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U.S. National Parks that contain designated wilderness areas are required by law to develop and maintain a wilderness stewardship plan. The Olympic National Park, containing some of the most biologically diverse wilderness in the United States, had a survey conducted in 2012 to collect information relevant to the development of such a plan. National Park Service staff visited 30 wilderness trailheads in moderate- to high-use areas over a 60-day period and asked visitors as they completed their hike to complete a questionnaire. The 1019 completed questionnaires, giving a response rate of \(50.4 \%\), provided each subject's opinions on the use and management of wilderness. In particular, there were 694 day users and 325 overnight users in the sample. \({ }^{25}\) (a) Why do you think the National Park staff only visited trailheads in moderate- to high-use areas to obtain the sample? (b) Assuming the 1019 subjects represent a random sample of users of the wilderness areas in the Olympic National Park, give a \(90 \%\) confidence interval for the proportion of day users. (c) The response rate was \(49 \%\) for day users and \(52 \%\) for overnight users. Does this lessen any concerns you might have regarding the effect of nonresponse on the interval you obtained in part (b)? Explain briefly. (d) (d) Do you think it would be better to refer to the interval in part (b) as a confidence interval for the proportion of day users or the proportion of day users on the most popular trails in the park? Explain briefly.

Short Answer

Expert verified
(a) High-use areas were chosen to efficiently gather diverse data from main users.\n(b) 90% CI for day users proportion is (0.657, 0.705).\n(c) Similar response rates lessen nonresponse concerns.\n(d) Interval is for most popular trails' day users.

Step by step solution

01

Understanding the Sampling Location Choice

The National Park staff targeted trailheads in moderate- to high-use areas to efficiently gather data where most visitors are found. This decision increases the likelihood of obtaining a sufficient sample size, ensures the diversity of opinions by interacting with a mixed crowd of park users, and likely provides insights that are more broadly representative of the primary users of the park's wilderness.
02

Calculate the Proportion of Day Users

To find the proportion of day users, divide the number of day users by the total number of questionnaires completed. This gives \( p = \frac{694}{1019} \approx 0.681 \).
03

Determine the Confidence Interval

To calculate the 90% confidence interval for the proportion of day users, use the formula for a confidence interval for proportions: \[ p \pm Z \times \sqrt{\frac{p(1-p)}{n}}\]Where \( Z = 1.645 \) for a 90% confidence interval. Plugging in the values, we get:\[0.681 \pm 1.645 \times \sqrt{\frac{0.681 \times (1 - 0.681)}{1019}} \approx 0.681 \pm 0.024\]This results in an interval of \( (0.657, 0.705) \).
04

Assess Impact of Nonresponse

The given response rates are 49% for day users and 52% for overnight users. Since these rates are fairly similar and neither group is disproportionately underrepresented, nonresponse is unlikely to significantly bias our confidence interval for day users.
05

Defining the Interval's Scope

We would refer to the interval as a confidence interval for the proportion of day users on the most popular trails, as the sample and data are skewed towards these high-use areas, not necessarily the whole park.

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Key Concepts

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

Wilderness Stewardship Plan
A Wilderness Stewardship Plan is vital for managing the delicate balance between preserving natural ecosystems and providing enjoyable experiences for visitors. The plan outlines policies and strategies for maintaining the integrity of wilderness areas within National Parks, ensuring they remain unspoiled for future generations.
In the context of a park like Olympic National Park, which boasts a rich array of biodiversity, such a plan ensures that both conservation goals and visitor needs are met.
  • It helps set guidelines for sustainable tourism practices.
  • Includes measures for protecting native species and landscapes.
  • Supports educational initiatives to foster visitor awareness and appreciation of the wilderness.
The survey conducted by the National Park Service serves as a foundational tool for crafting an effective Wilderness Stewardship Plan by capturing visitor opinions and preferences regarding park use and management.
Confidence Interval Calculations
Calculating a confidence interval is a statistical method used to estimate the degree of certainty around a specific measure from a sample, such as the proportion of day users in a park. This technique is essential for inferential statistics, allowing researchers to make predictions about a population based on sample data.
In our example, we calculated a 90% confidence interval for the proportion of day users. The formula applied was \[ p \pm Z \times \sqrt{\frac{p(1-p)}{n}} \]where \( p \) is the proportion of day users determined from the sample, \( Z \) is the Z-score corresponding to the confidence level (1.645 for 90%), and \( n \) is the total number of responses.
  • The calculated interval provides a range, suggesting that the true proportion of day users in the park falls between 65.7% and 70.5%.
  • The confidence level (90%) indicates that such an interval would contain the true proportion in 90 out of 100 similar studies.
This process helps park managers plan better by understanding day visitor dynamics more clearly.
Survey Response Analysis
Survey response analysis involves examining the completeness and accuracy of the data collected during a survey. It is crucial for verifying that the survey's findings are reliable and meaningfully reflect the studied population.
In the Olympic National Park survey, the response rates were 49% for day users and 52% for overnight users. Since these rates are similar, it implies a balanced representation, suggesting that nonresponse bias may have a minimal effect.
  • The response rate, being over 50%, indicates a decent level of engagement from park visitors.
  • By measuring response disparity between different user groups (day vs overnight), analysts can ensure their findings are not skewed.
  • This helps validate the confidence interval calculations by confirming the randomness and representativeness of the sample.
Effective survey response analysis ensures that statistical insights are robust and actionable.
Sampling Techniques
Sampling techniques are strategies used to select a portion of a larger population for study, aiming to make inferences about the whole. In the context of the National Park survey, understanding why specific sampling techniques were chosen can clarify the reliability and relevance of the data.
The National Park Service staff opted to survey at trailheads in moderate- to high-use areas to maximize efficiency and diversity in responses. This strategic location choice ensures:
  • A high yield of responses due to greater visitor traffic.
  • A mix of day and overnight users that mirrors actual park usage trends, enhancing the sample's representativeness.
Different sampling techniques offer various benefits and trade-offs, including:
  • Simple random sampling for equal representation chances.
  • Stratified sampling to ensure inclusion of key subgroups.
By understanding and applying the right sampling technique, surveyors can obtain data that truly represent the usage patterns and opinions of all park visitors.

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Most popular questions from this chapter

How many American adults must be interviewed to estimate the proportion of all American adults who actively try to avoid drinking regular soda or pop within \(\pm 0.01\) with \(99 \%\) confidence using the large-sample confidence interval? Use \(0.5\) as the conservative guess for \(p\). (a) \(n=6765\) (b) \(n=9604\) (c) \(n=16590\)

In Exercise 22.17, suppose we computed a large-sample 99\% confidence interval for the proportion of all American adults who actively try to avoid drinking regular soda or pop. This \(99 \%\) confidence interval (a) would have a smaller margin of error than the \(90 \%\) confidence interval. (b) would have a larger margin of error than the \(90 \%\) confidence interval. (c) could have either a smaller or a larger margin of error than the \(90 \%\) confidence interval. This varies from sample to sample.

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.

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.

The value of the \(z\) statistic for the Exercise \(22.22\) is \(2.53\). This test is (a) not significant at either \(\alpha=0.05\) or \(\alpha=0.01\). (b) significant at \(\alpha=0.05\) but not at \(\alpha=0.01\). (c) significant at both \(\alpha=0.05\) and \(\alpha=0.01\).

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