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Chapter 16: Problem 8

Campus sample For her final project, Stacy plans on surveying a random sample of 50 students on whether they plan to go to Florida for spring break. From past years, she guesses that about \(10 \%\) of the class goes. Is it reasonablefor her to use a Normal model for the sampling distribution of the sample proportion? Why or why not?

Short Answer

Expert verified
No, it is not reasonable for Stacy to use a Normal model for the sampling distribution of the sample proportion as the success-failure condition is not met because the number of successes (\(np = 5\)) is less than 10.

Step by step solution

01

Check Binary Condition

The outcome is binary - The question is whether they plan to go to Florida for spring break or not, so each student either plans to go (success) or does not plan to go (failure).
02

Check Independence Condition

Given that the sample size is 50 and assuming that this is less than 10% of all students, it can be assumed that the sampling is done randomly and independently.
03

Check Success-Failure Condition

The estimated proportion of success (students planning to go to Florida for spring break) is 10%, that is, \(p = 0.10\). Hence, the estimated proportion of failures \(q = 1 - p = 0.90\). Multiply these percentages by the sample size (50) to confirm if they're 鈮� 10. So, \(np = 50 * 0.10 = 5\) and \(nq = 50 * 0.90 = 45\). The success condition does not hold (5 < 10) while the failure condition does hold (45 > 10).

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Key Concepts

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

Understanding the Binary Condition
When talking about sampling in statistics, the term binary condition refers to scenarios where there are only two possible outcomes. In the context of Stacy's survey about spring break plans, these two outcomes are either a student plans to go to Florida (which we can call a 'success') or does not plan to go (a 'failure').

This distinction is critical for determining the type of statistical model we can apply, as certain models, like the binomial or the Normal model, require a binary setup to make sense of the data. It's like flipping a coin - you only have heads or tails, no in-betweens. Stacy has correctly identified that her survey question meets this binary condition.
Independence Condition In Sampling
The independence condition is a cornerstone for reliable statistical calculations. It means each selection in the sample does not influence another, which is often satisfied if each individual is randomly selected and forms less than 10% of the population.

Imagine plucking apples from a tree - if plucking one doesn't affect which apple you pluck next, then the selections are independent. For Stacy's survey, we assume random sampling and that her 50 students are a fraction of the total student body, thus satisfying the independence condition. This ensures that results from one student do not bias another's response.
Delving into the Success-Failure Condition
The success-failure condition is pivotal to ascertain the validity of using a Normal model for binomial situations. For a sample size 'n' and the probability of success 'p', this condition requires both 'np' and 'n(1-p)' to be at least 10.

Understand it like this - if too few 'successes' or 'failures' are expected, the distribution of sample proportion will be skewed rather than Normal. Since Stacy's calculation gave her 5 expected successes, which is less than the minimum of 10, the success-failure condition fails. With this condition unmet, the use of Normal distribution is questionable.
Normal Model: When to Use It
A Normal model is a bell-shaped curve representing data distribution and it's very handy when dealing with probabilities and statistics. However, it hinges on certain assumptions like having a large enough number of successes and failures (as addressed in the success-failure condition) and the sample observations being independent.

Although popular for its simplicity and ease of use, particularly in predicting probabilities, the Normal model isn't the go-to for all scenarios. Like Stacy's survey, if expected successes (or failures) are too limited, the model might not accurately represent the data distribution, leading to incorrect conclusions.
Sample Proportion and Its Importance
In Stacy's context, sample proportion represents the part of her survey participants planning to go to Florida for spring break. Mathematically, it's determined by dividing the number of 'successes' in the sample by the total sample size.

It's a key piece of information as it provides an estimate of the actual proportion in the entire population - think of it like a snapshot from a bigger picture. A well-calculated sample proportion is vital for making predictions or inferences about the larger group, making it a fundamental concept in sample surveys and the field of inferential statistics.

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Most popular questions from this chapter

Teenage drivers An insurance company checks police records on 582 accidents selected at random and notes that teenagers were at the wheel in 91 of them. a. Create a \(95 \%\) confidence interval for the percentage of all auto accidents that involve teenage drivers. b. Explain what your interval means. c. Explain what "95\% confidence" means. d. A politician urging tighter restrictions on drivers' licenses issued to teens says, "In one of every five auto accidents, a teenager is behind the wheel." Does your confidence interval support or contradict this statement? Explain.

Pilot study A state's environmental agency worries that many cars may be violating clean air emissions standards. The agency hopes to check a sample of vehicles in order to estimate that percentage with a margin of error of \(3 \%\) and \(90 \%\) confidence. To gauge the size of the problem, the agency first picks 60 cars and finds 9 with faulty emissions systems. How many should be sampled for a full investigation?

More conditions Consider each situation described. Identify the population and the sample, explain what \(p\) and \(\hat{p}\) represent, and tell whether the methods of this chapter can be used to create a confidence interval. a.A consumer group hoping to assess customer experiences with auto dealers surveys 167 people who recently bought new cars; \(3 \%\) of them expressed dissatisfaction with the salesperson. b. What percent of college students have cell phones? 2883 students were asked as they entered a football stadium, and 2430 said they had phones with them. c. Two hundred forty potato plants in a field in Maine are randomly checked, and only 7 show signs of blight. How severe is the blight problem for the U.S. potato industry? d. Twelve of the 309 employees of a small company suffered an injury on the job last year. What can the company expect in future years?

Conditions For each situation described below, identify the population and the sample, explain what \(p\) and \(\hat{p}\) represent, and tell whether the methods of this chapter can be used to create a confidence interval. a. Police set up an auto checkpoint at which drivers are stopped and their cars inspected for safety problems. They find that 14 of the 134 cars stopped have at least one safety violation. They want to estimate the percentage of all cars that may be unsafe. b. A TV talk show asks viewers to register their opinions on prayer in schools by logging on to a website. Of the 602 people who voted, 488 favored prayer in schools. We want to estimate the level of support among the general public. c. A school is considering requiring students to wear uniforms. The PTA surveys parent opinion by sending a questionnaire home with all 1245 students; 380 surveys are returned, with 228 families in favor of the change. d. A college admits 1632 freshmen one year, and four years later, 1388 of them graduate on time. The college wants to estimate the percentage of all their freshman enrollees who graduate on time.

34\. Still living online The Pew Research poll described in Exercise 5 ? found that \(56 \%\) of a sample of 1060 teens go online several times a day. (Treat this as a Simple Random Sample.) a. Find the margin of error for this poll if we want \(95 \%\) confidence in our estimate of the percent of American mteens who go online several times a day. b. Explain what that margin of error means. c. If we only need to be \(90 \%\) confident, will the margin of error be larger or smaller? Explain. d. Find that margin of error. e. In general, if all other aspects of the situation remain the same, would smaller samples produce smaller or larger margins of error?
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