Finding and interpreting a confidence interval for a population standard deviation

Intro

Howdy!  I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we’re going to learn how to find and interpret a confidence interval for a population standard deviation.  Here's our problem statement: A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in older subjects.  After treatment with the drug, 16 subjects had a mean wake time of 98.9 minutes and a standard deviation of 44.1 minutes.  Assume that the 16 sample values appear to be from a normally distributed population, and construct a 90% confidence interval estimate of the standard deviation of the wake times for a population with the drug treatments.  Does the result indicate whether the treatment is effective?

Part 1

OK, this first part asks us to find the confidence estimate as described in the problem statement.  To do that, I'm going to bring up StatCrunch.  And using StatCrunch we can actually get this but we have to put in a small caveat, and that is StatCrunch cannot directly calculate the confidence interval estimate for standard deviation.  But it can do so for the variance.  Because the standard deviation that we want is the square root of the variance, we have to make some adjustments to the numbers we put into StatCrunch and take out of StatCrunch.  And I’m going to show you how this is done, so we’re going to get to this through the variance.

To start with, I’m going to come up to Stat –> Variance Stats (because were going through the variance), we've only got one sample, and then I'm to select With Summary because we don't have actual data, just summary statistics.  Here in my office window, it asks me for the sample variance.  Notice we’re not given the variance directly in the problem statement, but we are given the standard deviation, 44.1.  So I can get the variance from the standard deviation; the standard deviation is just the square root of the variance.  So the variance is the square of the standard deviation.

So in my calculator I’m going to put 44.1 (that's the standard deviation), I’m going to square it, and now this value here is the variance.  If I copy and paste that here using Ctrl-V on my keyboard, I’ll copy that value right in there.  That’s the variance.  The sample size is 16.  We want a confidence interval, and we want to construct the confidence interval estimate with a 90% confidence level, so I need to change this default value from 95 to 90%.  I hit Compute! and here we have the limits of the lower and upper limits for a confidence interval on the variance.

We want one for the standard deviation, so to do this we’re just going to take the square root of those limits, since the square root of the variance is the standard deviation.  I’m going to bring my calculator back up, I’m going to take these values here, I’m going to put them back into my calculator, take the square root, round to two decimal places.  So then my lower limit will be 34.16.  And I do the same thing with the upper limit, and I get 63.39.  Nice work!

Part 2

And now the second part of this problem asks, “Does the result indicate whether the treatment is effective?”  Well, the result we have is a confidence interval estimate on the standard deviation.  Because standard deviation is the square root of the variance, it's measuring the variation that we have in the data values for sample set.  So looking at the variation isn’t going to tell us very much about whether or not the treatment is effective.  What tells us whether or not the treatment is effective is we look at, say, like the mean value and compare it with some standard or some threshold value, some requirement that we need to meet.

We don't have that threshold value given in the problem statement, a standard that we can use for comparison.  And so there’s nothing that we can say from the actual confidence interval on the standard deviation.  They’re just looking at variation with that. We don’t have any idea whether the treatment is effective looking solely at the confidence interval estimate on the standard deviation.  So I’m going to look through my answer options.  There’s only one that says it's not effective.  I check my answer.  Good job!

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