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Math 3680 Lecture #16 Confidence Intervals: Part 2

Math 3680 Lecture #16 Confidence Intervals: Part 2. One-Sided Confidence Intervals. In all of the previous problems, the confidence intervals for m were centered on the sample average.

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Math 3680 Lecture #16 Confidence Intervals: Part 2

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  1. Math 3680 Lecture #16 Confidence Intervals:Part 2

  2. One-Sided Confidence Intervals

  3. In all of the previous problems, the confidence intervals for m were centered on the sample average. In some circumstances, it is preferable to have a confidence interval which gives either an upper bound or a lower bound on the population mean.

  4. Example:A leakage tank was conducted to determine the effectiveness of a seal designed to keep the inside of a plug air-tight. An air needle was inserted in the plug and this was placed under water. The pressure was then increased until leakage was observed. The following ten measurements were made: 3.1 3.3 4.5 2.8 3.5 3.5 3.7 4.2 3.9 3.3 Find a 95% one-sided confidence interval for m that provides an upper bound for m.

  5. Solution: As always, we’ll need the sample mean and sample standard deviation. This time, however, we need to find an noncentered area under the Student’s t-distribution with 9 degrees of freedom to find the upper bound for m.

  6. 95% Therefore, the 95% confidence interval is

  7. Example:Eighteen samples of “12-ounce” buttermilk biscuits are weighed (in pounds), with these results: 0.75 0.78 0.77 0.75 0.77 0.76 0.78 0.76 0.78 0.78 0.75 0.74 0.78 0.75 0.74 0.79 0.75 0.77 Find a 90% one-sided confidence interval for m that provides a lower bound for m.

  8. Confidence Intervals: Proportions

  9. Example:The Postmaster of Atlanta found that 44 packages out of a random sample of 650 had insufficient postage. Find a 95%-confidence interval for the probability that a randomly selected package has insufficient postage. Solution: For starters, the sample proportion is p = 44 / 650  0.067692.

  10. Box Model: Is it or 0 1 1 - p p 0 1 93.23% 6.77%

  11. Recall that the standard deviation of a dichotomous population is Naturally, since p is unknown, we cannot compute s exactly. However, we can use the estimate

  12. 95% Therefore, the 95% confidence interval is or 4.84-8.70%.

  13. Observations: 1. We do NOT say, “There is a 95% chance that the population parameter lies between 4.84% and 8.70%.” The population parameter (currently unknown) either lies in this range or it does not. Hence, the word “confident” instead of chance. 2. Correct interpretation: If several people run this experiment and they all find a 95%-confidence interval, then the true population parameter will lie in roughly 95% of these intervals.

  14. 3. The normal approximation has been used. Remember that a large number of draws is required for this assumption to hold. Question: How many draws are necessary? Answer: It depends: if the sample percentage is near 50%, then only 100 or so will be sufficient. More are needed if the sample percentage is close to 0% or 100%. 4. There is no such thing as a 100%-confidence interval.

  15. 5. For dichotomous populations, we will always use the normal distribution and NOT the Student’s t-distribution.

  16. Example: In a survey conducted the week before the 2006 Connecticut primary, challenger Ned Lamont led Senate incumbent Joe Lieberman. A poll of likely Democratic voters showed 399 of the sample of 784 supported Lamont. Find a 95% confidence interval for the proportion of Lamont voters in the subpopulation of Connecticut Democratic voters.  Note: In the context of political polling, the popular press often labels the radius of a 95% confidence interval as “the” margin of error.

  17. Conservative Estimate for s

  18. In the postage problem, we had to use This is an approximation, of course. If we wanted to be conservative, we could have used the inequality

  19. Notice that if the true proportion is in the ballpark of 0.5 (like for most political polls), then In these scenarios, the conservative estimate costs almost nothing.

  20. Example:In a political poll, how large should the sample be to ensure that the margin of error is 0.03?

  21. Example:In its January 29, 1996 issue, Time reported that 48% of Americans “like the principle of a flat tax.” This was based on a telephone poll of 800 Americans. The poll takers claimed a margin of error of 3%. What was their confidence level?

  22. Drawing without Replacement

  23. Example:According to tax records, a state has 5,000 manufacturing establishments. A simple random sample of size 400 is taken from all manufacturing establishments in the state. Of these, 16 had 250 employees or more. Find a 90% confidence interval for the percentage of manufacturing establishments with 250 employees or more.

  24. Question:For political polling, we typically draw without replacement. So why isn’t the finite population correction used in this scenario?

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