Calculating Sample Size for Yarn Tenacity Confidence Limits

This problem requires finding the new sample size ($n_2$) needed to achieve a narrower confidence interval for the mean yarn tenacity, given an initial sample size and its confidence interval.

Understanding Confidence Intervals and Sample Size

The margin of error ($E$) in a confidence interval is related to the sample size ($n$), the population standard deviation ($\sigma$), and the z-score ($z$) corresponding to the confidence level by the formula:

$E = z \times (\sigma / \sqrt{n})$

Since the confidence level (95%) and thus the z-score remain constant, and the population standard deviation ($\sigma$) is assumed to be the same, the margin of error ($E$) is inversely proportional to the square root of the sample size ($\sqrt{n}$).

This relationship can be expressed as: $E \propto 1/\sqrt{n}$, or $E \sqrt{n} = \text{constant}$.

Step-by-Step Calculation

1. Identify Initial Values:
   • Initial margin of error ($E_1$): $1.5$
   • Initial sample size ($n_1$): $100$
2. Identify Desired Values:
   • Desired margin of error ($E_2$): $0.5$
   • Desired sample size ($n_2$): ?
3. Apply the Relationship: Since $E \sqrt{n}$ is constant, we can set up the equation:

   $E_1 \sqrt{n_1} = E_2 \sqrt{n_2}$

4. Substitute Known Values:

   $1.5 \times \sqrt{100} = 0.5 \times \sqrt{n_2}$

5. Solve for $\sqrt{n_2}$:

   $1.5 \times 10 = 0.5 \times \sqrt{n_2}$

   $15 = 0.5 \times \sqrt{n_2}$

   $\sqrt{n_2} = 15 / 0.5$

   $\sqrt{n_2} = 30$

6. Calculate the Final Sample Size ($n_2$):

   $n_2 = (30)^2$

   $n_2 = 900$

Therefore, 900 test samples are required to obtain 95% confidence limits of $30 \pm 0.5$.