Solution to Quiz 9

The full quiz is here.  The answers appear below.  Comments, which are not part of the answers, are italicized.

Time:  20 minutes.  This quiz is open book, open notes.

An agency purports to have developed a 90% lower confidence limit (LCL) of the mean arsenic concentration in water as measured in 100 wells throughout southern Bangladesh.  These wells were randomly selected from all those used to supply water to southern Bangladesh (the "country").  The value of the LCL is 235 parts per billion (micrograms per liter).

The phrase "people in the country" below will mean "people who routinely drink untreated water from arsenic-contaminated wells like those surveyed by the agency."  (This apparently is about half the people living in Bangladesh.)

1.    Indicate which of the following statements are correct and which are incorrect.  Provide reasons for each.  Credit is given only for answers accompanied by reasons.

  1. At least 90% of the people in the country drink water with more than 235 ppb arsenic.
  2. There is a 90% chance that all the people in the country drink water with more than 235 ppb arsenic.
  3. The LCL procedure was chosen so that, regardless of the underlying distribution, there would be a 90% chance that the LCL would exceed the true mean concentration.
  4. The LCL procedure was chosen so that, regardless of the underlying distribution, there would be a 10% chance that the LCL would exceed the true mean concentration.
  5. The LCL procedure was chosen so that, regardless of the underlying distribution, there would be a 90% chance that the LCL would be less than 235 ppb.
  6. There is a 10% chance that the true mean equals 235 ppb.
  7. Any glass of water drawn at random from one of these 100 wells has a 90% chance of containing a mean arsenic concentration greater than 235 ppb.
  8. At least 90% of all arsenic-contaminated wells in the country have arsenic concentrations exceeding 235 ppb.

a.    Incorrect.  An LCL tells us about the mean, not about a proportion (or percentile).  This statement describes a 90% coverage tolerance interval (with no confidence specified), not a confidence interval.  Also, it  refers to people, not to well water.  The commonest mistake was to refer to this as a prediction interval.  A prediction interval refers to a definite number of future measurements, not to a proportion of future measurements.

b.    Incorrect.  Indeed, it is almost certainly true that some people drink water with less than 235 ppb arsenic.  This statement describes a 90% confidence, 100% coverage tolerance interval.  The value of 100% may seem impossible, and it is when we assume the underlying distribution can have infinite support.  (The only 100% interval in that case extends from -infinity to +infinity.)  In situations where we assume the underlying distribution has finite support, such as a uniform or beta distribution, then a 100% interval is possible.

c.    Incorrect.  A 90% chance of exceeding the mean concentration implies only a 10% chance that the interval will cover the mean.  Therefore, this statement describes a 10% LCL.  It does not describe an upper CL.  The difference is subtle: there is an implicit "+infinity" in the assertion that a number is a lower CL; that is, the LCL describes an interval from 235 to +infinity.

d.    Correct.  This statement says there is a 90% chance of the interval covering the mean, which is the defining property of a confidence interval procedure.

e.    Incorrect.  How could we choose a procedure in advance based on the result (235 ppb) of the procedure itself?  In fact, further replications of this procedure would seem just as likely to produce an LCL above 235 as one below 235 ppb.

f.    Incorrect.   There is a 0% chance that the true mean exactly equals 235 ppb.

g.    Incorrect.  This is a statement of a prediction interval, not a confidence interval.

h.    Incorrect.  This is a statement about a proportion and is therefore a tolerance interval, not a confidence interval.

2.    To develop the LCL in problem 1, the agency assumed the sampling distribution of the mean would be Normal, but of unknown mean and standard deviation.  It also reports that the mean of the 100 concentrations was 320 ppb.   Compute the 90% upper confidence limit (UCL) of the mean.

Normal confidence limits are symmetrically distributed about the mean.  Therefore UCL - 320 = 320 - LCL = 320 - 235 = 85, implying UCL = 320 + 85 = 405 ppb.

Extra credit:  (a)  Compute the standard deviation of the data.  (The 10th percentile of Student's t distribution with 99 degrees of freedom is -1.29).  (b)  Using this information, explain why the UCL (as computed in problem 2) has likely been substantially underestimated.

Use LCL = Mean - t * SD / Sqrt(N) and plug in the known values: 235 = 320 - 1.29 * SD / Sqrt(100).  Solve for SD = (320 - 235) * 10 / 1.29 = 660 ppb, approximately.

The CV = SD/Mean = 660/320 = about 2 is very high.  Therefore the underlying distribution is far from a Normal distribution.  Even 100 values may not be enough for the Normal approximation to the sampling distribution of the mean to apply.  We see from the high CV evidence of strong positive skewness.  Therefore the uncertainty at the high end is larger than suggested by the Normal theory.

Scoring: The passing score is 88.

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This page was created 23 March 2001.