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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.

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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 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.

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Most popular questions from this chapter

Refer to Exercise 8.5 . What effect does an increased sample size have on the margin of error?

An experiment was conducted to compare two diets \(\mathrm{A}\) and \(\mathrm{B}\) designed for weight reduction. Two groups of 30 overweight dieters each were randomly selected. One group was placed on diet \(\mathrm{A}\) and the other on diet \(\mathrm{B},\) and their weight losses were recorded over a 30 -day period. The means and standard deviations of the weight-loss measurements for the two groups are shown in the table. Find a \(95 \%\) confidence interval for the difference in mean weight loss for the two diets. Interpret your confidence interval. $$\begin{array}{ll}\text { Diet } A & \text { Diet B } \\\\\hline \bar{x}_{A}=21.3 & \bar{x}_{B}=13.4 \\\s_{A}=2.6 & s_{B}=1.9\end{array}$$

Suppose you wish to estimate the mean \(\mathrm{pH}\) of rainfalls in an area that suffers heavy pollution due to the discharge of smoke from a power plant. You know that \(\sigma\) is in the neighborhood of \(.5 \mathrm{pH},\) and you wish your estimate to lie within .1 of \(\mu\), with a probability near .95. Approximately how many rainfalls must be included in your sample (one pH reading per rainfall)? Would it be valid to select all of your water specimens from a single rainfall? Explain.

State the Central Limit Theorem. Of what value is the Central Limit Theorem in large-sample statistical estimation?

Independent random samples were selected from populations 1 and 2 . The sample sizes, means, and variances are as follows: $$\begin{array}{lcc} & \multicolumn{2}{c} {\text { Population }} \\\\\cline { 2 - 3 } & 1 & 2 \\\\\hline \text { Sample Size } & 35 & 49 \\\\\text { Sample Mean } & 12.7 & 7.4 \\\\\text { Sample Variance } & 1.38 & 4.14\end{array}$$ a. Find a \(95 \%\) confidence interval for estimating the difference in the population means \(\left(\mu_{1}-\mu_{2}\right) .\) b. Based on the confidence interval in part a, can you conclude that there is a difference in the means for the two populations? Explain.
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