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Calculate the margin of error in estimating a binomial proportion \(p\) using samples of size \(n=100\) and the values of p given in Exercises \(15-19 .\) What value of p produces the largest margin of error? \(p=.7\)

Short Answer

Expert verified
Answer: The margin of error for p = 0.7 is approximately 0.08946.

Step by step solution

01

Understanding the Binomial Proportion Estimation Process

Given a binomial proportion estimation problem, we have a sample size \(n\) and a target proportion \(p\). The margin of error is the amount by which the true proportion may differ from the sample proportion.
02

Find the Standard Deviation of the Proportion

To find the standard deviation of the proportion, we use the following formula: \(\sigma_p = \sqrt{\frac{p \times (1-p)}{n}}\)
03

Calculate Standard Deviation for p = 0.7

Substitute the given value of \(p=0.7\) and sample size \(n=100\) into the standard deviation formula: \(\sigma_p = \sqrt{\frac{0.7 \times (1-0.7)}{100}}\) \(\sigma_p = \sqrt{\frac{0.7 \times 0.3}{100}}\) \(\sigma_p \approx 0.04564\)
04

Calculate the Margin of Error

The margin of error for a binomial proportion estimation is typically calculated as follows using \(Z\)-score for \(95\%\) confidence level (\(1.96\)): \(Margin\ of\ Error = Z \times \sigma_p\) Using the calculated standard deviation (\(\sigma_p = 0.04564\)) and a Z-score of \(1.96\) for a \(95\%\) confidence interval, we can calculate the margin of error: \(Margin\ of\ Error = 1.96 \times 0.04564 \approx 0.08946\) The margin of error for p = 0.7 is approximately 0.08946. To determine the value of p that produces the largest margin of error, we'll have to calculate and compare the margin of error for other values of p (in Exercises 15-19).

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Key Concepts

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

Margin of Error
In statistical analysis, the concept of the 'margin of error' is crucial when it comes to understanding the precision of an estimate. Simply put, the margin of error tells us how much we can expect our survey or sample results to differ from the true population value. It acts as a buffer around our estimate, providing a range that the true value is likely to fall within. The concept gets particularly interesting when dealing with proportions, like predicting election results or estimating the percentage of a population with specific characteristics.

For a binomial proportion, the margin of error increases as the variability in the sampling distribution increases. This variability is captured by the standard deviation of the estimated proportion, which depends on both the true proportion (\(p\)) and the sample size (\(n\)). Larger sample sizes generally result in a smaller margin of error, as they provide more information and thus reduce uncertainty. A common way to express the margin of error is in conjunction with a confidence level—for example, a 95% confidence level means that if we could repeat the study many times, we would expect the true proportion to fall within the margin of error in 95 out of 100 cases.

When calculating the margin of error using the standard deviation and the desired confidence level, a Z-score is used. This Z-score corresponds to the chosen confidence level and is multiplied by the standard deviation to obtain the margin of error. The Z-score is a statistical measure that tells us how many standard deviations away from the mean a particular value lies, given a normal distribution.
Standard Deviation
Understanding 'standard deviation' is essential for interpreting the variability of data in statistics. In the context of a binomial proportion, standard deviation provides insight into the spread of the data around the proportion. \( \sigma_p \) is the standard deviation of the estimated proportion, which quantifies how much we expect the sample proportion to vary from sample to sample, assuming random selection. The formula \( \sigma_p = \sqrt{\frac{p \times (1-p)}{n}} \) is used to calculate the standard deviation for a binomial distribution.

In a practical sense, if you were conducting a survey, a smaller standard deviation would mean that the responses are tightly clustered around the average response, indicating that there is less variability in the data. Conversely, a larger standard deviation suggests that the responses are more spread out, showing greater variability. When estimating a binomial proportion, the maximum standard deviation occurs when \( p = 0.5 \) since the product \( p \times (1-p) \) is largest in this case. This is because the outcomes are most uncertain when the probability of success and failure are equal, leading to the greatest possible variability in sample proportions.
Confidence Interval
A 'confidence interval' is a range of values, derived from the sample data, that is likely to contain the value of an unknown population parameter. In other words, it gives an estimated range of values which is likely to include the parameter we're interested in, such as a population proportion, with a certain level of confidence. For the binomial proportion estimation, the 95% confidence interval is the most commonly used.

The 95% confidence interval is calculated by taking the estimated proportion (from our sample) and adding and subtracting the margin of error, which we obtained by multiplying the standard deviation by the Z-score corresponding to our chosen confidence level. This results in a range: \( \text{estimated proportion} \pm \text{margin of error} \). The '95%' in the 95% confidence interval implies that if we were to take 100 different samples and compute a confidence interval for each sample, then approximately 95 of the 100 confidence intervals will contain the true population proportion.

To use confidence intervals effectively, it's important to understand that they do not provide a probability that the true parameter is within the interval. Instead, they offer a method for making repeated sampling a reliable process for estimating parameters. The actual calculated interval from a particular study either contains the true parameter or it doesn't, but the level of confidence relates to the long-run success rate of the method, not the particular interval. When reporting the results of a binomial proportion estimation, the confidence interval provides much more information than a simple point estimate, as it reflects the precision of the estimate and the uncertainty associated with it.

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Most popular questions from this chapter

Independent random samples were selected from two binomial populations, with sample sizes and the number of successes given in Exercises \(1-2 .\) Construct a \(95 \%\) and a \(99 \%\) confidence interval for the difference in the population proportions. What does the phrase "95\% confident" or "99\% confident" mean? $$\begin{array}{lcc}\hline & \text { Population } \\\\\cline { 2 - 3 } & 1 & 2 \\\\\hline \text { Sample Size } & 800 & 640 \\\\\text { Number of Successes } & 337 & 374 \\\\\hline\end{array}$$

Find the z-values needed to calculate large-sample confidence intervals for the confidence levels given. A \(90 \%\) confidence interval

You want to estimate the mean hourly yield for a process that manufactures an antibiotic. You observe the process for 100 hourly periods chosen at random, with the results \(\bar{x}=1020\) grams per hour and \(s=90 .\) Estimate the mean hourly yield for the process 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 a random sample of 30 adults was selected from each of two regions of the United States and that each person's daily selenium intake was recorded. The means and standard deviations of the selenium daily intakes for the two groups are shown in the table. Find a \(95 \%\) confidence interval for the difference in the mean selenium intakes for the two regions. Interpret this interval.

What \(i s\) normal, when it comes to people's body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker \(^{11}\) in the Journal of Statistical Education, had a mean of \(98.25^{\circ}\) Fahrenheit and a standard deviation of \(0.73^{\circ}\) Fahrenheit. a. Construct a \(99 \%\) confidence interval for the average body temperature of healthy people. b. Does the confidence interval constructed in part a contain the value \(98.6^{\circ}\) Fahrenheit, the usual average temperature cited by physicians and others? If not, what conclusions can you draw?

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