Computing Confidence Interval Proportion

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

Proportion RedSuppose 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

ColorncolorpcolorBlue20pblue= (20/50) = .40Green15pgreen= (15/50) = .30Red15pred= (15/50) = .30Total5050/50 = 1

Sample Proportion Rednred = 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

Confidence LevelOur 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 0.95450Our multiplier is 2.00.

Z(k) PROBRT PROBCENT0.00 0.50000 0.000000.05 0.48006 0.039880.10 0.46017 0.079660.15 0.44038 0.119240.20 0.42074 0.158520.25 0.40129 0.197410.30 0.38209 0.235820.35 0.36317 0.273660.40 0.34458 0.310840.45 0.32636 0.347290.50 0.30854 0.382920.55 0.29116 0.417680.60 0.27425 0.451490.65 0.25785 0.484310.70 0.24196 0.516070.75 0.22663 0.546750.80 0.21186 0.576290.85 0.19766 0.604670.90 0.18406 0.631880.95 0.17106 0.657891.00 0.15866 0.68269Z(k) PROBRT PROBCENT1.05 0.14686 0.706281.10 0.13567 0.728671.15 0.12507 0.749861.20 0.11507 0.769861.25 0.10565 0.788701.30 0.09680 0.806401.35 0.088508 0.822981.40 0.080757 0.838491.45 0.073529 0.852941.50 0.066807 0.866391.55 0.060571 0.878861.60 0.054799 0.890401.65 0.049471 0.901061.70 0.044565 0.910871.75 0.040059 0.919881.80 0.035930 0.928141.85 0.032157 0.935691.90 0.028717 0.942571.95 0.025588 0.948822.00 0.022750 0.95450Z(k) PROBRT PROBCENT2.05 0.020182 0.959642.10 0.017864 0.964272.15 0.015778 0.968442.20 0.013903 0.972192.25 0.012224 0.975552.30 0.010724 0.978552.35 0.009387 0.981232.40 0.008198 0.983602.45 0.007143 0.985712.50 0.006210 0.987582.55 0.005386 0.989232.60 0.004661 0.990682.65 0.004025 0.991952.70 .0034670 0.993072.75 .0029798 0.994042.80 .0025551 0.994892.85 .0021860 0.995632.90 .0018658 0.996272.95 .0015889 0.996823.00 .0013499 0.99730

Lower Confidence Bound pred = .30sdp .06481Z = 2

lower bound = pred Z*sdp =.30 Z*sdp =.30 2*sdp .30 2*.06481 .30 2*.06481 .1703

Upper Confidence Bound pred = .30sdp .06481Z = 2

upper bound = pred + Z*sdp =.30 + Z*sdp =.30 + 2*sdp .30 + 2*.06481 .30 + 2*.06481 .4296

Write the IntervalWe write the approximate interval as

[.1703, .4296 ].

Confidence Estimation Schematic

PopulationPredObtainSample Size = n Compute nredpredsdpComputelower = pred Z*sdpupper = pred + Z*sdp

Interpretation Population and ProportionWe 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

[pred 2*sdp, pred + 2*sdp]; wherepred = nred/n, andsdp=sqrt(pred*(1- pred)/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 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.