MHB Are the statements about the confidence interval correct?

AI Thread Summary
The discussion centers on the accuracy of statements regarding a 90% confidence interval. It is clarified that doubling the sample size reduces the width of the confidence interval but does not change the confidence level, which remains at 90%. The assertion that a larger standard error leads to a smaller confidence interval is corrected; a larger standard error actually results in a wider confidence interval. The conversation also highlights the need to use a t-score instead of a z-score when the population standard deviation is unknown, emphasizing the distinction between standard error and standard deviation. Overall, the participants confirm their understanding of these statistical concepts.
mathmari
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Hey! :o

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? (Wondering)

What about the first statement? (Wondering)
 
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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.
 
Ackbach said:
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! (Smile)
 
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? (Wondering)

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? (Wondering)
 
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