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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?

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    $\begingroup$ Your result is correct. $\endgroup$ – callculus Apr 17 '15 at 13:57
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    $\begingroup$ In this example, sigma is the population standard deviation, and thus is independent of sample size. So your answer is correct. Technically however your working should be 1.96sigma/sqrt(n) is the margin error, but the 1.96 cancels out anyway. $\endgroup$ – Kenshin Apr 17 '15 at 13:59

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