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Independent ram 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 required from each population is approximately 543 observations.

Step by step solution

01

Find the Z-score corresponding to the desired confidence level

To accomplish this, we need to find the Z-score that corresponds to a probability of 0.98. Using a Standard Normal Distribution table or calculator, we find that the Z-score for a 0.98 probability is approximately 2.33.
02

Write down the formula for sample size

The formula for finding the sample size in comparing two proportions is: \(n = \frac{(Z^2)(p_1(1-p_1) + p_2(1-p_2))}{E^2}\) where \(Z\) is the Z-score, \(p_1\) and \(p_2\) are the population proportions, and \(E\) is the margin of error.
03

Substitute the values into the formula

Since we don't have any prior information about \(p_1\) and \(p_2\), we have to assume the worst-case scenario, which is when both \(p_1\) and \(p_2\) are equal to \(1/2\). Also, the margin of error is 0.05. So, substitute those values into the formula: \(n = \frac{(2.33^2)((1/2)(1-1/2) + (1/2)(1-1/2))}{0.05^2}\)
04

Calculate the result

Now simplify and calculate the result: \(n = \frac{(5.4289)(1/4)}{0.0025}\) \(n ≈ 542.89\) Since we cannot have a fraction of an observation, we will round up to the nearest whole number.
05

Report the final result

The sample size required from each population to estimate the difference in their proportions with a margin of error of 0.05 and a probability of 0.98 is approximately 543 observations.

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Key Concepts

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

Binomial Distribution
The Binomial Distribution is a statistical distribution that summarizes the number of successes in a fixed number of trials. Each trial is independent, and there are only two possible outcomes: success and failure. This distribution is characterized by two parameters: the number of trials, denoted by \( n \), and the probability of success on a single trial, denoted by \( p \).
The formula for the binomial probability is given by:
  • \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where \( P(X = k) \) represents the probability of getting exactly \( k \) successes in \( n \) trials.
The Binomial Distribution is particularly useful when analyzing proportion data, like in our exercise where we aim to estimate the difference between two population proportions. This distribution helps us model the outcomes of each population under study.
Z-score
A Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is measured in terms of standard deviations from the mean.
When determining the sample size or assessing the significance of findings, Z-scores are often used to reflect the standard difference between two points of data.
  • The formula for a Z-score is:\[Z = \frac{X - \mu}{\sigma}\]
where \( X \) is the value being measured, \( \mu \) is the mean of the population, and \( \sigma \) is the standard deviation of the population.
In the context of sample size determination, like in our exercise, the Z-score is used to match the desired confidence level. For example, a Z-score of approximately 2.33 correlates with a 98% confidence level, indicating high certainty in the estimate being computed.
Confidence Interval
A Confidence Interval is a range of values within which we can say with a specific level of confidence that the true parameter lies. It quantifies the uncertainty of an estimated value, like the difference between the two proportions in our exercise.
Confidence Intervals are often expressed as a percentage, with common levels being 90%, 95%, and 99%. The width of the interval depends on the desired level of confidence and the variability in the data.
  • The formula for a confidence interval is typically:\[ \text{Confidence Interval} = \text{estimate} \pm (\text{Z-score}) \times \frac{\sigma}{\sqrt{n}}\]
where "estimate" refers to the data point under consideration, and \( \sigma \) is the standard error.
In our exercise, we wanted to estimate the interval of the difference in population proportions with a confidence level of 98%, which is why we calculated sample size with this probability in mind.
Population Proportion
Population Proportion is the fraction of a whole population with a particular attribute. In statistics, it is estimated using sample data, especially when the entire population cannot be surveyed.
This proportion is an essential element when evaluating outcomes from binomial experiments.
  • The population proportion is denoted as:\[ p = \frac{X}{N}\]
where \( X \) is the number of individuals with the attribute, and \( N \) is the total number of individuals in the population.
When we lack prior information about specific proportions \( p_1 \) and \( p_2 \), as in our exercise, we assume a worst-case scenario where these probabilities are 0.5. This is because maximum variability or uncertainty about an outcome occurs at \( p = 0.5 \), allowing for a more conservative estimate of sample size.

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Most popular questions from this chapter

Born between 1980 and 1990 , Generation Next is engaged with technology, and the vast majority is dependent upon it. \({ }^{17}\) Suppose that in a survey of 500 female and 500 male students in Generation Next, 345 of the females and 365 of the males reported that they decided to attend college in order to make more money. a. Construct a \(98 \%\) confidence interval for the difference in the proportions of female and male students who decided to attend college in order to make more money. b. What does it mean to say that you are "98\% confident"? c. Based on the confidence interval in part a, can you conclude that there is a difference in the proportions of female and male students who decided to attend college in order to make more money?

Does Mars, Incorporated use the same proportion of red candies in its plain and peanut varieties? A random sample of 56 plain M\&M'S contained 12 red candies, and another random sample of 32 peanut M\&M'S contained 8 red candies. a. Construct a \(95 \%\) confidence interval for the difference in the proportions of red candies for the plain and peanut varieties. b. Based on the confidence interval in part a, can you conclude that there is a difference in the proportions of red candies for the plain and peanut varieties? Explain.

Calculate the margin of error in estimating a binomial proportion for each of the following values of \(n\). Use \(p=.5\) to calculate the standard error of the estimator. a. \(n=30\) b. \(n=100\) c. \(n=400\) d. \(n=1000\)

A sampling of political candidates- 200 randomly chosen from the West and 200 from the East-was classified according to whether the candidate received backing by a national labor union and whether the candidate won. In the West, 120 winners had union backing, and in the East, 142 winners were backed by a national union. Find a \(95 \%\) confidence interval for the difference between the proportions of union-backed winners in the West versus the East. Interpret this interval.

A random sample of \(n=500\) observations from a binomial population produced \(x=450\) successes. Estimate the binomial proportion \(p\) and calculate the margin of error.

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