Calculating Confidence Intervals With Standard Error
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normal distribution calculator to find the value of z to use for a confidence interval Compute a confidence interval on the mean when σ is known Determine whether to use a t distribution or a normal distribution Compute a confidence interval on the
Calculating Confidence Intervals For Proportions And Their Differences
mean when σ is estimated View Multimedia Version When you compute a confidence interval calculating confidence intervals in minitab on the mean, you compute the mean of a sample in order to estimate the mean of the population. Clearly, if you calculating confidence intervals in stata already knew the population mean, there would be no need for a confidence interval. However, to explain how confidence intervals are constructed, we are going to work backwards and begin by assuming characteristics of the population. Then http://onlinelibrary.wiley.com/doi/10.1002/9781444311723.oth2/pdf we will show how sample data can be used to construct a confidence interval. Assume that the weights of 10-year-old children are normally distributed with a mean of 90 and a standard deviation of 36. What is the sampling distribution of the mean for a sample size of 9? Recall from the section on the sampling distribution of the mean that the mean of the sampling distribution is μ and the standard error http://onlinestatbook.com/2/estimation/mean.html of the mean is For the present example, the sampling distribution of the mean has a mean of 90 and a standard deviation of 36/3 = 12. Note that the standard deviation of a sampling distribution is its standard error. Figure 1 shows this distribution. The shaded area represents the middle 95% of the distribution and stretches from 66.48 to 113.52. These limits were computed by adding and subtracting 1.96 standard deviations to/from the mean of 90 as follows: 90 - (1.96)(12) = 66.48 90 + (1.96)(12) = 113.52 The value of 1.96 is based on the fact that 95% of the area of a normal distribution is within 1.96 standard deviations of the mean; 12 is the standard error of the mean. Figure 1. The sampling distribution of the mean for N=9. The middle 95% of the distribution is shaded. Figure 1 shows that 95% of the means are no more than 23.52 units (1.96 standard deviations) from the mean of 90. Now consider the probability that a sample mean computed in a random sample is within 23.52 units of the population mean of 90. Since 95% of the distribution is within 23.52 of 90, the probability that the mean from any given sample will be within 23.52 of 90 is 0.95. This means tha
the standard error can be calculated as SE = (upper limit – lower limit) / 3.92. http://handbook.cochrane.org/chapter_7/7_7_7_2_obtaining_standard_errors_from_confidence_intervals_and.htm For 90% confidence intervals divide by 3.29 rather than 3.92; for 99% confidence intervals divide by 5.15. Where exact P values are quoted alongside http://www.milefoot.com/math/stat/ci-variances.htm estimates of intervention effect, it is possible to estimate standard errors. While all tests of statistical significance produce P values, different tests use different confidence interval mathematical approaches to obtain a P value. The method here assumes P values have been obtained through a particularly simple approach of dividing the effect estimate by its standard error and comparing the result (denoted Z) with a standard normal distribution (statisticians often refer to this as a Wald test). calculating confidence intervals Where significance tests have used other mathematical approaches the estimated standard errors may not coincide exactly with the true standard errors. The first step is to obtain the Z value corresponding to the reported P value from a table of the standard normal distribution. A standard error may then be calculated as SE = intervention effect estimate / Z. As an example, suppose a conference abstract presents an estimate of a risk difference of 0.03 (P = 0.008). The Z value that corresponds to a P value of 0.008 is Z = 2.652. This can be obtained from a table of the standard normal distribution or a computer (for example, by entering =abs(normsinv(0.008/2) into any cell in a Microsoft Excel spreadsheet). The standard error of the risk difference is obtained by dividing the risk difference (0.03) by the Z value (2.652), which gives 0.011.
we can estimate a population standard deviation from a sample standard deviation, and when the original population is normally distributed, we can construct confidence intervals of the standard deviation as well. The Theory Variances and standard deviations are a very different type of measure than an average, so we can expect some major differences in the way estimates are made. We know that the population variance formula, when used on a sample, does not give an unbiased estimate of the population variance. In fact, it tends to underestimate the actual population variance. For that reason, there are two formulas for variance, one for a population and one for a sample. The sample variance formula is an unbiased estimator of the population variance. (Unfortunately, the sample standard deviation is still a biased estimator.) Also, both variance and standard deviation are nonnegative numbers. Since neither can take on a negative value, the domain of the probability distribution for either one is not $(-\infty, \infty)$, thus the normal distribution cannot be the distribution of a variance or a standard deviation. The correct PDF must have a domain of $[0, \infty)$. It can be shown that if the original population of data is normally distributed, then the expression $\dfrac{(n-1)s^2}{\sigma^2}$ has a chi-square distribution with $n-1$ degrees of freedom. The chi-square distribution of the quantity $\dfrac{(n-1)s^2}{\sigma^2}$ allows us to construct confidence intervals for the variance and the standard deviation (when the original population of data is normally distributed). For a confidence level $1 - \alpha$, we will have the inequality $\chi_{1-\alpha/2}^2 \le \dfrac{(n-1)s^2}{\sigma^2} \le \chi_{\alpha/2}^2$. Solving this inequality for the population variance $\sigma^2$, and then the population standard deviation $\sigma$, leads us to the following pair of confidence intervals. $\dfrac{(n-1)s^2}{\chi_{\alpha/2}^2} \le \sigma^2 \le \dfrac{(n-1)s^2}{\chi_{1-\alpha/2}^2}$ $\sqrt{ \dfrac{(n-1)s^2}{\chi_{\alpha/2}^2}} \le \sigma \le \sqrt{ \dfrac{(n-1)s^2}{\chi_{1-\alpha/2}^2}}$ It is worth noting that since the chi-square distribution is not symmetric, we will be obtaining confidence intervals that are not symmetric about the point estimate. Example