/*! 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 29 Biomass, the total amount of veg... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 29

Biomass, the total amount of vegetation held by the earth's forests, is important in determining the amount of unabsorbed carbon dioxide that is expected to remain in the earth's atmosphere. \(^{2}\) Suppose a sample of 75 one-square-meter plots, randomly chosen in North America's boreal (northern) forests, produced a mean biomass of 4.2 kilograms per square meter \(\left(\mathrm{kg} / \mathrm{m}^{2}\right)\), with a standard deviation of \(1.5 \mathrm{~kg} / \mathrm{m}^{2}\) Estimate the average biomass for the boreal forests of North America and find the margin of error for your estimate.

Short Answer

Expert verified
Question: Estimate the average biomass for the boreal forests of North America and find the margin of error for our estimate based on the given sample information. Answer: The estimated average biomass for the boreal forests of North America is around \(4.2\,\mathrm{kg/m^2}\), and the margin of error is approximately \(0.339\,\mathrm{kg/m^2}\). We are \(95\%\) confident that the true average biomass is within the range of \(\approx \,3.861\,\mathrm{kg/m^2}\) to \(\approx \,4.539\,\mathrm{kg/m^2}\).

Step by step solution

01

Identify the sample statistics

We are given the following sample statistics: - Sample size: \(n = 75\) - Sample mean: \(\bar{x} = 4.2\,\mathrm{kg/m^2}\) - Sample standard deviation: \(s = 1.5\,\mathrm{kg/m^2}\)
02

Choose a confidence level

In the problem statement, we are not explicitly given a confidence level. In such cases, we typically choose a \(95\%\) confidence level, which means we can be \(95\%\) confident that our estimate contains the true population mean.
03

Determine the critical value

Since we are using a \(95\%\) confidence level, we will utilize the \(z\)-scores for a standard normal distribution. The \(z\)-score corresponds to the confidence level such that the proportion of the area under the curve is \(1 - \alpha\) where \(\alpha\) is the chosen level of significance (\(5\%\) in this case). For a \(95\%\) confidence interval, the \(z\)-score is approximately \(1.96\).
04

Calculate the standard error

In order to estimate the average biomass and the margin of error, we need to calculate the standard error. The standard error is calculated as follows: Standard Error (SE) \(= \frac{s}{\sqrt{n}}\) Using the given data, we get the standard error as: SE \(= \frac{1.5\,\mathrm{kg/m^2}}{\sqrt{75}} \approx 0.173\,\mathrm{kg/m^2}\)
05

Compute the margin of error

Now we will compute the margin of error (ME) using the critical value and the standard error: ME \(= z \times \mathrm{SE}\) Substituting the values, we get: ME \(= 1.96 \times 0.173\,\mathrm{kg/m^2} \approx 0.339\,\mathrm{kg/m^2}\)
06

Estimate the average biomass and create the confidence interval

To estimate the average biomass for the boreal forests of North America, we can use the sample mean as our point estimate: Estimated average biomass \(= \bar{x} = 4.2\,\mathrm{kg/m^2}\) Now, we will create a confidence interval using the estimated average biomass and the margin of error: Confidence interval \(= (\bar{x} - \mathrm{ME},\, \bar{x} + \mathrm{ME})\) Substituting the values, we get the confidence interval as: Confidence interval \(\approx (4.2 - 0.339\,\mathrm{kg/m^2},\, 4.2 + 0.339\,\mathrm{kg/m^2})\) Confidence interval \(\approx (3.861\,\mathrm{kg/m^2},\, 4.539\,\mathrm{kg/m^2})\) Thus, we can estimate the average biomass for the boreal forests of North America to be around \(4.2\,\mathrm{kg/m^2}\), and we are \(95\%\) confident that this estimate is within the range of \(\approx \,3.861\,\mathrm{kg/m^2}\) to \(\approx \,4.539\,\mathrm{kg/m^2}\).

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 91影视!

Key Concepts

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

Understanding Sample Statistics
When conducting research, we don't always have access to data from an entire population. Instead, we often take a smaller, manageable segment known as a sample. From this sample, we derive sample statistics鈹�numeric descriptions of the sample's attributes. In the context of estimating the average biomass in North America's boreal forests, we focus on two key statistics: the sample mean and the sample standard deviation.

The sample mean, symbolized by \( \bar{x} \) in our given problem, represents the average biomass per square meter (4.2 kg/m虏) based on the selected sample plots. It serves as a point estimate for the true population mean. The sample standard deviation, denoted by \( s \) and equating to 1.5 kg/m虏, measures the variation or spread of biomass weights from the mean. By gathering these stats from our sample of 75 plots, we can infer information about the larger forest area with reasonable accuracy.
Deciphering Standard Error
A crucial step in confidence interval estimation is understanding the standard error (SE). Standard error is an indication of how much our sample mean (our estimate) is expected to vary from the true population mean. The less the standard error, the more precise our estimate is likely to be.

To compute the standard error, we use the equation SE = \( \frac{s}{\sqrt{n}} \) where \( s \) is the sample standard deviation and \( n \) is the sample size. For the boreal forests study, a standard error of approximately 0.173 kg/m虏 indicates that if we were to take many samples and calculate the mean for each, the means would typically deviate from the true population mean by about 0.173 kg/m虏.
Margin of Error: Measuring Estimate Precision
The margin of error (ME) plays a pivotal role in expressing the confidence we have in our sample's estimates. It signifies how much we can reasonably expect our estimate to differ from the actual population parameter. In other words, it defines the range within which the true value is likely to fall.

The formula for the margin of error is ME = \( z \times SE \), with \( z \) being the z-score associated with our desired confidence level (often 95%), and SE being the standard error. For the boreal forest biomass study, the margin of error of approximately 0.339 kg/m虏 tells us that the true average biomass per square meter for the entire forest is likely to be within 0.339 kg/m虏 of our sample mean, with a confidence of 95%. This helps quantify the uncertainty in our estimation, giving us a more informed understanding of our results' reliability.

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

Does It Pay to Haggle? In Exercise \(16,\) a survey done by Consumer Reports indicates that you should always try to negotiate for a better deal when shopping or paying for services. \({ }^{20}\) In fact, based on the survey, \(37 \%\) of the people under age 34 were more likely to "haggle," while only \(13 \%\) of those 65 and older. Suppose that this group included 72 people under the age of 34 and 55 people who are 65 or older. a. What are the values of \(\hat{p}_{1}\) and \(\hat{p}_{2}\) for the two groups in this survey? b. Find a \(95 \%\) confidence interval for the difference in the proportion of people who are more likely to "haggle" in the "under 34 " versus " 65 and older" age groups. c. What conclusions can you draw regarding the groups compared in part b?

Find the z-values needed to calculate large-sample confidence intervals for the confidence levels given. Confidence coefficient \(1-\alpha=.95\)

Using the sample information given in Exercises \(22-23,\) give the best point estimate for the binomial proportion \(p\) and calculate the margin of error. A random sample of \(n=500\) observations from a binomial population produced \(x=450\) successes.

Independent random samples are drawn from two quantitative populations, producing the sample information shown in the table. Find a \(95 \%\) upper confidence bound for the difference in the two population means. $$\begin{array}{lcc}\hline & \multicolumn{2}{c} {\text { Population }} \\\\\cline { 2 - 3 } & 1 & 2 \\\\\hline \text { Sample Size } & 50 & 50 \\\\\text { Sample Mean } & 12 & 10 \\\\\text { Sample Standard Deviation } & 5 & 7 \\\\\hline\end{array}$$

You want to estimate the mean hourly yield for a process that manufactures an antibiotic. You observe the process for 100 hourly periods chosen at random, with the results \(\bar{x}=1020\) grams per hour and \(s=90 .\) Estimate the mean hourly yield for the process and calculate the margin of error.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation
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.