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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 \(\approx \,4.539\,\mathrm{kg/m^2}\).

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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─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/m²) 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/m², 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/m² 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/m².
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/m² tells us that the true average biomass per square meter for the entire forest is likely to be within 0.339 kg/m² 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.

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Most popular questions from this chapter

To determine whether there is a significant difference in the average weights of boys and girls beginning kindergarten, random samples of 50 5-year-old boys and 50 5-year-old girls produced the following information. \(^{17}\) $$\begin{array}{llcc}\hline & & \text { Standard } & \text { Sample } \\\& \text { Mean } & \text { Deviation } & \text { Size } \\\\\hline \text { Boys } & 19.4 \mathrm{~kg} & 2.4 & 50 \\\\\text { Girls } & 17.0 \mathrm{~kg} & 1.9 & 50 \\\\\hline\end{array}$$ Find a \(99 \%\) lower confidence bound for the difference in the average weights of 5 -year-old boys and girls. Can you conclude that on average 5 -year-old boys weigh more than 5-year-old girls?

M\&M'S Does Mars, Incorporated use the same proportion of red candies in its plain and peanut varieties? A random sample of 56 plain M\&M'S contained 12 red candies, and another random sample of 32 peanut M\&M'S contained 8 red candies. a. Construct a \(95 \%\) confidence interval for the difference in the proportions of red candies for the plain and peanut varieties. b. Based on the confidence interval in part a, can you conclude that there is a difference in the proportions of red candies for the plain and peanut varieties? Explain.

Explain what is meant by "margin of error" in point estimation.

Can you afford the price of an NBA ticket during the regular season? The website Www.answers.com indicates that the low prices are around \(\$ 10\) for the high up seats while the court-side seats are around \(\$ 2000\) to \(\$ 5000\) per game and the average price of a ticket is \(\$ 75.50\) a game. \(^{7}\) Suppose that we test this claim by selecting a random sample of \(n=50\) ticket purchases from a computer database and find that the average ticket price is \(\$ 82.50\) with a standard deviation of \(\$ 75.25 .\) a. Do you think that \(x\), the price of an individual regular season ticket, has a mound-shaped distribution? If not, what shape would you expect? b. If the distribution of the ticket prices is not normal, you can still use the standard normal distribution to construct a confidence interval for \(\mu,\) the average price of a ticket. Why? c. Construct a \(95 \%\) confidence interval for \(\mu,\) the average price of a ticket. Does this confidence interval cause you support or question the claimed average price of \(\$ 75.50 ?\) Explain.

Use the information given in Exercises 9-15 to find the necessary confidence interval for the population mean \(\mu .\) Interpret the interval that you have constructed. \(\alpha=.05, n=89, \bar{x}=66.3, s^{2}=2.48\)

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