/*! 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 22 Ecology: Sand Dunes At wind spee... [FREE SOLUTION] | 91Ó°ÊÓ

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Ecology: Sand Dunes At wind speeds above 1000 centimeters per second \((\mathrm{cm} / \mathrm{sec})\), significant sand-moving events begin to occur. Wind speeds below \(1000 \mathrm{~cm} / \mathrm{sec}\) deposit sand, and wind speeds above \(1000 \mathrm{~cm} / \mathrm{sec}\) move sand to new locations. The cyclic nature of wind and moving sand determines the shape and location of large dunes (Reference: Hydraulic, Geologic, and Biologic Research at Great Sand Dunes National Monument and Vicinity, Colorado, Proceedings of the National Park Service Research Symposium). At a test site, the prevailing direction of the wind did not change noticeably. However, the velocity did change. Sixty wind speed readings gave an average velocity of \(\bar{x}=1075 \mathrm{~cm} / \mathrm{sec} .\) Based on long-term experience, \(\sigma\) can be assumed to be \(265 \mathrm{~cm} / \mathrm{sec} .\) (a) Find a \(95 \%\) confidence interval for the population mean wind speed at this site. (b) Interpretation Does the confidence interval indicate that the population mean wind speed is such that the sand is always moving at this site? Explain.

Short Answer

Expert verified
The 95% confidence interval is (1007.92, 1142.08) cm/sec, indicating the sand is generally moving.

Step by step solution

01

Identify Given Information

We are given the following information: the sample mean wind speed \( \bar{x} \) is 1075 cm/sec, the population standard deviation \( \sigma \) is 265 cm/sec, and the sample size \( n \) is 60. We need to find the 95% confidence interval for the population mean wind speed.
02

Determine the Z-Score for 95% Confidence

For a 95% confidence interval, the Z-score corresponding to the critical value is approximately 1.96. This value can be found from a Z-table or standard normal distribution table.
03

Calculate the Standard Error

The standard error \( SE \) of the mean is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} \] Substitute the known values to get: \[ SE = \frac{265}{\sqrt{60}} \approx 34.22 \text{ cm/sec} \]
04

Calculate the Confidence Interval

The confidence interval for the population mean \( \mu \) is given by: \[ \bar{x} \pm Z \times SE \] Substituting the known values yields: \[ 1075 \pm 1.96 \times 34.22 \] This calculates to: \[ 1075 \pm 67.08 \] Hence, the 95% confidence interval is (1007.92, 1142.08) cm/sec.
05

Interpret the Confidence Interval

The confidence interval (1007.92, 1142.08) cm/sec supports that the average wind speed is above 1000 cm/sec. Since the entire interval is above this threshold, it indicates that, on average, the wind speed is sufficient for sand movement.

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Key Concepts

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

Understanding Population Mean
The population mean is often denoted by the symbol \( \mu \) and represents the average of all possible observations of a particular measurement in the entire population.
In this context, it would mean the average wind speed of all possible wind speed readings at this site over a long period.
Calculating this exact value would be nearly impossible without assessing every single wind speed reading since the population could be massive.
Instead, we can infer about the population mean using a sample mean, which is the average of observations from a finite sample taken from the population.
In our example, the sample mean \( \bar{x} \) is 1075 cm/sec based on the 60 measured readings.
The inference about the population mean is made more accurate by considering both the sample mean and the standard deviation.
Grasping Standard Deviation
Standard deviation, denoted as \( \sigma \) for populations, is a measure of the dispersion or spread of data points in a data set.
It tells us how much individual readings are likely to differ from the mean, which in this case is the average wind speed.
A small standard deviation indicates the wind speed readings are close to the mean, while a large standard deviation indicates a wider spread of readings.
  • Standard deviation provides valuable insights into the variability of wind speeds at the site.
For this exercise, the population standard deviation of wind speeds is given as 265 cm/sec.
This means, in reality, wind speed can considerably vary from the average of 1075 cm/sec.
Insight Into Standard Error
While standard deviation informs us about how data points are spread out, standard error (abbreviated as SE) provides insight into the accuracy of the sample mean as an estimate of the population mean.
It shows how much the sample mean is expected to fluctuate from sample to sample if many samples were drawn.
The standard error is computed with the formula \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the standard deviation and \( n \) is the sample size.
  • In our exercise, the standard error is approximately 34.22 cm/sec, representing how much the 1075 cm/sec average could vary.
The smaller the standard error, the closer the sample mean is expected to be to the population mean, which reflects high precision in estimating the wind speed.
Decoding the Z-Score
The Z-score is a numerical measurement used in statistics that describes a value's relation to the mean of a group of values.
It is measured in terms of standard deviations from the mean.
A Z-score can tell us if a measurement is typical or falls outside the norm. In confidence interval calculations, a Z-score value corresponds to the desired confidence level. For instance, a 95% confidence level correlates with a Z-score of approximately 1.96.
  • This means if you were to repeatedly take samples and compute their means, 95% of those means would fall within 1.96 standard errors of the population mean.
This is crucial for constructing confidence intervals, as seen here in the calculation of wind speed intervals.
The confidence interval, built upon this Z-score, gives us a range where we can be relatively sure (95% confidence) the true population mean lies.

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Most popular questions from this chapter

Brain Teaser A requirement for using the normal distribution to approximate the \(\hat{p}\) distribution is that both \(n p>5\) and \(n q>5 .\) Since we usually do not know \(p\), we estimate \(p\) by \(\hat{p}\) and \(q\) by \(\hat{q}=1-\hat{p} .\) Then we require that \(n \hat{p}>5\) and \(n \hat{q}>5\). Show that the conditions \(n \hat{p}>5\) and \(n \hat{q}>5\) are equivalent to the condition that out of \(n\) binomial trials, both the number of successes \(r\) and the number of failures \(n-r\) must exceed \(5 .\) Hint \(:\) In the inequality \(n \hat{p}>5\), replace \(\hat{p}\) by \(r / n\) and solve for \(r .\) In the inequality \(n \hat{q}>5\), replace \(\hat{q}\) by \((n-r) / n\) and solve for \(n-r\).

Lorraine was in a hurry when she computed a confidence interval for \(\mu\). Because \(\sigma\) was not known, she used a Student's \(t\) distribution. However, she accidentally used degrees of freedom \(n\) instead of \(n-1\). Was her confidence interval longer or shorter than one found using the correct degrees of freedom \(n-1 ?\) Explain.

Statistical Literacy As the degrees of freedom increase, what distribution does the Student's \(t\) distribution become more like?

If a \(90 \%\) confidence interval for the difference of proportions contains some positive and some negative values, what can we conclude about the relationship between \(p_{1}\) and \(p_{2}\) at the \(90 \%\) confidence level?

Confidence Intetvals: Values of \(\sigma\) A random sample of size 36 is drawn from an \(x\) distribution. The sample mean is \(100 .\) (a) Suppose the \(x\) distribution has \(\sigma=30\). Compute a \(90 \%\) confidence interval for \(\mu\). What is the value of the margin of error? (b) Suppose the \(x\) distribution has \(\sigma=20\). Compute a \(90 \%\) confidence interval for \(\mu\). What is the value of the margin of error? (c) Suppose the \(x\) distribution has \(\sigma=10\). Compute a \(90 \%\) confidence interval for \(\mu\). What is the value of the margin of error? (d) Compare the margins of error for parts (a) through (c). As the standard deviation decreases, does the margin of error decrease? (e) Critical Thinking Compare the lengths of the confidence intervals for parts (a) through (c). As the standard deviation decreases, does the length of a \(90 \%\) confidence interval decrease?

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