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Chapter 8: Problem 7

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. \(n=1000\)

Short Answer

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Question: Calculate the margin of error in estimating a binomial proportion with a probability of success of 0.5 for the following sample sizes: a) 30, b) 100, c) 400, and d) 1000. Answer: a) For a sample size of 30, the margin of error is approximately 1.96 脳 鈭�(0.25/30). b) For a sample size of 100, the margin of error is approximately 1.96 脳 鈭�(0.25/100). c) For a sample size of 400, the margin of error is approximately 1.96 脳 鈭�(0.25/400). d) For a sample size of 1000, the margin of error is approximately 1.96 脳 鈭�(0.25/1000).

Step by step solution

01

a. Calculating the margin of error for \(n = 30\)

First, calculate the standard error of the estimator using the formula \(SE = \sqrt{\frac{p(1-p)}{n}}\). In this case, \(p = 0.5\) and \(n = 30\). \(SE = \sqrt{\frac{0.5(1-0.5)}{30}} = \sqrt{\frac{0.25}{30}}\) Next, calculate the margin of error using the formula \(MOE = 1.96 \times SE\). \(MOE = 1.96 \times \sqrt{\frac{0.25}{30}}\)
02

b. Calculating the margin of error for \(n = 100\)

First, calculate the standard error of the estimator using the formula \(SE = \sqrt{\frac{p(1-p)}{n}}\). In this case, \(p = 0.5\) and \(n = 100\). \(SE = \sqrt{\frac{0.5(1-0.5)}{100}} = \sqrt{\frac{0.25}{100}}\) Next, calculate the margin of error using the formula \(MOE = 1.96 \times SE\). \(MOE = 1.96 \times \sqrt{\frac{0.25}{100}}\)
03

c. Calculating the margin of error for \(n = 400\)

First, calculate the standard error of the estimator using the formula \(SE = \sqrt{\frac{p(1-p)}{n}}\). In this case, \(p = 0.5\) and \(n = 400\). \(SE = \sqrt{\frac{0.5(1-0.5)}{400}} = \sqrt{\frac{0.25}{400}}\) Next, calculate the margin of error using the formula \(MOE = 1.96 \times SE\). \(MOE = 1.96 \times \sqrt{\frac{0.25}{400}}\)
04

d. Calculating the margin of error for \(n = 1000\)

First, calculate the standard error of the estimator using the formula \(SE = \sqrt{\frac{p(1-p)}{n}}\). In this case, \(p = 0.5\) and \(n = 1000\). \(SE = \sqrt{\frac{0.5(1-0.5)}{1000}} = \sqrt{\frac{0.25}{1000}}\) Next, calculate the margin of error using the formula \(MOE = 1.96 \times SE\). \(MOE = 1.96 \times \sqrt{\frac{0.25}{1000}}\)

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Key Concepts

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

Binomial Proportion
The binomial proportion involves a statistical experiment with two possible outcomes: success or failure. We denote it often as \( p \), representing the probability of success in a trial. In calculations, like margin of error, choosing a value for \( p \) is crucial. A common practice when specifics are unknown is to use \( p = 0.5 \) because it shows maximum variability. This maximizes the estimate's uncertainty, providing a conservative measure for safety. Using this value helps ensure our calculated margin of error encompasses the real-world proportions.
Standard Error
Standard Error (SE) quantifies the variability or spread of a sampling distribution. In our context, the standard error of a binomial proportion tells us how much the proportion \( p \) might vary from one sample to another. We calculate it using the formula:
  • \( SE = \sqrt{\frac{p(1-p)}{n}} \)
Here, \( p \) is the probability of success (here often set at 0.5 for caution), \( 1-p \) is the probability of failure, and \( n \) is the number of observations or trials. A smaller SE indicates that the sample proportion is likely to be close to the true population proportion. Thus, it influences the margin of error by providing the base for its computation.
Estimator
An estimator refers to a rule or a formula we use to make inferences about our population of interest based on sample data. In terms of binomial proportions, we use the sample proportion (\( \hat{p} \)) as an estimator of the true population proportion \( p \). A good estimator should be unbiased and have low variability. The standard error assists here by informing us about the variability, telling us how good our estimator might be. Lower variability implies greater reliability of our estimator.
Confidence Interval
A confidence interval is a range of values, derived from sample data, that likely contains the population parameter of interest, in this case, the true binomial proportion \( p \). The margin of error (MOE) helps construct this interval. The formula for a confidence interval (CI) around a proportion is:
  • \( \hat{p} \pm \text{MOE} \)
This approach guarantees that if we were to repeat our study many times, the true proportion would lie within this interval in a certain percentage of those studies (commonly 95% for a standard \( 1.96 \)). Knowing the CI helps express our uncertainty about the estimator and provides a clear picture of where the true proportion can fall.

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Most popular questions from this chapter

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?

In a study to establish the absolute threshold of hearing, 70 male college freshmen were asked to participate. Each subject was seated in a soundproof room and a \(150 \mathrm{H}\) tone was presented at a large number of stimulus levels in a randomized order. The subject was instructed to press a button if he detected the tone; the experimenter recorded the lowest stimulus level at which the tone was detected. The mean for the group was \(21.6 \mathrm{db}\) with \(s=2.1\). Estimate the mean absolute threshold for all college freshmen and calculate the margin of error.

A small amount of the trace element selenium, \(50-200\) micrograms ( \(\mu \mathrm{g}\) ) per day, is considered essential to good health. Suppose that random samples of \(n_{1}=n_{2}=30\) adults were selected from two regions of the United States and that a day's intake of selenium was recorded for each person. The mean and standard deviation of the selenium daily intakes for the 30 adults from region 1 were \(\bar{x}_{1}=\) 167.1 and \(s_{1}=24.3 \mu \mathrm{g},\) respectively. The corresponding statistics for the 30 adults from region 2 were \(\bar{x}_{2}=\) 140.9 and \(s_{2}=17.6 \mu \mathrm{g} .\) Find a \(95 \%\) confidence interval for the difference in the mean selenium intakes for the two regions. Interpret this interval.

A random sample of \(n\) measurements is selected from a population with unknown mean \(\mu\) and known standard deviation \(\sigma=10\). Calculate the width of a \(95 \%\) confidence interval for \(\mu\) for these values of \(n\) : a. \(n=100\) b. \(n=200\) c. \(n=400\)

Based on repeated measurements of the iodine concentration in a solution, a chemist reports the concentration as \(4.614,\) with an "error margin of .006." a. How would you interpret the chemist's "error margin"? b. If the reported concentration is based on a random sample of \(n=30\) measurements, with a sample standard deviation \(s=.017,\) would you agree that the chemist's "error margin" is .006?
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