Determining the Minimum Sample Size Required: Videos & Practice Problems

To determine the minimum sample size $n$ for a desired maximum margin of error $e$, use the formula $n={\frac{z{\alpha }_{/2}\times s}{e}}^2$, where $z{\alpha }_{/2}$ is the critical z-value for the confidence level and $s$ is the estimated standard deviation. When $s$ is unknown, estimate it using the range rule of thumb: range divided by four. Always round up $n$ to ensure the margin of error does not exceed the maximum allowed, supporting accurate confidence interval estimation.

Determining the Minimum Sample Size Required

When constructing confidence intervals for population means, the margin of error (e) plays a crucial role in determining the width of the interval around the sample mean. Typically, the margin of error is calculated using the sample size (n), but in some cases, there is a restriction on how large the margin of error can be. This leads to the need to find the minimum sample size required to ensure the margin of error does not exceed a specified maximum.

The margin of error formula is given by:

\[ e = z_{\alpha/2} \times \frac{s}{\sqrt{n}} \]

where \( z_{\alpha/2} \) is the critical z-value for the desired confidence level, \( s \) is the sample standard deviation, and \( n \) is the sample size. Since the sample size \( n \) is in the denominator under the square root, increasing \( n \) decreases the margin of error. To find the minimum sample size for a given maximum margin of error, this formula can be rearranged to solve for \( n \):

\[ n = \left( \frac{z_{\alpha/2} \times s}{e} \right)^2 \]

It is important to note that the sample size must be a whole number, so if the calculation results in a decimal, always round up to the next whole number. This ensures the margin of error stays within the desired limit, as rounding down could lead to a margin of error that is too large.

For example, if a researcher wants to estimate the average daily exercise time of college students with a 95% confidence interval and a margin of error no greater than 3 minutes, and a previous study suggests the standard deviation is approximately 12 minutes, the minimum sample size can be calculated as follows. The critical z-value for a 95% confidence level is 1.96. Plugging these values into the formula:

\[ n = \left( \frac{1.96 \times 12}{3} \right)^2 = (7.84)^2 = 61.47 \]

Rounding up, the minimum sample size required is 62.

When the sample size is unknown, the degrees of freedom needed to find the t-critical value cannot be determined. In such cases, the z-critical value is used as an approximation, which is generally acceptable because the sample size tends to be large enough for the t-distribution to closely approximate the normal distribution.

In situations where the standard deviation is not known from prior studies, it can be estimated using the range rule of thumb:

\[ s \approx \frac{\text{Range}}{4} = \frac{\text{Maximum} - \text{Minimum}}{4} \]

This provides a reasonable estimate of the standard deviation based on the expected spread of the data.

Understanding how to calculate the minimum sample size for a given margin of error is essential for designing studies that achieve desired precision in estimating population means. This process involves manipulating the margin of error formula, applying critical values from the normal distribution, and making informed estimates of variability when necessary.

A puzzle company is interested in the average number of pieces in their jigsaw puzzles, so they plan to create a confidence interval for the true mean. The easiest puzzle in their line is $100$ pieces, and the largest is $1000$ pieces. Use the range rule of thumb to estimate the sample standard deviation.

• A. $275$
• B. $550$
• C. $450$
• D. $225$

Answer: (D)

A bookstore is interested in the average length of books in its adult fiction section. Find the minimum sample size required to create a $99\%$ confidence interval with a margin of error no more than $30$ pages if...
(A) The standard deviation is known to be about $96$ pages.

• A. $2039$
• B. $8$
• C. $68$
• D. $45$

Answer: (C)

A bookstore is interested in the average length of books in its adult fiction section. Find the minimum sample size required to create a $99\%$ confidence interval with a margin of error no more than $30$ pages if...
(B) The standard deviation is unknown, but the longest book is $697$ pages and the shortest is $298$ pages.

• A. $74$
• B. $9$
• C. $2201$
• D. $47$

Answer: (A)

Here’s what students ask on this topic:

How do you calculate the minimum sample size required for a given margin of error in confidence intervals?

To calculate the minimum sample size $n$ required for a desired maximum margin of error $e$ in confidence intervals, use the formula:

$$n={(}^{\frac{z}{{\alpha }_2}}2$$

Here, $z_{\mathrm{\alpha /2}}$ is the critical z-value corresponding to the confidence level (e.g., 1.96 for 95%), and $s$ is the estimated standard deviation of the population. This formula ensures the margin of error does not exceed the specified maximum. After calculating $n$, always round up to the next whole number because sample size must be an integer and rounding down could result in a margin of error larger than allowed.

What do you do if the population standard deviation is unknown when determining minimum sample size?

If the population standard deviation $\sigma$ is unknown when determining the minimum sample size, you can estimate the standard deviation $s$ using the range rule of thumb. This rule estimates the standard deviation as:

$$s\approx \frac{\mathrm{range}}{4}$$

where the range is the difference between the maximum and minimum values of the data. This estimation is useful when prior data or studies are unavailable. Using this estimated $s$ in the sample size formula allows you to calculate a reasonable minimum sample size to achieve the desired margin of error in your confidence interval.

Why do we use the z critical value instead of the t critical value when calculating minimum sample size?

When calculating the minimum sample size, we use the z critical value $z_{\mathrm{\alpha /2}}$ instead of the t critical value $t_{\mathrm{\alpha /2}}$ because the degrees of freedom, which depend on the sample size $n$, are unknown at this stage. Since $n$ is what we are trying to find, we cannot determine the exact t critical value. The z critical value is a good approximation, especially for large sample sizes, because as $n$ increases, the t distribution approaches the standard normal distribution. This approximation simplifies calculations and provides a reliable estimate for the minimum sample size needed.

How does increasing the sample size affect the margin of error in confidence intervals?

Increasing the sample size $n$ decreases the margin of error $e$ in confidence intervals. This is because the margin of error formula includes the sample size in the denominator under a square root, so as $n$ grows, the margin of error shrinks. Specifically, the margin of error is inversely proportional to the square root of the sample size, meaning that to reduce the margin of error by half, you need to quadruple the sample size. This relationship helps researchers determine how large a sample they need to achieve a desired precision in their estimates.

Why must the minimum sample size always be rounded up when calculating it?

The minimum sample size must always be rounded up because sample size must be a whole number, and rounding down could result in a sample size that is too small to achieve the desired margin of error. For example, if the calculation yields 61.47, rounding down to 61 might produce a margin of error larger than the maximum allowed. Rounding up to 62 ensures the margin of error stays within the specified limit, maintaining the accuracy and reliability of the confidence interval.