GATE (TF) Textile 2018 Question Paper Solution | GATE/2018/TF/54

Question 54 (Textile Technology & Fibre Science)

The standard deviation of population P is two times the standard deviation of population Q. The size of a random sample from population P is four times the size of a random sample from population Q. If $e_P$ and $e_Q$ denote the standard error of means of the samples from P and Q, respectively, then the ratio of $e_P$ to $e_Q$ is __1__.

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Given in the question-
Let, standard deviation of the population Q = $\rho_Q$

Standard deviation of the population P $(\rho_p)=2\times\rho_Q$

Let, the size of a random sample from population Q= $N_q$
Then , the size of a random sample from population P= $4\times N_q$

Standard error of the mean of the sample P= $e_p$

Standard error of the mean of the sample Q= $e_q$

$e_p:e_q=?$

Formula:

$N=(\frac{1.98 or 2.56\times CV}{SE})^2$

Where, $CV%=\frac{\sigma} {X}\times100$ and N =number of samples

Where, X=mean value

$N=(\frac{1.98 or 2.56\times\frac{\sigma}{X}\times100}{SE})^2$

i.e. $N \alpha (\frac{\sigma}{SE})^2$

$N \alpha (\frac{\sigma}{e})^2$

$\frac{N_p}{N_q}=(\frac{\frac{\sigma_p}{e_p}}{\frac{\sigma_q}{e_q}})^2$

$\frac{N_p}{N_q}=(\frac{\sigma_p\times e_q}{\sigma_q\times e_p})^2$

$\frac{4\times N_q}{N_q}=(\frac{2\times\sigma_q\times e_q}{\sigma_q\times e_p})^2$

$\frac{4\times N_q}{N_q}=(\frac{2\times e_q}{e_p})^2$

$4=(\frac{2\times e_q}{e_p})^2$

$4=\frac{4\times e_q^2}{e_p^2}$

$4=\frac{e_q^2}{e_p^2}$

$1=(\frac{e_q}{e_p})^2$

$\sqrt1=\frac{e_q}{e_p}$

$\frac{e_q}{e_p}=1$

$\frac{e_p}{e_q}=1$

$e_p:e_q=1$ (Answer)