Textbook Question
In Exercises 9–12, construct the indicated confidence intervals for (a) the population variance and (b) the population standard deviation . Assume the sample is from a normally distributed population.
c = 0.90, s^2 = 35, n = 18
Ch. 6 - Confidence Intervals
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Introduction to Statistics87
- Ch. 2 - Descriptive Statistics176
- Ch. 3 - Probability184
- Ch. 4 - Discrete Probability Distributions102
- Ch. 5 - Normal Probability Distributions183
- Ch. 6 - Confidence Intervals154
- Ch. 7 - Hypothesis Testing with One Sample165
- Ch. 8 - Hypothesis Testing with Two Samples134
- Ch. 9 - Correlation and Regression87
- Ch. 10 - Chi-Square Tests and the F-Distribution87
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Not the one you use?Change textbook
Chapter 6, Problem 6.1.24
In Exercises 21–24, construct the indicated confidence interval for the population mean μ.
c = 0.80, xbar = 20.6, σ = 4.7, n = 100.
Verified step by step guidance1
Identify the key components of the problem: the confidence level (c = 0.80), sample mean (x̄ = 20.6), population standard deviation (σ = 4.7), and sample size (n = 100).
Determine the z-score corresponding to the given confidence level (c = 0.80). For an 80% confidence level, the critical z-value can be found using a z-table or statistical software. This z-value corresponds to the middle 80% of the standard normal distribution.
Calculate the standard error of the mean (SE) using the formula: . Substitute σ = 4.7 and n = 100 into the formula.
Compute the margin of error (ME) using the formula: . Use the z-value from step 2 and the SE from step 3.
Construct the confidence interval for the population mean μ using the formula: . Substitute x̄ = 20.6 and the ME from step 4 into the formula to find the interval.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Confidence Interval
A confidence interval is a range of values, derived from sample statistics, that is likely to contain the population parameter (like the mean) with a specified level of confidence. For example, a confidence level of 80% means that if we were to take many samples and construct intervals, approximately 80% of those intervals would contain the true population mean.
Recommended video:
06:33
Introduction to Confidence Intervals
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of confidence intervals, it helps quantify the uncertainty around the sample mean. A smaller standard deviation indicates that the data points tend to be closer to the mean, leading to a narrower confidence interval.
Recommended video:
Guided course
08:45Calculating Standard Deviation
Sample Size
Sample size refers to the number of observations in a sample. It plays a crucial role in determining the width of the confidence interval; larger sample sizes generally lead to more precise estimates of the population mean, resulting in narrower confidence intervals. In this case, a sample size of 100 provides a solid basis for estimating the population mean.
Recommended video:
05:11Sampling Distribution of Sample Proportion
Related Practice
Textbook Question
In Exercises 9–12, construct the indicated confidence interval for the population mean μ using the t-distribution. Assume the population is normally distributed.
c = 0.99, xbar = 24.7, s = 4.6, n = 50
Textbook Question
In Exercises 9–12, construct the indicated confidence intervals for (a) the population variance and (b) the population standard deviation . Assume the sample is from a normally distributed population.
c = 0.98, s^2 = 278.1, n =41
Textbook Question
"Finding p^ and q^ In Exercises 3–6, let p be the population proportion for the situation. Find point estimates of p and q.
Tax Fraud In a survey of 1040 U.S. adults, 62 have had someone impersonate them to try to claim tax refunds. (Adapted from Pew Research Center)"
Textbook Question
Translating Statements In Exercises 29–34, translate the statement into a confidence interval. Approximate the level of confidence.
In a survey of 880 unmarried U.S. adults who are living with a partner, 73% say love was a major reason why they decided to move in together. The survey’s margin of error is ±4.8%. (Source: Pew Research Center)
Textbook Question
Does a population have to be normally distributed to use the chi-square distribution?