/*! 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 71 Independent random samples of \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 71

Independent random samples of \(n_{1}=n_{2}=n\) observations are to be selected from each of two binomial populations 1 and \(2 .\) If you wish to estimate the difference in the two population proportions correct to within . 05 , with probability equal to .98 , how large should \(n\) be? Assume that you have no prior information on the values of \(p_{1}\) and \(p_{2},\) but you want to make certain that you have an adequate number of observations in the samples.

Short Answer

Expert verified
Answer: The sample size should be 388 for each of the two binomial populations.

Step by step solution

01

Find the standard deviation of the difference in proportions under the worst-case scenario

Since we have no information about the values of \(p_1\) and \(p_2\), we will assume the worst-case scenario, where the difference in proportions is the largest. This occurs when \(p_1 = 0.5\) and \(p_2 = 0.5\). In this case, the standard deviation of the difference in proportions is given by: \(SE_{p_1-p_2} = \sqrt{\frac{p_1(1-p_1)}{n} + \frac{p_2(1-p_2)}{n}}\) For the worst-case scenario, we have: \(SE_{p_1-p_2} = \sqrt{\frac{0.5(1-0.5)}{n} + \frac{0.5(1-0.5)}{n}}\)
02

Calculate the z-score corresponding to the desired confidence level

We want to estimate the difference in population proportions with a probability of 0.98. This means that we need to find the z-score that corresponds to the area under the standard normal distribution that contains 98% of the total area. Using a z-score table or calculator, we find that the z-score is approximately 2.33, since the area to the left of z = 2.33 is approximately 0.99 (remember that since it's two-tailed, we consider the area up to 0.99).
03

Calculate the margin of error

We want the difference in proportions to be accurate within 0.05. The margin of error is given by: \(ME = z \cdot SE_{p_1-p_2}\) Since we want the margin of error to be 0.05, we have: \(0.05 = 2.33 \cdot SE_{p_1-p_2}\)
04

Calculate the sample size, n

Now, we need to solve for \(n\) in the equation from Step 3: \(0.05 = 2.33 \cdot SE_{p_1-p_2}\) Substituting the worst-case scenario standard deviation from Step 1: \(0.05 = 2.33 \cdot \sqrt{\frac{0.5(1-0.5)}{n} + \frac{0.5(1-0.5)}{n}}\) Solve for n: \(n = \frac{4 \cdot (2.33)^2}{0.05^2} \approx 387.12\) Since the sample size must be an integer, we round up to the nearest whole number: \(n = 388\)
05

Interpret the result

The sample size, \(n\), should be 388 for each of the two binomial populations to estimate the difference in the population proportions with a margin of error of 0.05 and a probability of 0.98.

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 91影视!

Key Concepts

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

Binomial Distribution
The binomial distribution is a fundamental concept in statistics, especially useful when dealing with probabilities of fixed numbers of successes in several trials. Imagine you are flipping a coin - each flip can land on either heads or tails, much like a trial in a binomial distribution where success or failure is the outcome.
In a binomial distribution, you have:
  • Two possible outcomes (success or failure)
  • A fixed number of trials
  • An independent and constant probability of success
In our scenario, we have two distinct populations, each following a binomial distribution. The key goal is to estimate their differences in proportion. With no prior knowledge of these proportions, we assume the worst-case scenario, often using values like 0.5 for the probability of success, which maximizes variability and gives the most conservative estimate.
Confidence Interval
A confidence interval gives us a range wherein we expect the true parameter to lie. It's like casting a net to catch a fish in a pond - we aim to make sure the fish, or in this case, the true parameter, falls within our net.
Confidence intervals are often expressed with a confidence level, such as 98% in this problem, reflecting how sure we are that the true value is within our calculated range.
When dealing with proportions, we often calculate the confidence interval around the estimate of the difference in proportions. In this exercise, achieving a high confidence level is essential because it indicates greater reliability of the estimate.
Margin of Error
The margin of error represents the amount of error we can tolerate in our estimation of a population parameter. It acts as a boundary showing the span of values that the true population parameter might lie within, under the given confidence level.
For this problem, the desired margin of error is 0.05, meaning we aim to estimate the difference in proportions within plus or minus this value. It's crucial for ensuring accuracy as it reflects the precision of our estimate.
The smaller the margin of error, the more precise our estimate, but often this requires a larger sample size or lower confidence level. Here, the equation linking margin of error, z-score, and standard error helps compute the necessary sample size to meet the margin of error requirement.
Z-Score
A z-score, in statistics, is a measure that describes a value's position in relation to the mean of a set of values, expressed in terms of standard deviations. It tells us how many standard deviations an element is from the mean.
In confidence interval calculations, the z-score is used to determine how much the sample mean deviates from the population mean. For a confidence level of 98%, the corresponding z-score is derived to cover this probability in a two-tailed distribution.
In this exercise, the z-score of approximately 2.33 was calculated, covering the central 98% of the distribution. This z-score ensures the confidence interval encloses the true parameter difference within the set confidence level, aligning with our statistical goals.

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

Do well-rounded people get fewer colds? A study on the Chronicle of Higher \(E d u\) cation was conducted by scientists at Carnegie Mellon University, the University of Pittsburgh, and the University of Virginia. They found that people who have only a few social outlets get more colds than those who are involved in a variety of social activities. \(^{14}\) Suppose that of the 276 healthy men and women tested, \(n_{1}=96\) had only a few social outlets and \(n_{2}=105\) were busy with six or more activities. When these people were exposed to a cold virus, the following results were observed: $$\begin{array}{lcc} & \text { Few Social Outlets } & \text { Many Social Outlets } \\\\\hline \text { Sample Size } & 96 & 105 \\\\\text { Percent with Colds } & 62 \% & 35 \%\end{array}$$ a. Construct a \(99 \%\) confidence interval for the difference in the two population proportions. b. Does there appear to be a difference in the population proportions for the two groups? c. You might think that coming into contact with more people would lead to more colds, but the data show the opposite effect. How can you explain this unexpected finding?

Explain what is meant by "margin of error" in point estimation.

One of the major costs involved in planning a summer vacation is the cost of lodging. Even within a particular chain of hotels, costs can vary substantially depending on the type of room and the amenities offered. \(^{4}\) Suppose that we randomly select 50 billing statements from each of the computer databases of the Marriott, Radisson, and Wyndham hotel chains, and record the nightly room rates. $$\begin{array}{lccc} & \text { Marriott } & \text { Radisson } & \text { Wyndham } \\\\\hline \text { Sample average } & \$ 170 & \$ 145 & \$ 150 \\\\\text { Sample standard deviation } & 17.5 & 10 & 16.5\end{array}$$ a. Describe the sampled population(s). b. Find a point estimate for the average room rate for the Marriott hotel chain. Calculate the margin of error. c. Find a point estimate for the average room rate for the Radisson hotel chain. Calculate the margin of error. d. Find a point estimate for the average room rate for the Wyndham hotel chain. Calculate the margin of error. e. Display the results of parts \(\mathrm{b}, \mathrm{c},\) and d graphically, using the form shown in Figure \(8.5 .\) Use this display to compare the average room rates for the three hotel chains.

For a number of years, nearly all Americans say that they would vote for a woman for president IF she were qualified, and IF she were from their own political party. But is America ready for a female president? A CBS/New York Times poll asked this question of a random sample of 1229 adults, with the following results: 19 $$\begin{array}{lcc} & \text { \% Responding "Yes } \\\& \text { Now } & 1999 \\\\\hline \text { Total } & 55 \% & 48 \% \\\\\text { Men } & 60 & 46 \\\\\text { Women } & 51 & 49 \\\\\text { Republicans } & 48 & 47 \\\\\text { Democrats } & 61 & 44 \\\\\text { Independents } & 55 & 54\end{array}$$ a. Construct a \(95 \%\) confidence interval for the proportion of all Americans who now believe that America is ready for a female president. b. If there were \(n_{1}=610\) men and \(n_{2}=619\) women in the sample, construct a \(95 \%\) confidence interval for the difference in the proportion of men and women who now believe that America is ready for a female president. Can you conclude that the proportion of men who now believe that America is ready for a female president is larger than the proportion of women? Explain. c. Look at the percentages of "yes" responses for Republicans, Democrats and Independents now compared to the percentages in \(1999 .\) Can you think of a reason why the percentage of Democrats might have changed so dramatically?

An experimenter fed different rations, \(A\) and \(B\), to two groups of 100 chicks each. Assume that all factors other than rations are the same for both groups. Of the chicks fed ration \(\mathrm{A}, 13\) died, and of the chicks fed ration \(\mathrm{B}, 6\) died. a. Construct a \(98 \%\) confidence interval for the true difference in mortality rates for the two rations. b. Can you conclude that there is a difference in the mortality rates for the two rations?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation
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.