GATE (TF) Textile 2005 Question Paper Solution | GATE/2005/TF/69

A 36 cotton yarn has an average strength of 300 cN with a CV of 12.5%.

The 95% confidence range based on twice the number of sample tested will be approximately

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Given-

CV=12.5%

SE1=5%

SE2=?

Number of samples(N1)=24

Number of samples(N2)=2N1

By formula-
95% confidence range-

$N=(\frac{1.96 \times CV}{SE} )^2$

$N \alpha \frac{1}{SE^2}$

$\frac{N_1}{N_2} =\frac{SE_2^2}{SE_1^2}$

$\frac{N1}{2N_1} =\frac{SE_2^2}{5^2}$

$\frac{1}{2} =\frac{SE_2^2}{25}$

$\frac{25}{2} =\frac{SE_2^2}{1}$

$12.5=SE_2^2$

$SE_2=\sqrt 12.5$

SE2=3.53

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