Confidence Intervals that Contain the Mean: Designing an ActivityRoll a fair 6 sided dice 25 times. Take the sample mean of the face value. Using the standard deviation of the uniform probability distribution, construct confidence intervals with C = .90 , .95 , .99.I had 16 groups total doing this exercise. All 16 groups found their 90% confidence intervals containing the the mean of the probability distribution (3.5.).I thought I would have at least one confidence intervals that wouldn't contain the mean... considering that the probability that all 16 confidence intervals contains μ would be .90 16 ≃ .1853.Is my prediction completely flawed? Or was I just unlucky? Or is there something inherently wrong with my experiment?

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Confidence Intervals that Contain the Mean: Designing an ActivityRoll a fair 6 sided dice 25 times. Take the sample mean of the face value. Using the standard deviation of the uniform probability distribution, construct confidence intervals with   C  =  .90  ,  .95  ,  .99.I had 16 groups total doing this exercise. All 16 groups found their 90% confidence intervals containing the the mean of the probability distribution (3.5.).I thought I would have at least one confidence intervals that wouldn&#039;t contain the mean... considering that the probability that all 16 confidence intervals contains   μ would be       .90          16        ≃  .1853.Is my prediction completely flawed? Or was I just unlucky? Or is there something inherently wrong with my experiment?

Nelson Santana · Accepted Answer

Step 1The standard deviation of the uniform probability distribution for 1,2,3,4,5,6 is   σ  =      105        /    6. The standard deviation of the mean of a sample of size 25 is then   σ      /    5. If (as I suspect) they are constructing confidence intervals for a normal distribution with this standard deviation, they will use approximately   (  μ  −  1.6449  σ      /    5  ,  μ  +  1.6449  σ      /    5  ) for the confidence interval, where μ is their sample mean. This confidence interval contains 3.5 iff   2.93816  &amp;lt;  μ  &amp;lt;  4.06184. The actual distribution is not normal, but it&#039;s not far enough from normal to explain the discrepancy: from a simulation, the probability of the confidence interval containing 3.5 seems to be quite close to 0.9.Step 2Of course, not having seen the students&#039; actual work, it&#039;s possible that they were doing something other than what I wrote above. Maybe some group got an interval that didn&#039;t contain 3.5, found this upsetting, and repeated the experiment to get a better result. But 0.1853 is not all that small a probability; it could easily have arisen by chance.

Jimena Hatfield · Accepted Answer

Step 1This seems an excellent classroom experiment to reinforce the practical meaning of the confidence level of an interval estimate, and I wish more instructors would do this kind of thing.First, let&#039;s just check to make sure we&#039;re on the same page about the experiment. Each of the 16 groups rolls a fair die   n  =  25 times. The sum of the 25 values is divided by 25 to get             X      ¯      .For a single roll of a die, we get the value       X    i   which has   μ  =  E  (      X    i    )  =  3.5 and       σ    2    =  V  (      X    i    )  =  35      /    12. Thus   V  (            X      ¯        )  =      σ    2        /    n  =  35      /    300  =  7      /    60.. Then you assume             X      ¯       is close enough to normal to use the 90% z-interval             X      ¯        ±  1.645  σ      /        n   or             X      ¯        ±  0.5619..Step 2You have 16 of these 90% CIs, and consider each of them to be a Success if it includes 3.5 and Failure if it does not. You were surprised to get no Failures because you think the chances of that are about 18% or 19%. (Just to make sure I&#039;m not having a &#039;senior moment&#039; here, I did a simulation in R and with a million runs of what you did, and I got around 18.5% of such 25-die experiments with 16 groups got no Failures.)If that is the scenario, then you were moderately unlucky. But you might also have considered yourself unlucky if you had gotten less than 13 successes (even a little more likely). I suppose you would have been very pleased with 14 or 15 Successes (the most likely two results, to be sure), but the probability of that is only about 60%. And the THIRD most likely result is 16.

Roll a fair 6 sided dice 25 times. Take the sample mean of the face value. Using the standard deviation of the uniform probability distribution, construct confidence intervals with C=.90,.95,.99.

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