Time: 20 minutes. Open book, open notes.
The full quiz is here. The answers appear below. Comments, which are not part of the answers, are italicized.
A statistician has computed several statistical intervals from the same set of data, assuming a Normal distribution, but he forgot to write down how each interval was computed. Help him out by matching each interval with its description. Explain your choices.
| Interval or limit | Description | ||
| 1. | [27.5, 34] | F. | 80% confidence interval for the mean The center of this interval and the centers of #5 and #6 coincide, indicating they are the confidence interval for the mean and the prediction intervals. The estimated mean is therefore about 30.7. This interval, being the narrowest of #1, #5, and #6, must be the confidence interval about the mean. |
| 2. | [6.5, 11.5] | C. | 80% confidence interval for the standard deviation. Both values are low in this interval, but none of the available choices corresponds to an interval around a lower percentile. Therefore this interval is likely to contain the standard deviation. Inspection of the remaining intervals shows they are all consistent with a standard deviation in this range. (See the "overall check" below.) |
| 3. | [40.4, 50.8] | A. | 80% confidence interval for the 95th percentile. Both values are high in this interval and the next, but their range is moderate. Therefore #3 and #4 must be confidence intervals for upper percentiles. The interval that has higher endpoints and a wider range must correspond to the higher percentile, given that they have the same confidence. |
| 4. | [34.5, 42.2] | B. | 80% confidence interval for the 80th percentile. See the remark to #3. |
| 5. | [24.3, 37.2] | E. | 80% prediction interval for the mean of four future values. See the remarks to #1 and #6. We know this interval must be wider than the confidence interval of the mean, because it must account for the additional uncertainty in the future average. |
| 6. | [11.9, 49.5] | D. | 80% prediction interval to contain four future values. This interval is the widest of #1, #5, and #6, and so must be the one containing all four future values. |
Extra credit. Determine how many numbers are in the data set.
Let n be the data set size. The length of the confidence interval of the mean, [27.5, 34], is a multiple of sqrt(1/n). The length of the prediction interval for the mean of four future values, [24.3, 37.2], is the same multiple of sqrt(1/n + 1/4). Divide the latter value by the former to get the equation
2 = (37.2 - 24.3) / (34 - 27.5) = sqrt(1 + n/4).
Solve for n, giving n=12.
As a check of this result, figure the standard deviation estimate is near (but less than) the middle of its interval, [6.5, 11.5], and therefore is about 8 or 9. The standard error is therefore around 8/sqrt(12) = 2.3. We already know the mean is (27.5 + 34)/2 = (24.3 + 37.2)/2 = (11.9 + 49.5)/2 = 30.7 (from #1, #5, and #6, respectively). The confidence limits for the mean should be roughly 1.5 standard errors from the mean, or about 3.5 units away. Indeed, 30.7 + 3.5 = 34.2 and 30.7 - 3.5 = 27.2, in close agreement with #1.
As an overall check, estimate the 80th percentile as mean + Z0.80 * sd = 30.7 + Z0.80 * 8, where Z0.80 is the 80th percentile of N(0, 1) and is therefore slightly less than 1, giving an answer slightly less than 40: just in the middle of interval #4. Similarly, estimate the 95th percentile as 30.7 + Z0.95 * 8 = 30.7 + 1.65 * 8 (more or less) = about 45, right in the middle of interval #5. Everything is consistent.
The data set used to create this quiz consists of the values 25.34 34.28 39.51 33.36 35.06 37.84 10.99 25.49 40.62 25.34 28.68 and 32.31.
Hint: Consider the midpoint (center) of each interval.
Scoring: The passing score is 90 (one incorrect answer).
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This page was created 2 April 2001.