Suppose the weight of n primary one students has sample mean of 20KG. If n = 40, a certain percentage of confidence interval for the population mean is (15.5,24.5). Find the confidence interval if we decrease the sample size to 30.
Am I doing it right this way?
Method 1
Given margin error = $\frac{\sigma}{\sqrt n}$
It means if sample size is quadrupled, margin error will be halved.
Since $n = 40, 0.75n = 30$
Margin error now is $4.5$
New margin error = $\frac{4.5}{\sqrt {0.75}}$ = $3\sqrt 3$
Therefore, new interval = $20-3\sqrt 3$ to $20+3\sqrt3$ = $14.804 to 25.196$
Method 2
I solve for $\sigma$ first.
Given margin error = $\frac{\sigma}{\sqrt n}$ = $4.5$ and $n = 40$
$\sigma = 4.5\sqrt{40}$
Sub new $n$ to find new margin error
New margin error = $\frac{4.5\sqrt{40}}{\sqrt{30}}$ = $3\sqrt 3$
Which turns out the be the same. But I have one question for this method. Wouldn't the sample deviation, $\sigma$ changes whenever our sample size changes(which means sampled data will be different)? Why can I still use method 2 which assumed $\sigma$ remained unchanged?