To determine the minimum sample size needed for a 95% confidence level where the sample standard deviation is within 30% of the population standard deviation, we use the concept of the chi-squared distribution. The formula relates the sample size, confidence level, and allowable error (in this case, 30% of the population standard deviation).

Key Variables:

• Confidence level: 95%
• Allowable margin of error: 30% of the population standard deviation
• Population standard deviation ($\sigma$): unknown, but our goal is to make the sample's standard deviation close to this value
• $\alpha$: Significance level, which is $1 - 0.95 = 0.05$
• Degrees of freedom: $n - 1$, where $n$ is the sample size
• Chi-squared distribution values: We'll use $\chi^2_{\alpha/2}$ and $\chi^2_{1-\alpha/2}$ based on the degrees of freedom and the desired confidence interval.

Formula:

The formula for the confidence interval of the population standard deviation is:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

Where:

• $s$ is the sample standard deviation
• $\sigma$ is the population standard deviation
• $n$ is the sample size
• $\chi^2_{\alpha/2}$ and $\chi^2_{1-\alpha/2}$ are chi-squared critical values for the given confidence level and degrees of freedom.

To ensure that the sample standard deviation is within 30% of the population standard deviation, the inequality:

$0.7 \leq \frac{s}{\sigma} \leq 1.3$

must hold, which leads us to solve for the sample size $n$.

Steps to Calculate the Sample Size:

1. Set confidence level: We have a 95% confidence level, so we use $\alpha = 0.05$, meaning $\alpha/2 = 0.025$.
2. Find chi-squared values: We need to find the chi-squared critical values for $\alpha/2 = 0.025$ and $1 - \alpha/2 = 0.975$ for different sample sizes.
3. Apply margin of error: Use the requirement that the sample standard deviation is within 30% of the population standard deviation.
4. Iterate to find the smallest $n$ that satisfies this condition.

The required sample size $n$ can be found through iterative methods, typically by checking values of $n$ until the relationship holds, using tables or a statistical calculator to compute the chi-squared critical values for different sample sizes.

Approximation Method:

For practical purposes, the approximate sample size $n$ can be calculated using the following simplified rule-of-thumb formula:

$n \approx \left(\frac{Z_{\alpha/2} \times (1 + M)}{M}\right)^2$

Where:

• $Z_{\alpha/2}$ is the z-score corresponding to the desired confidence level (1.96 for 95%)
• $M$ is the margin of error (30% or 0.30 in this case).

Plugging the values in:

$n \approx \left(\frac{1.96 \times (1 + 0.30)}{0.30}\right)^2 = \left(\frac{1.96 \times 1.30}{0.30}\right)^2 = \left(\frac{2.548}{0.30}\right)^2 \approx 72$

Conclusion:

The minimum sample size needed to be 95% confident that the sample's standard deviation is within 30% of the population's standard deviation is approximately 72.

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. How does changing the confidence level affect the sample size requirement?
2. What happens if the allowable margin of error changes to 20%?
3. How can we calculate sample size for different types of data distributions?
4. What is the role of chi-squared distribution in confidence intervals for variance?
5. How does the sample size affect the accuracy of the sample standard deviation?

Tip: Always consider the trade-off between confidence level and margin of error—the more precise your estimate (smaller error), the larger your required sample size!