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# Computing Confidence Interval Proportion. A Three Color Bowl Suppose we have a bowl containing marbles, each identical in size, texture and weight, in.

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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 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.

Sample Proportion Red Colorn color p color Blue20p blue = (20/50) =.40 Green15p green = (15/50) =.30 Red15p red = (15/50) =.30 Total5050/50 = 1

Sample Proportion Red n red = 15 n = 20 + 15 + 15 = 50 p red = 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

Confidence Level 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 http://www.mindspring.com/~cjalverson/ztable.htm Our multiplier is 2.00.

Z(k) PROBRT PROBCENT 0.00 0.50000 0.00000 0.05 0.48006 0.03988 0.10 0.46017 0.07966 0.15 0.44038 0.11924 0.20 0.42074 0.15852 0.25 0.40129 0.19741 0.30 0.38209 0.23582 0.35 0.36317 0.27366 0.40 0.34458 0.31084 0.45 0.32636 0.34729 0.50 0.30854 0.38292 0.55 0.29116 0.41768 0.60 0.27425 0.45149 0.65 0.25785 0.48431 0.70 0.24196 0.51607 0.75 0.22663 0.54675 0.80 0.21186 0.57629 0.85 0.19766 0.60467 0.90 0.18406 0.63188 0.95 0.17106 0.65789 1.00 0.15866 0.68269 Z(k) PROBRT PROBCENT 1.05 0.14686 0.70628 1.10 0.13567 0.72867 1.15 0.12507 0.74986 1.20 0.11507 0.76986 1.25 0.10565 0.78870 1.30 0.09680 0.80640 1.35 0.088508 0.82298 1.40 0.080757 0.83849 1.45 0.073529 0.85294 1.50 0.066807 0.86639 1.55 0.060571 0.87886 1.60 0.054799 0.89040 1.65 0.049471 0.90106 1.70 0.044565 0.91087 1.75 0.040059 0.91988 1.80 0.035930 0.92814 1.85 0.032157 0.93569 1.90 0.028717 0.94257 1.95 0.025588 0.94882 2.00 0.022750 0.95450 Z(k) PROBRT PROBCENT 2.05 0.020182 0.95964 2.10 0.017864 0.96427 2.15 0.015778 0.96844 2.20 0.013903 0.97219 2.25 0.012224 0.97555 2.30 0.010724 0.97855 2.35 0.009387 0.98123 2.40 0.008198 0.98360 2.45 0.007143 0.98571 2.50 0.006210 0.98758 2.55 0.005386 0.98923 2.60 0.004661 0.99068 2.65 0.004025 0.99195 2.70.0034670 0.99307 2.75.0029798 0.99404 2.80.0025551 0.99489 2.85.0021860 0.99563 2.90.0018658 0.99627 2.95.0015889 0.99682 3.00.0013499 0.99730

Lower Confidence Bound p red =.30 sdp .06481 Z = 2 lower bound = p red – Z*sdp =.30 – Z*sdp =.30 ─ 2*sdp ≈.30 ─ 2*.06481 ≈.1703

Upper Confidence Bound p red =.30 sdp .06481 Z = 2 upper bound = p red + Z*sdp =.30 + Z*sdp =.30 + 2*sdp ≈.30 + 2*.06481 ≈.4296

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

Confidence Estimation Schematic Population P red Obtain Sample Size = n Compute n red p red sdp Compute lower = p red – Z*sdp upper = p red + Z*sdp

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

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.

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

Interpretation ─ Confidence Approximately 95% of the members of the Family of Intervals cover P red, 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.

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