Want this question answered?

Q: When calculating the confidence interval why is the sample standard deviation used to derive the standard error of the mean?

Write your answer...

Submit

Still have questions?

Continue Learning about Math & Arithmetic

It goes up.

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.

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.

You probably mean the confidence interval. When you construct a confidence interval it has a percentage coverage that is based on assumptions about the population distribution. If the population distribution is skewed there is reason to believe that (a) the statistics upon which the interval are based (namely the mean and standard deviation) might well be biased, and (b) the confidence interval will not accurately cover the population value as accurately or symmetrically as expected.

Assuming var is variance, simply square the standard deviation and the result is the variance.

Related questions

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

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 goes up.

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.

You probably mean the confidence interval. When you construct a confidence interval it has a percentage coverage that is based on assumptions about the population distribution. If the population distribution is skewed there is reason to believe that (a) the statistics upon which the interval are based (namely the mean and standard deviation) might well be biased, and (b) the confidence interval will not accurately cover the population value as accurately or symmetrically as expected.

b-a/6