Quiz 9 (retest)
Time: 20
minutes. This quiz is open book, open notes.
An agency purports to have developed a symmetric 90% two-sided
confidence limit 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 "water supply"). The
interval ranges from 210 parts per billion to 430 ppb.
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.
- The mean arsenic concentration in the water supply must be
between 210 and 430 ppb .
- 90% of the water supply contains between 210 and 430 ppb
arsenic.
- There is a 95% chance that the average arsenic
concentration in the water supply is less than 430 ppb and a 95% chance that
the average is greater than 210 ppb.
- There is a 90% chance that the average arsenic
concentration in the water supply is less than 430 ppb and a 90% chance that
the average is greater than 210 ppb.
- There is an 80% chance that the average arsenic
concentration in the water supply is less than 430 ppb and an 80% chance
that the average is greater than 210 ppb.
- If another agency attempts to replicate these results by
randomly selecting and measuring 100 wells for arsenic, there is a 90%
chance its mean will be between 210 and 430 ppb.
- If another agency attempts to replicate these results by
randomly selecting and measuring 100 wells for arsenic, there is a 90%
chance its confidence interval will cover the true mean.
- If another agency attempts to replicate these results by
randomly selecting and measuring 100 wells for arsenic, there is a 90%
chance its confidence interval will overlap with the 210 to 430 ppb range.
2. To develop the confidence interval in problem 1, the
agency assumed the sampling distribution of the mean would be Normal, but of
unknown mean and standard deviation. Estimate the 80% upper confidence
limit (UCL) of the mean by using the Normal distribution instead of the (more
accurate) Student t distribution.
Scoring: The passing score is 88.

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