Answer to Question #283003 in Statistics and Probability for anmol

Question

The 95% confidence limits of mean yarn tenacity based on 100 test samples is 30±1.5.The number of tes samples required to obtain 95% confidence limits of 30±0.5 is..

Answer 1

The 95% confidence limits of mean yarn tenacity based on 100 test samples is 30±1.5

n = 100

Margin of error (ME) = 1.5

Z value at 95% confidence = 1.96

First, we need to calculate standard deviation using formula of ME.

$ME=Z\times\frac{\sigma}{\sqrt{n}}$

$1.5=1.96\times\frac{\sigma}{\sqrt{100}}$

$\sigma=\frac{1.5\times\sqrt{100}}{1.96}=7.65$

The number of test samples required to obtain 95% confidence limits of 30±0.5 can be calculated using this standard deviation.

$n=(\frac{Z\times \sigma}{ME})^2$

$n=(\frac{1.96\times 7.65}{0.5})^2$

$n=(30)^2=900$

This means 900 test samples are required to obtain 95% confidence limits of 30±0.5.