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Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther (Reference: Hummingbirds by \(\mathrm{K}\). Long and W. Alther). A small group of 15 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is \(\bar{x}=3.15\) grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with \(\sigma=0.33\) gram. (a) Find an \(80 \%\) confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (b) What conditions are necessary for your calculations? (c) Interpret Compare your results in the context of this problem. (d) Sample Size Find the sample size necessary for an \(80 \%\) confidence level with a maximal margin of error \(E=0.08\) for the mean weights of the hummingbirds.

Short Answer

Expert verified
(a) Confidence interval: (3.041, 3.259), Margin of Error: 0.109. (b) Normal distribution, known \(\sigma\). (c) True mean weight: 3.041-3.259 grams. (d) Required sample size: 28 for \(E=0.08\).

Step by step solution

01

Determine the Z-Score for 80% Confidence Interval

The confidence level of 80% corresponds to a Z-score that cuts off 10% in each tail of the standard normal distribution because it's a two-tailed test. You can look up these values in a Z-table or use statistical software to find that the Z-score for 80% confidence is approximately 1.28.
02

Calculate the Margin of Error

The margin of error (ME) is calculated using the formula:\[ME = Z \times \frac{\sigma}{\sqrt{n}}\]where \(Z\) is the Z-score, \(\sigma = 0.33\) is the population standard deviation, and \(n = 15\) is the sample size.\[ME = 1.28 \times \frac{0.33}{\sqrt{15}} \approx 0.109\]
03

Compute the Confidence Interval

The confidence interval is calculated using the formula:\[(\bar{x} - ME, \bar{x} + ME)\]Where \(\bar{x} = 3.15\) grams and \(ME \approx 0.109\).\[(3.15 - 0.109, 3.15 + 0.109) = (3.041, 3.259)\]
04

State the Necessary Conditions

The necessary conditions for the above calculations include having a normally distributed population and a known population standard deviation. Additionally, even though the sample size is less than 30, the normality condition suffices here because the population is normally distributed.
05

Interpret the Results

This means we are 80% confident that the true average weight of Allen's hummingbirds in this study region lies between 3.041 and 3.259 grams. The margin of error is approximately 0.109 grams.
06

Determine Sample Size for Given Margin of Error

To find the sample size \(n\)necessary for a given margin of error \(E = 0.08\), use the formula:\[n = \left(\frac{Z \times \sigma}{E}\right)^2\]Substituting, we have:\[n = \left(\frac{1.28 \times 0.33}{0.08}\right)^2 \approx 27.62\]Rounding up to the nearest whole number, the sample size required is 28.

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Key Concepts

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

Confidence Interval
When analyzing data, especially in the field of statistics, the concept of a confidence interval is crucial. It's a way of understanding how reliable an estimate might be. A confidence interval gives you a range of values that you can be a certain percentage confident the population parameter lies within. For this example, we're looking at an 80% confidence interval for the average weights of Allen's hummingbirds. This interval is calculated using the sample mean and the margin of error, allowing researchers to infer that the true mean weight of the entire population is likely within this range. In practical terms, the confidence interval is found using the formula:\[(\bar{x} - ME, \bar{x} + ME)\]where \(\bar{x}\) is the sample mean, and ME is the margin of error.
For example, based on the calculations, we're 80% confident that the average weight of Allen's hummingbirds is between 3.041 and 3.259 grams. This type of information is particularly useful in scientific research and decision-making processes.
Margin of Error
The margin of error is an essential component in confidence intervals, representing the extent of the uncertainty in estimates. It shows how much the estimated value may differ from the true population value.The smaller the margin of error, the more precise the estimate. For Allen's hummingbirds, the margin of error was found to be 0.109 grams for an 80% confidence level. The margin of error is calculated with the following formula:\[ME = Z \times \frac{\sigma}{\sqrt{n}}\]where:
  • \(Z\) is the Z-score from the standard normal distribution table corresponding to the desired confidence level.
  • \(\sigma\) represents the population standard deviation.
  • \(n\) signifies the sample size.
By controlling factors like the sample size and the confidence level, researchers can influence the margin of error to achieve more precise results.
Sample Size
Sample size is a critical factor in determining the precision of your estimates in statistics. It not only affects the margin of error but also the reliability of the confidence interval.Generally, increasing the sample size will reduce the margin of error, making your confidence interval narrower and more precise. In this study, if we want the margin of error to be at most 0.08 with the same 80% confidence level, we would calculate the required sample size using:\[n = \left(\frac{Z \times \sigma}{E}\right)^2\]where:
  • \(E\) is the desired margin of error.
  • \(Z\) is the Z-score that corresponds to the 80% confidence level.
  • \(\sigma\) is the population standard deviation.
With these values, the calculation shows we would need a sample size of 28 birds to achieve this level of precision. This demonstrates how adjustments to sample size allow researchers to achieve specific accuracy and confidence in their results.
Normal Distribution
Normal distribution is a foundational concept in statistics that describes how the values of a variable are spread or distributed. In the case of the hummingbirds' weights, we're assuming a normal distribution for the analysis. A normal distribution is often referred to as a "bell curve" due to its characteristic shape. This distribution is symmetric, with most of the data clustering around a central mean and decreasing in frequency as you move away from the center.
The properties of normal distribution allow statisticians to make reliable inferences about the population. Knowing that the weights of hummingbirds are normally distributed allows us to use Z-scores effectively to determine confidence intervals and margins of error.

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Most popular questions from this chapter

(a) Suppose a \(95 \%\) confidence interval for the difference of means contains both positive and negative numbers. Will a \(99 \%\) confidence interval based on the same data necessarily contain both positive and negative numbers? Explain. What about a \(90 \%\) confidence interval? Explain. (b) Suppose a \(95 \%\) confidence interval for the difference of proportions contains all positive numbers. Will a \(99 \%\) confidence interval based on the same data necessarily contain all positive numbers as well? Explain. What about a \(90 \%\) confidence interval? Explain.

Expand Your Knowledge: Sample Size, Difference of Proportions What about the sample size \(n\) for confidence intervals for the difference of proportions \(p_{1}-p_{2}\) ? Let us make the following assumptions: equal sample sizes \(n=n_{1}=n_{2}\) and all four quantities \(n_{1} \hat{p}_{1}, n_{1} \hat{q}_{1}, n_{2} \hat{p}_{2}\), and \(n_{2} \hat{q}_{2}\) are greater than \(5 .\) Those readers familiar with algebra can use the procedure outlined in Problem 28 to show that if we have preliminary estimates \(\hat{p}_{1}\) and \(\hat{p}_{2}\) and a given maximal margin of error \(E\) for a specified confidence level \(c\), then the sample size \(n\) should be at least $$ n=\left(\frac{z_{c}}{E}\right)^{2}\left(\hat{p}_{1} \hat{q}_{1}+\hat{p}_{2} \hat{q}_{2}\right) $$ However, if we have no preliminary estimates for \(\hat{p}_{1}\) and \(\hat{p}_{2}\), then the theory similar to that used in this section tells us that the sample size \(n\) should be at least $$ n=\frac{1}{2}\left(\frac{z_{c}}{E}\right)^{2} $$ (a) In Problem 17 (Myers-Briggs personality type indicators in common for married couples), suppose we want to be \(99 \%\) confident that our estimate \(\hat{p}_{1}-\hat{p}_{2}\) for the difference \(p_{1}-p_{2}\) has a maximal margin of error \(E=0.04\). Use the preliminary estimates \(\hat{p}_{1}=289 / 375\) for the proportion of couples sharing two personality traits and \(\hat{p}_{2}=23 / 571\) for the proportion having no traits in common. How large should the sample size be (assuming equal sample size-i.e., \(n=n_{1}=n_{2}\) )? (b) Suppose that in Problem 17 we have no preliminary estimates for \(\hat{p}_{1}\) and \(\hat{p}_{2}\) and we want to be \(95 \%\) confident that our estimate \(\hat{p}_{1}-\hat{p}_{2}\) for the difference \(p_{1}-p_{2}\) has a maximal margin of error \(E=0.05 .\) How large should the sample size be (assuming equal sample size-i.e., \(n=n_{1}=n_{2}\) )?

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.

For all these problems, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. In order to use a normal distribution to compute confidence intervals for \(p\), what conditions on \(n p\) and \(n q\) need to be satisfied?

Archaeology: Cultural Affiliation "Unknown cultural affiliations and loss of identity at high elevations." These words are used to propose the hypothesis that archaeological sites tend to lose their identity as altitude extremes are reached. This idea is based on the notion that prehistoric people tended \(n o t\) to take trade wares to temporary settings and/or isolated areas (Source: Prehistoric New Mexico: Background for Survey, by D. E. Stuart and R. P. Gauthier, University of New Mexico Press). As elevation zones of prehistoric people (in what is now the state of New Mexico) increased, there seemed to be a loss of artifact identification. Consider the following information. $$ \begin{array}{lcc} \hline \text { Elevation Zone } & \text { Number of Artifacts } & \text { Number Unidentified } \\ \hline 7000-7500 \mathrm{ft} & 112 & 69 \\ 5000-5500 \mathrm{ft} & 140 & 26 \\ \hline \end{array} $$ Let \(p_{1}\) be the population proportion of unidentified archaeological artifacts at the elevation zone \(7000-7500\) feet in the given archaeological area. Let \(p_{2}\) be the population proportion of unidentified archaeological artifacts at the elevation zone \(5000-5500\) feet in the given archaeological area. (a) Check Requirements Can a normal distribution be used to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(99 \%\) confidence interval for \(p_{1}-p_{2}\). (c) Interpretation Explain the meaning of the confidence interval in the context of this problem. Does the confidence interval contain all positive numbers? all negative numbers? both positive and negative numbers? What does this tell you (at the \(99 \%\) confidence level) about the comparison of the population proportion of unidentified artifacts at high elevations \((7000-7500\) feet \()\) with the population proportion of unidentified artifacts at lower elevations (5000-5500 feet)? How does this relate to the stated hypothesis?

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