/*! 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 34 You are planning a survey of stu... [FREE SOLUTION] | 91Ó°ÊÓ

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You are planning a survey of students at a large university to determine what proportion favor an increase in student fees to support an expansion of the student newspaper. Using records provided by the registrar, you can select a random sample of students. You will ask each student in the sample whether he or she is in favor of the proposed increase. Your budget will allow a sample of 100 students. (a) For a sample of size 100, construct a table of the margins of error for \(95 \%\) confidence intervals when \(\mathrm{p}^{\wedge \hat{p}}\) takes the values \(0.1,0.3,0.5,0.7\), and \(0.9\). (b) A former editor of the student newspaper offers to provide funds for a sample of size 500 . Repeat the margin of error calculations in part (a) for the larger sample size. Then write a short thank-you note to the former editor describing how the larger sample size will improve the results of the survey.

Short Answer

Expert verified
Larger sample sizes reduce the margin of error, improving survey reliability.

Step by step solution

01

Understanding Margin of Error

The margin of error (ME) for a proportion is calculated using the formula \( ME = Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( Z \) is the Z-score for a 95% confidence interval (which is approximately 1.96), \( \hat{p} \) is the sample proportion, and \( n \) is the sample size.
02

Calculating Margin of Error for Sample Size 100

Using \( n = 100 \), calculate the margin of error for each given value of \( \hat{p} \): 1. For \( \hat{p} = 0.1 \), \( ME = 1.96 \times \sqrt{\frac{0.1 \times 0.9}{100}} = 0.0588 \).2. For \( \hat{p} = 0.3 \), \( ME = 1.96 \times \sqrt{\frac{0.3 \times 0.7}{100}} = 0.0918 \).3. For \( \hat{p} = 0.5 \), \( ME = 1.96 \times \sqrt{\frac{0.5 \times 0.5}{100}} = 0.098 \).4. For \( \hat{p} = 0.7 \), \( ME = 1.96 \times \sqrt{\frac{0.7 \times 0.3}{100}} = 0.0918 \).5. For \( \hat{p} = 0.9 \), \( ME = 1.96 \times \sqrt{\frac{0.9 \times 0.1}{100}} = 0.0588 \).
03

Table Construction for Margin of Error (n=100)

Construct a table for \( n = 100 \):\[\begin{array}{|c|c|}\hline\hat{p} & \text{Margin of Error} \\hline0.1 & 0.0588 \0.3 & 0.0918 \0.5 & 0.098 \0.7 & 0.0918 \0.9 & 0.0588 \\hline\end{array}\]
04

Calculating Margin of Error for Sample Size 500

Using \( n = 500 \), calculate the margin of error for each value of \( \hat{p} \): 1. For \( \hat{p} = 0.1 \), \( ME = 1.96 \times \sqrt{\frac{0.1 \times 0.9}{500}} = 0.0262 \).2. For \( \hat{p} = 0.3 \), \( ME = 1.96 \times \sqrt{\frac{0.3 \times 0.7}{500}} = 0.0409 \).3. For \( \hat{p} = 0.5 \), \( ME = 1.96 \times \sqrt{\frac{0.5 \times 0.5}{500}} = 0.0439 \).4. For \( \hat{p} = 0.7 \), \( ME = 1.96 \times \sqrt{\frac{0.7 \times 0.3}{500}} = 0.0409 \).5. For \( \hat{p} = 0.9 \), \( ME = 1.96 \times \sqrt{\frac{0.9 \times 0.1}{500}} = 0.0262 \).
05

Table Construction for Margin of Error (n=500)

Construct a table for \( n = 500 \):\[\begin{array}{|c|c|}\hline\hat{p} & \text{Margin of Error} \\hline0.1 & 0.0262 \0.3 & 0.0409 \0.5 & 0.0439 \0.7 & 0.0409 \0.9 & 0.0262 \\hline\end{array}\]
06

Writing the Thank-You Note

Dear Former Editor, Thank you very much for your generous contribution to our student survey project. With your support, we will be able to survey a larger sample of 500 students. This increase in sample size will significantly reduce the margin of error in our findings, leading to more precise and reliable results. This will greatly enhance the quality and impact of the survey on student opinions regarding the increase in student fees. Sincerely, [Your Name]

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Key Concepts

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

Confidence Intervals
Confidence intervals are a crucial concept in statistics that help us estimate how reliable our survey results are. Imagine you conducted a survey on a group of university students to find out what proportion of them favor an increase in student fees. However, your survey won’t always show the exact same result if you repeat it multiple times. That's where confidence intervals come into play.

A confidence interval gives a range of values within which the true proportion is likely to fall. For a 95% confidence interval, we can be 95% sure that the true population proportion lies within this range. It is like saying, "I am fairly confident that the true percentage of students who support the fee increase is between this lower number and this higher number."
  • The width of the confidence interval depends on the sample size and the variability in the data.
  • A wider interval suggests less certainty about the estimate.
  • A narrower interval indicates more precision and confidence in the results.
Understanding confidence intervals is key to interpreting survey results accurately in the context of the student fee survey.
Sample Size
When conducting surveys, the sample size is a critical element that affects the reliability of your results. In our example of surveying a university's student opinions on fee increases, the original sample size was 100. Later, a generous contribution allowed it to increase to 500.

The sample size determines how well the survey reflects the larger population. Generally, a larger sample size provides more accurate and reliable results, as it reduces the margin of error. A smaller margin of error means you can be more confident that your survey results are closer to the truth.
  • Smaller sample sizes lead to larger margins of error and less reliable estimates.
  • Larger sample sizes provide stronger statistical power, reducing the uncertainty in estimates.
  • Increasing sample size doesn't just lower margin of error, but also makes the survey findings more generalizable to the entire population.
When deciding on the appropriate sample size for a survey, consider the precision you need for your results and the resources available for conducting the survey.
Survey Statistics
Survey statistics involve the methods and techniques employed to gather, analyze, and interpret data collected from surveys. In our student survey, we use statistical methods to determine how students view fee increases, and how confident we can be in those results based on the data collected.

Survey statistics help ensure that the results are valid, reliable, and representative of the entire population. They guide decisions on how to design the survey, select respondents, and analyze responses. Below are some points to consider in survey statistics:
  • Random sampling techniques are essential for minimizing bias and ensuring each student has an equal chance of being chosen.
  • The statistical methods applied help in calculating vital metrics like the margin of error and confidence intervals.
  • Statistical analysis provides important insights by comparing different sample sizes and their impacts on survey outcomes.
Survey statistics are the backbone that guides the entire process from data collection to insights, ensuring the results are meaningful and actionable.
Proportions in Statistics
Proportions in statistics are used to compare quantities, often representing the relationship between a part and a whole. When conducting surveys, like the one about student opinions on fee increases, proportions play a vital role.

In this context, a proportion might be the fraction or percentage of students who favor the fee increase. It is an essential metric because it characterizes the main subject of your survey - whether there is enough support for the change.
  • Proportions are calculated as the number of favorable responses divided by the total number of responses.
  • They provide insights into the distribution of opinions within the sample.
  • Understanding proportions helps in making significant statistical inferences about the broader student population.
By focusing on proportions, researchers can effectively summarize and communicate the findings of their surveys, making it easier to interpret the level of support or opposition among the group studied.

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Most popular questions from this chapter

Students are reluctant to report cheating by other students. A student project put this question to an SRS of 172 undergraduates at a large university: "You witness two students cheating on a quiz. Do you go to the professor?" Only 19 answered Yes. \({ }^{20}\) Give a \(95 \%\) confidence interval for the proportion of all undergraduates at this university who would report cheating.

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.

Throughout Europe, more than 8000 pedestrians are killed each year in road accidents, with approximately \(25 \%\) of these dying when using a pedestrian crossing. Although failure to stop for pedestrians at a pedestrian crossing is a serious traffic offense in France, more than half of drivers do not stop when a pedestrian is waiting at a crosswalk. In this experiment, a female research assistant was instructed to stand at a pedestrian crosswalk and stare at the driver's face as a car approached the crosswalk. In 400 trials, the research assistant maintained a neutral expression, and in a second set of 400 trials, the research assistant was instructed to smile. The order of smiling or not smiling was randomized, and several pedestrian crossings were used in a town on the coast in the west of France. Research assistants were dressed in normal attire for their age (jeans/T-shirt/trainers/sneakers). \({ }^{23}\) (a) In the 400 trials in which the assistant maintained a neutral expression, the driver stopped in 229 out of the 400 trials. Find a \(95 \%\) confidence interval for the proportion of drivers who would stop when a neutral expression is maintained. (b) In the 400 trials in which the assistant smiled at the driver, the driver stopped in 277 out of the 400 trials. Find a 95\% confidence interval for the proportion of drivers who would stop when the assistant is smiling. (c) What do your results in parts (a) and (b) suggest about the effect of a smile on a driver stopping at a pedestrian crosswalk? Explain briefly. (In Chapter 23, we will consider formal methods for comparing two proportions.)

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.

\(\hat{p}\). Greenville County, South Carolina, has 461,299 adult residents, of which 59,969 are 65 years or older. A survey wants to contact \(n=689\) residents. \({ }^{5}\) (a) Find \(p\), the proportion of Greenville county adult residents who are 65 years or older. (b) If repeated simple random samples of 689 residents are taken, what would be the range of the sample proportion of adults over 65 in the sample according to the 95 part of the 68-95-99.7 rule? (c) The actual survey contacted 689 adults using random digit dialing of residential numbers using a database of exchanges, with no cell phone numbers contacted. The 689 respondents represent a response rate of approximately \(30 \%\). In the sample obtained, 253 of the 689 adults contacted were over 65 . Do you have any concerns treating this as a simple random sample from the population of adult residents of Greenville County? Explain briefly.

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