Question

Suppose a random sample of size \(n\) is required to produce a margin of error of \(\pm E\). Write an expression in terms of \(n\) for the sample size needed to reduce the margin of error to \(\pm \frac{1}{2} E\). How many times must the sample size be increased to cut the margin of error in half? Explain.

Short answer

To cut the margin of error in half, the sample size must be quadrupled. In other words, the sample size needs to be increased by four times.

Step-by-step solution

Express Initial Margin of Error

Write down the original formula for the margin of error: \(E = z\frac{\sigma}{\sqrt{n}}\).

Halve Margin of Error

Set the new margin of error to \(\frac{1}{2}E\). Substitute this into the margin of error formula to get: \(\frac{1}{2}E = z\frac{\sigma}{\sqrt{n_1}}\), where \(n_1\) is the new sample size we want to find.

Rearrange for New Sample Size

Rearrange the formula to solve for \(n_1\), which gives: \(n_1 = \left(z \cdot \frac{\sigma}{\frac{1}{2}E}\right)^2 = 4 \cdot \left(z\frac{\sigma}{E}\right)^2 = 4n\). Since \(n = z\frac{\sigma}{E}\)^2, we multiply by 4 to get the new sample size required to cut the margin of error in half.

Calculate Sample Size Increase

Based on the formula from Step 3, we can see that to cut the margin of error in half, the sample size needs to be quadrupled (\(4n\)), meaning the sample size must be increased by four times.

Key concepts

Sample Size Calculation

Understanding how to calculate the required sample size in statistics is essential for conducting research with precise and reliable results. To obtain a specific margin of error, researchers must determine the appropriate sample size using a formula that encapsulates the desired confidence level and population variability.

In the context of the provided exercise, the initial margin of error formula is expressed as:
\begin{align*}E &= z\frac{\sigma}{\sqrt{n}},\end{align*}
where

• \(E\) represents the margin of error,
• \(z\) is the z-score corresponding to the desired confidence level,
• \(\sigma\) is the population standard deviation, and
• \(n\) is the sample size.

Making changes to the margin of error, like halving it to \(\frac{1}{2}E\), impacts the required sample size. Resultantly, an important aspect of sample size calculation is understanding the relationship between margin of error and sample size—demonstrating that to reduce the margin of error, one must increase the sample size. As seen in the exercise, to halve the margin of error, the sample size needs to be quadrupled.

Statistics in Algebra

Algebra plays a crucial role in statistics by providing the necessary tools to manipulate formulas and solve for unknown variables. In statistical algebra, equations are used to represent real-world Variables, enabling the solving of problems about population characteristics from sample data.

For instance, the formula \(E = z\frac{\sigma}{\sqrt{n}}\)
is an algebraic expression where \(E\) is solved by rearranging terms and squaring both sides of the equation, a typical algebraic operation, to isolate \(n\).

Algebraic techniques were used in the exercise solution to solve for the new sample size, \(n_1\), showcasing how algebra helps in decoding relationships between the elements of a statistical formula. The steps involved include the setting of new conditions (halving the margin of error), rearranging the equation to solve for \(n_1\), squaring the whole equation to eliminate the square root, and finally determining the scale of increase in sample size needed.

Confidence Interval

A confidence interval gives an estimated range of values which is likely to include an unknown population parameter. It is calculated from the observed data and has an associated confidence level that quantifies the level of confidence that the parameter lies within the interval.

In a typical margin of error formula, such as \(E = z\frac{\sigma}{\sqrt{n}}\), the z-score corresponds to the confidence level. The higher the confidence level (e.g., 95% or 99%), the larger the z-score, which results in a wider confidence interval if the sample size \(n\) remains unchanged.

The concept of halving the margin of error is inherently connected to confidence intervals. By reducing the margin of error, one is in fact narrowing the confidence interval, thereby getting a more precise estimate of the population parameter. However, the trade-off requires increasing the sample size; as the exercise suggests, quadrupling the sample size is necessary to halve the margin of error.