Confidence Intervals Calculation and InterpretationCould someone verify if my reasoning is correct?1. Is it true that if a population is normally distributed, then the sample variance for any sample size is also normally distributed, so that we may use an interval of the form [(sample variance) ± (critical value)(standard deviation of sample variance)] to estimate the population variance?I understand that the form statistic ± (critical value) ∗ (standard dev. of statistic) is applicable to mean and proportion, but is it applicable to any statistic? I believe there is a more complex formula for variance, as it follows a chi-squared distribution and is not always normally distributed, even when the population is.2. A 90% confidence interval for the height, in meters, of adults in Switzerland is 1.78 ± 0.2. Are we 90% confident that the sample mean of the next sample of adults taken in Switzerland will be between 1.58 and 1.98 meters tall?Would this also be false? I understand that a particular confidence interval of 90% calculated from an experiment does not mean that there is a 90% probability of a sample mean from a repeat of the experiment falling within this interval.

Question

Confidence Intervals Calculation and InterpretationCould someone verify if my reasoning is correct?1. Is it true that if a population is normally distributed, then the sample variance for any sample size is also normally distributed, so that we may use an interval of the form [(sample variance)   ± (critical value)(standard deviation of sample variance)] to estimate the population variance?I understand that the form statistic   ± (critical value) ∗ (standard dev. of statistic) is applicable to mean and proportion, but is it applicable to any statistic? I believe there is a more complex formula for variance, as it follows a chi-squared distribution and is not always normally distributed, even when the population is.2. A 90% confidence interval for the height, in meters, of adults in Switzerland is 1.78   ± 0.2. Are we 90% confident that the sample mean of the next sample of adults taken in Switzerland will be between 1.58 and 1.98 meters tall?Would this also be false? I understand that a particular confidence interval of 90% calculated from an experiment does not mean that there is a 90% probability of a sample mean from a repeat of the experiment falling within this interval.

Layton Leach · Accepted Answer

Step 1The sample variance of a random sample       S    2   of n observations from a normal population is not normal. Its distribution can be described as follows:   (  n  −  1  )      S    2        /        σ    2    ∼  C  h  i  s  q  (  d  f  =  n  −  1  )  ,, where   σ is the population standard deviation. This information allows one to find a confidence interval for       σ    2    ,, using the chi-squared distribution.For such a normal sample, a confidence interval for the population mean   μ is             X      ¯        ±      t    ∗    S      /        n   if the population standard deviation σ is unknown and thus must be estimated by S. Here t∗ is a number that cuts 2.5% probability from the upper tail of Student&#039;s t distribution with   n  −  1 degrees of freedom.In this formula   S      /        n   is called the &#039;(estimated) standard error of the mean&#039;. That terminology arises because   S  D  (            X      ¯        )  =  σ      /        n    . which is estimated by   S      /        n    .Step 2If the population SD   σ is known, then a 95% CI for   μ is given by             X      ¯        ±  1.96  σ      /        n    , where 1.96 cuts 2.5% probability from the upper tail of a standard normal distribution.Unfortunately, the meaning of the word confidence in &#039;confidence interval&#039; has many different interpretations. What is considered correct in one text is scorned as misleading in another. So I am not going to venture a True/False answer to the question about statures of Swiss adults.

Is it true that if a population is normally distributed, then the sample variance for any sample size is also normally distributed, so that we may use an interval of the form [(sample variance) pm (critical value)(standard deviation of sample variance)] to estimate the population variance?

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