Sanderson M. Smith

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SAMPLE SIZE NEEDED FOR

SPECIFIED MARGIN OF ERRORThe general formula for a

confidence intervalis

estimateplus/minus(critical value)(standard deviation of the estimate)A

95% confidence intervalfor proportions has the formp(hat)

plus/minus1.96 ÷[((p(hat))(1-p(hat))/N]where N is the sample size and p(hat) is the sample proportion.

Since 1.96 is approximately 2, we will use 2 in what follows to simply computations.

If the population proportion parameter is p, the margin of error,

m, for a95% confidence intervalcan be calculated using the formula

m= 2 ÷[p(1-p)/N]When sampling, p is replaced by p(hat), the sample proportion, to compute

m.We now ask the question:

What sample size is needed if one wants a specific margin of error?Solving the above equation for N yields

m^{2}/4 = p(1-p)/N ==> N = 4p(1-p)/m^{2}.YIKE! We face a "Catch 22" situation. We want N, and we know

m, but we don't know a value for p, and we can't get such a value until we actually take a sample.We get around this dilemma by finding the value of p that will maximize N. Since 4 and

m^{2}are known constants, we need only maximize y = p(1-p) = p - p^{2}. This is simply a parabola that opens downward. We need only find the vertex. We can take a derivative and note that dy/dp = 1 - 2p which has value of 0 when p = 1/2. In other words, looking at the equationN = 4p(1-p)/

m^{2}we will get the largest possible value of N when we substitute p = 1/2. Note that is the substitution is made, we get N = 4(1/2)(1/2)/m

^{2}= 1/m^{2}, a very simple formula. In other words, if we want a 95% confidence interval and knowm, margin of error, we can determine the sample size needed for the specifiedm. For instance, if we want a margin of error = 2%, then the sample size required is 1/(.02)^{2}= 2,500.What is shown in the box below is a published survey related to the Persian Gulf War some years ago.

Would you support or oppose U.S. forces resuming action to force Saddam from power?54% Support

37% Oppose

For this Newsweek Poll, the Gallop Organization interviewed a national sample of 751 adults by telephone April 4-5. The margin of error is plus or minus 4 percentage points. Some "Don't Know" and other responses not shown.

Let's do some computations:

If we were to compute the margin of error using 54%, we would get 2 ÷[(.54)(.46)/751] = 0.0363736. Rounding "out" to the nearest integer percent, we would get the 4% stated in the survey results. If one calculates the margin of error using 37%, one obtains 2 ÷[(.37)(.63)/751] = 0.0352356. Again, if we round "out," we get 4%.

If we wanted a margin of error = 4%, the sample size needed would be 1/(.04)

^{2}= 625. A margin of error of 3% would require a sample size = 1/(.03)^{2}= 1,111. What is reported in the survey "jives" with these calculations.While published surveys such as the one above do not generally talk about a 95% confidence interval, the reported margin of error does relate to such an interval, as has been demonstrated. Using the information provided in the survey above, the 95% confidence interval for those

supportusing action to remove Saddam from power is [50%, 58%]. The corresponding 95% confidence interval for those whoopposeis [33%,41%]."Numbers rule the universe."-PYTHAGORAS (around 550 B.C.)

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