/*! 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 46 Multiple choice: Select the best... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Multiple choice: Select the best answer for Exercises 43 to \(46,\) which refer to the following setting. The magazine Sports Illustrated asked a random sample of 750 Division I college athletes, "Do you believe performance- enhancing drugs are a problem in college sports?" Suppose that 30\(\%\) of all Division I athletes think that these drugs are a problem. Let \(\hat{p}\) be the sample proportion who say that these drugs are a problem. The sampling distribution of \(\hat{p}\) is approximately Normal because (a) there are at least 7570 Division I college athletes. (b) \(n p=225\) and \(n(1-p)=525\) (c) a random sample was chosen. (d) a large sample size like \(n=750\) guarantees it. (e) the sampling distribution of \(\hat{\rho}\) always has this shape.

Short Answer

Expert verified
Option (b) is correct.

Step by step solution

01

Understand the Question

We need to determine the reason why the sampling distribution of the sample proportion \( \hat{p} \) is approximately normal based on the given options.
02

Recall Conditions for Normality

The sampling distribution of \( \hat{p} \) is approximately normal if certain conditions are met, typically this involves having a large enough sample size to satisfy the conditions \( np \geq 10 \) and \( n(1-p) \geq 10 \).
03

Calculate \( np \) and \( n(1-p) \)

Given \( n = 750 \) and \( p = 0.3 \), calculate \( np \) and \( n(1-p) \): - \( np = 750 \times 0.3 = 225 \)- \( n(1-p) = 750 \times 0.7 = 525 \). Both values are greater than 10.
04

Analyze the Options

Based on our calculations:- (a) at least 7570 athletes is not relevant to the normality.- (b) is true as \( np = 225 \) and \( n(1-p) = 525 \) are both greater than 10.- (c) random sampling ensures unbiasedness, not normality.- (d) a large \( n \) aids in normality but needs \( np \) and \( n(1-p) \).- (e) isn't always true since normal shape depends on \( np \) and \( n(1-p) \).
05

Select the Best Answer

Option (b) is the correct answer as it correctly identifies that both \( np \) and \( n(1-p) \) being greater than 10 ensures the shape is approximately normal.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Normality Conditions
In statistics, normality conditions are crucial for understanding the behavior of sampling distributions. The sampling distribution of a sample proportion, denoted as \( \hat{p} \), is approximately normal when certain conditions are fulfilled, primarily concerning the sample size.
A key criterion is that both \( np \) and \( n(1-p) \) should be greater than or equal to 10. This ensures that there is a sufficient number of successes and failures in the sample.
  • \( np \geq 10 \)
  • \( n(1-p) \geq 10 \)
These conditions help in stabilizing the distribution so that it can be approximated by a normal distribution. This approximation is very useful when performing tasks like hypothesis testing or constructing confidence intervals.
Sample Proportion
The sample proportion, denoted as \( \hat{p} \), represents the fraction of items in a sample that possess a certain attribute. In the context of the exercise, it's the proportion of college athletes in the sample who believe performance-enhancing drugs pose a problem.
Calculating \( \hat{p} \) is relatively straightforward. It is simply:\[ \hat{p} = \frac{x}{n} \]Where:
  • \( x \) = number of occurrences of the attribute in the sample
  • \( n \) = total sample size
Understanding the variability in \( \hat{p} \) is important because the sample proportion may differ from the true population proportion. This is why examining the sampling distribution of \( \hat{p} \) becomes significant, especially under the assumption of normality.
This allows researchers to draw inferences about the population with greater accuracy.
Performance-Enhancing Drugs
Performance-enhancing drugs are substances used by athletes to improve their physical capabilities. The issue of such drugs in sports is quite significant because of ethical, health, and fairness concerns. In the exercise's setting, the magazine seeks to determine how pervasive this problem is in college sports by surveying athletes.
Understanding the perceptions of athletes on this issue can shed light on the severity of the problem, potential deterrents, and the need for policy interventions. This information is essential for:
  • Ensuring fair competition
  • Protecting athletes' health
  • Maintaining the integrity of sports environments
By sampling athletes' views, researchers and policymakers can gauge the extent of awareness and concern regarding performance-enhancing drugs, guiding them to act accordingly.
Random Sampling
Random sampling is a fundamental method in statistics used to ensure that every individual in a population has an equal chance of being selected in a sample. This approach minimizes biases and enhances the representativeness of the sample, making it easier to generalize findings to the larger population.
In the exercise, Sports Illustrated employed random sampling of 750 Division I college athletes to ascertain views on performance-enhancing drugs.
Here are some reasons why random sampling is valuable:
  • Increases reliability and validity of results
  • Ensures diversity within the sample
  • Allows for unbiased estimates of population parameters
By ensuring a random selection, the study achieves a more accurate reflection of the athletes' opinions, mitigating the risks of skewed data due to sampling bias.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Bottling cola A bottling company uses a flling machine to fill plastic bottles with cola. The bottles are supposed to contain 300 milliters \((\mathrm{ml}) .\) In fact, the contents vary according to a Normal distribution with mean \(\mu=298 \mathrm{ml}\) and standard deviation \(\sigma=3 \mathrm{ml}\) (a) What is the probability that an individual bottle contains less than 295 \(\mathrm{ml}\) ? Show your work. (b) What is the probability that the mean contents of six randomly selected bottles is less than 295 \(\mathrm{ml}\) ? Show your work.

Run a mile During World War II, \(12,000\) able bodied male undergraduates at the University of Illinois participated in required physical training. Each student ran a timed mile. Their times followed the Normal distribution with mean 7.11 minutes and standard deviation 0.74 minute. An SRS of 100 of these students has mean time \(\overline{x}=7.15\) minutes. A second SRS of size 100 has mean \(\overline{x}=6.97\) minutes. After many SRSs, the values of the sample mean \(\overline{x}\) follow the Normal distribution with mean \(7.11 \mathrm{minutes}-\) and standard deviation 0.074 minute. (a) What is the population? Describe the population distribution. (b) Describe the sampling distribution of \(\overline{x} .\) How is it different from the population distribution?

Increasing the sample size of an opinion poll will (a) reduce the bias of the poll result. (b) reduce the variability of the poll result. (c) reduce the effect of nonresponse on the poll. (d) reduce the variability of opinions. (e) all of the above.

For each boldface number in Exercises 5 to \(8,(1)\) state whether it is a parameter or a statistic and (2) use appropriate notation to describe each number; for example, \(p=0.65\) Unlisted numbers A telemarketing firm in Los Angeles uses a device that dials residential telephone numbers in that city at random. Of the first 100 numbers dialed, 48\(\%\) are unlisted. This is not surprising because 52\(\%\) of all Los Angeles residential phones are unlisted.

The candy machine Suppose a large candy machine has 15\(\%\) orange candies. Imagine taking an SRS of 25 candies from the machine and observing the sample proportion \(\hat{p}\) of orange candies. (a) What is the mean of the sampling distribution of \(\hat{p} ?\) Why? (b) Find the standard deviation of the sampling distribution of \(\hat{p}\) . Check to see if the 10\(\%\) condition is met. (c) Is the sampling distribution of \(\hat{p}\) approximately Normal? Check to see if the Normal condition is met. (d) If the sample size were 75 rather than \(25,\) how would this change the sampling distribution of \(\hat{p} ?\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.