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Chapter 22: Problem 43

Some shrubs have the useful ability to resprout from their roots after their tops are destroyed. Fire is a particular threat to shrubs in dry climates because it can injure the roots as well as destroy the aboveground material. One study of resprouting took place in a dry area of Mexico. \({ }^{32}\) The investigators clipped the tops of samples of several species of shrubs. In some cases, they also applied a propane torch to the stumps to simulate a fire. Of 12 specimens of the shrub Krameria cytisoides, five resprouted after fire. Estimate with \(90 \%\) confidence the proportion of all shrubs of this species that will resprout after fire.

Short Answer

Expert verified
The 90% confidence interval is approximately [0.1833, 0.6501].

Step by step solution

01

Define the Problem

We need to estimate the proportion of the shrub species Krameria cytisoides that can resprout after a fire, using a confidence level of 90%. We're given that out of 12 specimens, 5 resprouted after being subjected to simulated fire.
02

Calculate Sample Proportion

First, calculate the sample proportion of shrubs that resprouted. We denote this proportion with \( p \). The formula for sample proportion is given by: \[ \hat{p} = \frac{x}{n} \] where \( x \) is the number of resprouting shrubs and \( n \) is the total number of shrubs surveyed. Here, \( x = 5 \) and \( n = 12 \). Thus, \[ \hat{p} = \frac{5}{12} \approx 0.4167 \].
03

Find the Standard Error

Calculate the standard error \( SE \) of the sample proportion \( \hat{p} \). The standard error is calculated using the formula: \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \] Substituting \( \hat{p} = 0.4167 \) and \( n = 12 \), we get: \[ SE = \sqrt{\frac{0.4167 (1 - 0.4167)}{12}} \approx 0.1419 \].
04

Determine the Z-Score for 90% Confidence

For a 90% confidence level, we use a Z-score of approximately 1.645. This Z-score corresponds to the critical value that covers the central 90% of the normal distribution.
05

Calculate the Confidence Interval

Use the formula for the confidence interval, which is: \[ \hat{p} \pm Z \times SE \] Substituting \( \hat{p} = 0.4167 \), \( Z = 1.645 \), and \( SE = 0.1419 \): \[ 0.4167 \pm 1.645 \times 0.1419 \] This results in an interval: \[ 0.4167 \pm 0.2334 \] Thus, the 90% confidence interval is approximately \( [0.1833, 0.6501] \).
06

Conclusion

We can be 90% confident that the true proportion of Krameria cytisoides shrubs that can resprout after a fire is between 18.33% and 65.01%.

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Key Concepts

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

Sample Proportion
The concept of sample proportion is a foundational aspect of statistical analysis, especially when estimating characteristics of a population. In simple terms, the sample proportion is the ratio of the number of times an event occurs to the total number of observations or trials. In the example of Krameria cytisoides shrubs, we need to determine how many shrubs resprout after a simulated fire.
The formula for sample proportion is:
  • \( \hat{p} = \frac{x}{n} \)
where \( x \) represents the number of successful observations (shrubs that resprouted), and \( n \) denotes the total number of observations (total shrubs).
For this study, \( x = 5 \) and \( n = 12 \) giving us the sample proportion \( \hat{p} \approx 0.4167 \). This means approximately 41.67% of the shrubs resprouted in the sample provided.
Standard Error
Understanding the standard error (SE) is crucial for making accurate predictions and drawing meaningful conclusions from your data. The standard error provides an estimate of how much the sample proportion might differ from the true population proportion.
To calculate SE, use the formula:
  • \( SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \)
By substituting our values (\( \hat{p} \approx 0.4167 \) and \( n = 12 \)), we find:
  • \( SE \approx 0.1419 \)
This tells us the variability or uncertainty around our sample proportion is about 14.19% in this specific study. The standard error helps in constructing the confidence interval, a critical step in estimation.
Z-Score
A Z-score is a statistical measure that helps determine how a sample data point relates to a population mean. When calculating confidence intervals, Z-scores are used to find the distance in standard error units from the mean to include a specified percentage of the data in the interval.
For a 90% confidence level, the Z-score used is 1.645. This value indicates the extent to which data varies within a 90% probability range from the estimated mean sample proportion. The Z-score is essential to ensure that the calculated confidence interval accurately reflects the median cluster of the population data.
Resprouting Shrubs
Resprouting shrubs have an incredible ability to recover and grow again after significant disturbance, such as being burnt or clipped. The study of Krameria cytisoides shrubs provides scientists with insights into how these plants may survive fires, which are common in certain climates. Resprouting is a survival mechanism for many plant species, allowing them to withstand natural disasters and regenerate even after their above-ground portions are destroyed.
In the context of the study, understanding the ability of these shrubs to resprout is critical for ecological management and the conservation of biodiversity in dry, fire-prone areas.
Estimation
Estimation is the statistical process of making predictions or inferring information about a population based on sample data. In the case of the shrubs surveyed in Mexico, estimation allows us to infer the resprouting proportion for all shrubs of the species Krameria cytisoides following a fire.
Steps in this process include calculating the sample proportion, determining the standard error, applying the Z-score, and arriving at a confidence interval. Estimation gives a range鈥攈ere between approximately 18.33% and 65.01%鈥攚ithin which we expect the true proportion to fall. The use of estimation techniques is widespread across scientific research, healthcare, business, and beyond to make informed decisions based on observable data.

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Most popular questions from this chapter

Does the order in which wine is presented make a difference? Several choices of wine are presented one at a time and in sequence, and the subject is then asked to choose the preferred wine at the end of the sequence. In this study, subjects were asked to taste two wine samples in sequence. Both samples given to a subject were the same wine, although subjects were expecting to taste two different samples of a particular variety. Of the 32 subjects in the study, 22 selected the wine presented first, when presented with two identical wine samples. 29 (a) Do the data give good reason to conclude that the subjects are not equally likely to choose either of the two positions when presented with two identical wine samples in sequence? (b) The subjects were recruited in Ontario, Canada, via advertisements to participate in a study of "attitudes and values toward wine." Can we generalize our conclusions to all wine tasters? Explain

Although an increasing share of Americans are reading ebooks on tablets and smartphones rather than dedicated e-readers, print books continue to be much more popular than books in digital format. (Digital format includes both e-books and audio books.) A Pew Research Center survey of 1520 adults nationwide conducted March 7-April 14, 2016, found that 1125 of those surveyed read a book in either print or digital format in the last 12 months. \({ }^{27}\) (a) What can you say with \(95 \%\) confidence about the percent of all adults who have read a book in either print or digital format in the last 12 months? (b) Of the 1125 surveyed who have read a book in the last 12 months, 91 had read only digital books. Among those adults who have read a book in the last 12 months, find a \(95 \%\) confidence interval for the proportion that have read digital books exclusively. (You may regard the 1125 adults in the survey who have read a book in the last 12 months as a random sample of readers.)

Canada has much stronger gun control laws than the United States, and Canadians support gun control more strongly than do Americans. A sample survey asked a random sample of 1505 adult Canadians, "Do you agree or disagree that all firearms should be registered?" Of the 1505 people in the sample, 1288 answered either "Agree strongly" or "Agree somewhat."9 (a) The survey dialed residential telephone numbers at random in all 10 Canadian provinces (omitting the sparsely populated northern territories). Based on what you know about sample surveys, what is likely to be the biggest weakness in this survey? (b) Nonetheless, act as if we have an SRS from adults in the Canadian provinces. Give a \(95 \%\) confidence interval for the proportion who support registration of all firearms.

Sample surveys usually contact large samples, so we can use the large-sample confidence interval if the sample design is close to an SRS. Scientific studies often use smaller samples that require the plus four method. For example, Familial Adenomatous Polyposis (FAP) is a rare inherited disease characterized by the development of an extreme number of polyps early in life and colon cancer in virtually \(100 \%\) of patients before the age of 40 . A group of 14 people suffering from FAP being treated at the Cleveland Clinic drank black raspberry powder in a slurry of water every day for nine months. The numbers of polyps were reduced in 11 out of 14 of these patients. \({ }^{18}\) (a) Why can't we use the large-sample confidence interval for the proportion \(p\) of patients suffering from FAP that will have the number of polyps reduced after nine months of treatment? (b) The plus four method adds four observations-two successes and two failures. What are the sample size and the number of successes after you do this? What is the plus four estimate \(\mathrm{p}^{-\bar{p}}\) of \(p\) ? (c) Give the plus four \(90 \%\) confidence interval for the proportion of patients suffering from FAP who will have the number of polyps reduced after nine months of treatment.

A high-school teacher in a lowincome urban school in Worcester, Massachusetts, used cash incentives to encourage student learning in his AP statistics class. \({ }^{28}\) In 2010,15 of the 61 students enrolled in his class scored a 5 on the AP statistics exam. Worldwide, the proportion of students who scored a 5 in 2010 was \(0.15\). Is this evidence that the proportion of students who would score a 5 on the AP statistics exam when taught by the teacher in Worcester using cash incentives is higher than the worldwide proportion of \(0.15\) ? (a) State hypotheses, find the \(P\)-value, and give your conclusions in the context of the problem. Do you have any reservations about using the \(z\) test for proportions for this data? (b) Does this study provide evidence that cash incentives cause an increase in the proportion of \(5 \mathrm{~s}\) on the AP statistics exam? Explain your answer.
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