1 / 15

Computing Confidence Interval Proportion

Computing Confidence Interval Proportion. A Three Color Bowl. Suppose we have a bowl containing marbles, each identical in size, texture and weight, in three colors: Red , Green , Blue . Proportion Red.

hunter
Download Presentation

Computing Confidence Interval Proportion

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Computing Confidence Interval Proportion

  2. A Three Color Bowl Suppose we have a bowl containing marbles, each identical in size, texture and weight, in three colors: Red, Green, Blue.

  3. Proportion Red Suppose we have a large population containing marbles, each identical in size, texture and weight, in three colors: Red, Green, Blue. Suppose further that we wish to estimate the population proportion of red, but that examining the population directly and exhaustively is impractical.

  4. Sample Proportion Red

  5. Sample Proportion Red nred = 15 n = 20 + 15 + 15 = 50 pred = 15 / n = 15 / 50 = .30 sdp = sqrt(p*(1-p)/n) = sqrt(.30*(1-p)/n) = sqrt(.30*.70/n) = sqrt(.30*.70/50) = sqrt(.210/50) ≈ sqrt(.0042)  .06481

  6. Our next step is to select a confidence level this number will provide a level of confidence in our estimation process. A standard choice is 95% confidence. Using the table @ http://www.mindspring.com/~cjalverson/ztable.htm, we obtain the following row: 2.00 0.022750 0.95450 Our multiplier is 2.00. Confidence Level

  7. Lower Confidence Bound pred = .30 sdp  .06481 k = 2 lower bound = pred – k*sdp = .30 – k*sdp = .30 ─ 2*sdp ≈ .30 ─ 2*.06481 ≈ .30 ─ 2*.06481 ≈ .1703

  8. Upper Confidence Bound pred = .30 sdp  .06481 k = 2 upper bound = pred + k*sdp = .30 + k*sdp = .30 +2*sdp ≈ .30 + 2*.06481 ≈ .30 + 2*.06481 ≈ .4296

  9. Write the Interval We write the approximate interval as [.1703, .4296].

  10. Confidence Estimation Schematic Compute lower = pred – k*sdp upper = pred + k*sdp Compute nred pred sdp Population Pred Obtain Sample n = 50

  11. Interpretation ─ Population and Proportion We have a large population of marbles. We seek the true population proportion of red marbles for this population.

  12. Interpretation ─ Family of Samples We obtain random samples of n=50 marbles per sample. Each marble is drawn from the population with replacement. Our Family of Samples consists of every possible random sample as described above.

  13. Interpretation ─ Family of Intervals From each member of the Family of Samples we comupute the interval [pred─ 2*sdp, pred+ 2*sdp]; where pred = nred/n, and sdp=sqrt(pred*(1- pred)/n). Our Family of Intervals consists of every possible interval computed as above.

  14. Interpretation ─ Confidence Approximately 95% of the members of the Family of Intervals cover Pred, the true population proportion of red marbles. The remaining 5% or so fail. We view our single interval, [.1703, .4296], as being drawn at random from the Family of Intervals. If our interval is drawn from the 95% supermajority, then between 17.03% and 42.96% of the marbles are red.

More Related