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\(m\) and \(n\) Consider the sample size formula \(n=\left[\hat{p}(1-\hat{p}) z^{2}\right] / m^{2}\) for estimating a proportion. When \(\hat{p}\) is close to 0.50 , for \(95 \%\) confidence explain why this formula gives roughly \(n=1 / m^{2}\).

Short Answer

Expert verified
When \(\hat{p} = 0.5\) and 95% confidence is used, the formula simplifies to \(n \approx 1/m^2\) due to \(0.9604\) being close to 1.

Step by step solution

01

Understand the Formula Components

The formula provided is \(n=\left[\hat{p}(1-\hat{p}) z^{2}\right] / m^{2}\). It calculates the sample size \(n\) needed to estimate a population proportion \(\hat{p}\) with a margin of error \(m\). Here, \(z\) is the z-score corresponding to the confidence level.
02

Determine the Value of \(z\) for 95% Confidence

For 95% confidence, the z-score is typically 1.96 because it corresponds to the middle 95% of the standard normal distribution.
03

Analyze the Expression with \(\hat{p} = 0.5\)

Substitute \(\hat{p} = 0.5\) into the formula: \(n = \left[0.5(1-0.5)z^2\right]/m^2\). This simplifies the numerator to \(0.25z^2\) because \(0.5 \times 0.5 = 0.25\).
04

Substitute \(z = 1.96\) into the Formula

Now plug in the z-score: \(n = \left[0.25(1.96)^2\right]/m^2 = \left[0.25 \times 3.8416\right]/m^2\). After calculating, the expression becomes \(n = 0.9604/m^2\).
05

Approximate \(0.9604\) to 1

Since \(0.9604\) is very close to 1, it is approximated as \(n \approx 1/m^2\), especially when considering simplicity in preliminary sample size estimations.

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Key Concepts

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

Confidence Interval
The confidence interval is a statistical concept that helps us estimate the range within which a population parameter, such as a proportion or mean, likely falls. A 95% confidence interval means that if we were to draw 100 different random samples and compute the confidence interval for each sample, we would expect about 95 of these intervals to contain the true population parameter.
One key component of a confidence interval is its width, which is influenced by the sample size and variability in the data. A larger sample size results in a narrower confidence interval, indicating more precise estimates of the population parameter. In contrast, a smaller sample size leads to a wider interval, implying less precision.
This interval gives researchers and statisticians confidence in their sample estimates and helps in making informed decisions based on data analysis.
Population Proportion
The population proportion (\(\hat{p}\)) refers to the fraction of the entire population that possesses a particular characteristic. For example, if we are interested in the proportion of voters in a city who support a certain policy, \(\hat{p}\) would represent this proportion.
To estimate \(\hat{p}\) accurately, statisticians rely on sample data collected from a smaller portion of the population. A good sample should be random and sufficiently large to represent the population well. By using the provided formula,\(n = \left[\hat{p}(1-\hat{p}) z^{2}\right] / m^{2}\), we can determine the necessary sample size needed to estimate \(\hat{p}\) with a specified margin of error and confidence level.
Understanding the concept of the population proportion is crucial for making inferences about larger groups based on smaller samples.
Margin of Error
The margin of error (\(m\)) is a crucial element in statistical sampling that quantifies the degree of uncertainty associated with a sample estimate. It defines the range within which the true population proportion is expected to lie. For instance, if the margin of error is 5%, it means the actual population proportion could vary by 5% above or below the estimated sample proportion.
In the formula \(n = \left[\hat{p}(1-\hat{p}) z^{2}\right] / m^{2}\), the margin of error is squared and placed in the denominator, showcasing its significant influence on determining sample size. A smaller margin of error implies greater precision but requires a larger sample size, while a larger margin of error allows for a smaller sample size but decreases precision.
Balancing the margin of error with practical considerations like cost and time is often key in designing effective studies.
z-score
The z-score is a statistical measurement that expresses the number of standard deviations an element is from the mean. In the context of sample size estimation, the z-score reflects how confident we want to be about our sample estimates.
For a 95% confidence level, the z-score used is typically 1.96. This means that the interval we construct should capture the true population parameter 95% of the time. By plugging the z-score into the sample size formula, \(n = \left[\hat{p}(1-\hat{p}) z^{2}\right] / m^{2}\), it adjusts the size of the interval accordingly.
A higher confidence level, such as 99%, would require a higher z-score, leading to a wider confidence interval and a larger sample size. Understanding how z-scores work is essential for controlling the accuracy and reliability of statistical estimates.

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Most popular questions from this chapter

British monarchy In February 2002, the Associated Press quoted a survey of 3000 British residents conducted by YouGov.com. It stated, "Only \(21 \%\) wanted to see the monarchy abolished, but \(53 \%\) felt it should become more democratic and approachable. No margin of error was given." If the sample was random, find the \(95 \%\) margin of error for each of these estimated proportions.

Using \(t\) -table Using Table \(B\) or software or a calculator, report the \(t\) -score which you multiply by the standard error to form the margin of error for a a. \(95 \%\) confidence interval for a mean with 5 observations. b. \(95 \%\) confidence interval for a mean with 15 observations. c. \(99 \%\) confidence interval for a mean with 15 observations.

How often feel sad? Refer to the previous exercise. This question was asked of 10 students in a class at the University of Wisconsin recently. The responses were $$ 0,0,1,0,7,2,1,0,0,3 $$ Find and interpret a \(90 \%\) confidence interval for the population mean, and indicate what you would have to assume for this inference to apply to the population of all University of Wisconsin students.

Watching TV In response to the GSS question in 2008 about the number of hours daily spent watching \(\mathrm{TV}\), the responses by the five subjects who identified themselves as Hindu were 3,2,1,1,1 . a. Find a point estimate of the population mean for Hindus. b. The margin of error at the \(95 \%\) confidence level for this point estimate is 0.7 . Explain what this represents.

\(99.9999 \%\) confidence Explain why confidence levels are usually large, such as 0.95 or \(0.99,\) but not extremely large, such as \(0.999999 .\) (Hint: What impact does the extremely high confidence level have on the margin of error?)

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