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  1. MHB Master
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    #1
    Hey!!

    We have a 90%-confidence interval. I want to check if the following statements are correct.

    1. If double the sample, the possibility that the value that we are looking for is out of the confidence interval is smaller.

    2. The bigger the standard error, the smaller the confidence interval.


    Since the confidence interval is $\left (\overline{x}- Z_{a/2}\cdot s_x, \overline{x}+ Z_{a/2}\cdot s_x\right )$, where $s_x$ is the standard error, I think that the second statement is wrong and it should be that the bigger the standard error, the bigger the confidence interval.
    Is this correct?

    What about the first statement?
    Last edited by mathmari; January 11th, 2017 at 12:07.

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    #2
    It's still a $90\%$ confidence interval, right? With double the sample size, your confidence interval will shrink in absolute width, but it will still be calculated on the basis of $90\%$ confidence. The probabilities will not change, just the interval size.

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    #3 Thread Author
    Quote Originally Posted by Ackbach View Post
    It's still a $90\%$ confidence interval, right? With double the sample size, your confidence interval will shrink in absolute width, but it will still be calculated on the basis of $90\%$ confidence. The probabilities will not change, just the interval size.
    I see!! Thank you very much!!

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    #4
    Erm...
    We're applying a z-value, which can only apply if the standard deviation $\sigma_x$ is given.
    But apparently it's not, since we have an $s_x$, which would typically be calculated from a sample.
    In that case I think we're supposed to apply a $t$-score instead of a $z$-score, aren't we?

    Furthermore, you mention a so called standard error, but that means we're talking about the standard deviation of the mean of a sample, typically defined as $SE=s_{\bar x}=\frac{s_x}{\sqrt n}$.

    This is about the standard deviation of the sample mean. Are we on the same page here?

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