \%\) confidence interval for the proportion of adults who say they always wash their hands in public restrooms. (e) In the study where hand-washing behavior was observed, what is the variable of interest? Is it qualitative or quantitative? (f) We are told that 6076 adults were inconspicuously observed, but were not told how these adults were selected. We know randomness is a key ingredient in statistical studies that allows us to generalize results from a sample to a population. Suggest some ways randomness might have been used to select the individuals in this study. (g) Verify the requirements for constructing a confidence interval for the population proportion of adults who actually washed their hands while in a public restroom. (h) Using the results from the observational study, construct a \(95 \%\) confidence interval for the proportion of adults who wash their hands in public restrooms. (i) Based on your findings in parts (a) through (h), what might you conclude about the proportion of adults who say they always wash their hands versus the proportion of adults who actually wash their hands in public restrooms? (j) Cite some sources of variability in both studies.

A researcher for the U.S. Department of the Treasury wishes to estimate the percentage of Americans who support abolishing the penny. What size sample should be obtained if he wishes the estimate to be within 2 percentage points with \(98 \%\) confidence if (a) he uses a 2006 estimate of \(15 \%\) obtained from a Coinstar National Currency Poll? (b) he does not use any prior estimate?
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