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Short answer, complex. I presume you're in a basic stats class so your dealing with something like a normal distribution however (or something else very standard). You can think of it this way...

A confidence interval re-scales margin of likely error into a range. This allows you to say something along the lines, "I can say with 95% confidence that the mean/variance/whatever lies within whatever and whatever" because you're taking into account the likely error in your prediction (as long as the distribution is what you think it is and all stats are what you think they are).

This is because, if you know all of the things I listed with absolute certainty, you are able to accurately predict how erroneous your prediction will be. It's because central limit theory allow you to assume statistically relevance of the sample, even given an infinite population of data.

The main idea of a confidence interval is to create and interval which is likely to include a population parameter within that interval. Sample data is the source of the confidence interval. You will use your best point estimate which may be the sample mean or the sample proportion, depending on what the problems asks for. Then, you add or subtract the margin of error to get the actual interval. To compute the margin of error, you will always use or calculate a standard deviation.

An example is the confidence interval for the mean. The best point estimate for the population mean is the sample mean according to the central limit theorem.

So you add and subtract the margin of error from that. Now the margin of error in the case of confidence intervals for the mean is za/2 x Sigma/ Square root of n where a is 1- confidence level. For example, confidence level is 95%, a=1-.95=.05 and a/2 is .025. So we use the z score the corresponds to .025 in each tail of the standard normal distribution. This will be. z=1.96. So if Sigma is the population standard deviation, than Sigma/square root of n is called the standard error of the mean. It is the standard deviation of the sampling distribution of all the means for every possible sample of size n take from your population ( Central limit theorem again). So our confidence interval is the sample mean + or - 1.96 ( Population Standard deviation/ square root of sample size.

If we don't know the population standard deviation, we use the sample one but then we must use a t distribution instead of a z one. So we replace the z score with an appropriate t score.

In the case of confidence interval for a proportion, we compute and use the standard deviation of the distribution of all the proportions. Once again, the central limit theorem tells us to do this. I will post a link for that theorem.

It is the key to really understanding what is going on here!

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Q: What is the relationship between confidence interval and standard deviation?

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no

The increase in sample size will reduce the confidence interval. The increase in standard deviation will increase the confidence interval. The confidence interval is not based on a linear function so the overall effect will require some calculations based on the levels before and after these changes. It would depend on the relative rates at which the change in sample size and change in standard deviation occurred. If the sample size increased more quickly than then standard deviation, in some sense, then the size of the confidence interval would decrease. Conversely, if the standard deviation increased more quickly than the sample size, in some sense, then the size of the confidence interval would increase.

It will make it wider.

Never!

is variance the square of the standard deviation

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The standard deviation is used in the numerator of the margin of error calculation. As the standard deviation increases, the margin of error increases; therefore the confidence interval width increases. So, the confidence interval gets wider.

no

It goes up.

The increase in sample size will reduce the confidence interval. The increase in standard deviation will increase the confidence interval. The confidence interval is not based on a linear function so the overall effect will require some calculations based on the levels before and after these changes. It would depend on the relative rates at which the change in sample size and change in standard deviation occurred. If the sample size increased more quickly than then standard deviation, in some sense, then the size of the confidence interval would decrease. Conversely, if the standard deviation increased more quickly than the sample size, in some sense, then the size of the confidence interval would increase.

There is absolutely no relationship to what you've asked. I'm pretty sure you simply framed the question in the wrong way, but to literally answer your question... none. Zero relationship. There's no such thing. There is however a relationship between standard deviation and a CI, but a CI can in no shape way or form influence a standard deviation.

It will make it wider.

Never!

Confidence intervals may be calculated for any statistics, but the most common statistics for which CI's are computed are mean, proportion and standard deviation. I have include a link, which contains a worked out example for the confidence interval of a mean.

The confidence intervals will increase. How much it will increase depends on whether the underlying probability model is additive or multiplicative.

Did you mean, "How do you calculate the 99.9 % confidence interval to a parameter using the mean and the standard deviation?" ? The parameter is the population mean μ. Let xbar and s denote the sample mean and the sample standard deviation. The formula for a 99.9% confidence limit for μ is xbar - 3.08 s / √n and xbar + 3.08 s / √n where xbar is the sample mean, n the sample size and s the sample standard deviation. 3.08 comes from a Normal probability table.

Confidence interval considers the entire data series to fix the band width with mean and standard deviation considers the present data where as prediction interval is for independent value and for future values.

is variance the square of the standard deviation

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