Determining Sample Size


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Suppose we wish to approximate either a mean or a proportion with level of confidence r. However we wish to have a reasonably small margin of error in order to obtain a useful interval. For a specified level of confidence, what random sample size n will guarantee a margin of error no more than e regardless of the measurement or proportion we are studying?



Sample Size For Means
("Large" Populations)

If we know the range of the possible measurements under study, then we can determine the required sample size. Suppose the measurements X are bounded by c and d. For example, suppose a minimum GPA of 2.5 on a 4-point scale is required of applicants into the Business College. Then 2.5 <= X <= 4.0, where X is the GPA of all successful applicants.

For "large" populations, a confidence interval for µ is given by Xbar +/- z s / Sqrt(n), where where s is the true population standard deviation, n is the sample size, and z is the z-score such that P(-z <= Z <= z) = r. If we want the error to be no more than e, then we can set z s / Sqrt(n) <= e and solve for n. In doing so, we obtain (z s / e)^2 <= n.

However, we still do not know the true standard deviation s. We do not want to use the sample deviation S at this point because S may be an underestimate of s. In that case, we would also underestimate the required sample size n. Instead, we should replace s with an upper bound for s.

Since c <= X <= d, we can find an upper bound for s. Among all random variables with range [c, d], the one with the largest variance is the random variable Y that equals c with probability 1 / 2 and equals d with probability 1 / 2. In that case Var(Y) = (d - c)^2 / 4. Thus for our measurement X, Var(X) <= (d - c)^2 / 4 and s <= (d - c) / 2. We can then replace s by (d - c) / 2 in the formula for n.

Thus, for any measurement X with range [c, d], the required sample size must satisfy

(z (d - c) / ( 2 e))^2 <= n,

where n is always rounded up to the nearest integer.


Example. Suppose we want a maximum margin of error of 0.05 when finding a 95% confidence interval for the average GPA of Business College applicants, where GPA's range from 2.5 to 4.0. Find the required random sample size for such a confidence interval.


Solution. Using a z-score of 1.96 for a 95% confidence interval, we must choose n at least as large as [1.96* (4.0 - 2.5) / (2 *.05 ) ] ^ 2 = 864.36. So we require a random sample of size 865.



Sample Size For Means
("Finite" Populations)

Suppose now that the population under study has known finite size N. Then a confidence interval for µ is given by Xbar +/- z* s / Sqrt(n)* Sqrt( (N-n)/(N-1) ). If we want the error to be no more than e, then we can set z* s / Sqrt(n)* Sqrt( (N-n)/(N-1) ) <= e and solve for n. However, we first replace s by its upper bound (d-c)/2. Then solving for n, we obtain

N(z*(d-c) / (2e))^2 / [N - 1 +(z* (d-c) / (2e))^2 ] <= n


Example. Suppose there are 900 applicants to the business college. We want to find a 95% confidence interval for the GPA that has no more than a 0.05 margin of error. Find the required random sample size for such a confidence interval.


Solution. Using a z-score of 1.96 for a 95% confidence interval and the range of [2.5, 4.0], we must choose n at least as large as 900(1.96*1.5 / (2*.05))^2 / [899 +(1.96* 1.5 / (2*.05))^2 ] = 441.16. So we require a random sample of size 442 from the 900 applicants.



Using the SAMPSIZE Program


We can quickly find these required sample sizes using the SAMPSIZE program. When prompted, first enter either 1 or 2 to specify a "large" or a finite population. Then enter the parameters to obtain the sample size.

(Note: The program will be more accurate since it will not round off the z-scores. Thus the results of the program may differ by +/- 1 compared to working the problem "by hand.")



Sample Size For Proportions
("Large" Populations)

A special case of the mean is that of a proportion p. In this case, all measurements are either 0 (for "No") or 1 (for "Yes"). Thus, 0 <= X <= 1. Replacing c with 0 and d with 1, we obtain the required sample size for any proportion:

(z / ( 2 e))^2 <= n,

where again n is always rounded up to the nearest integer. Here we are assuming a relatively large population which does not require the finite population correction factor for the standard deviation s.


Example. What sample size will guarantee a maximum margin of error of 0.035 for any 90% confidence interval of a proportion?

Solution. Using a z-score of 1.645 for a 90% confidence interval, we see that n must be at least [1.645 / (2 *.035 ) ] ^ 2 = 552.25. So we require a random sample of size 553.



Sample Size For Proportions
("Finite" Populations)

Suppose now that the population under study has known finite size N. Then a confidence interval for p is given by pbar +/- z* s / Sqrt(n)* Sqrt( (N-n)/(N-1) ). If we want the margin of error to be no more than e, then we can set z* s / Sqrt(n)* Sqrt( (N-n)/(N-1) ) <= e and solve for n. However, we first replace s by its upper bound 1/2. Then solving for n, we obtain

N(z / (2e))^2 / [N - 1 +(z / (2e))^2 ] <= n


Example. From a population of size 600, what random sample size is required to have a margin of error of no more than 0.035 with 98% confidence for any proportion estimate?


Solution. Using a z-score of 2.326 for a 98% confidence interval, we must choose n at least as large as 600*(2.326 / (2*.035))^2 / [599 +(2.326 / (2*.035))^2 ] = 388.98. So we require a random sample of size 389 from the 600 members of the population.



Using the PSAMPSZE Program


We also can find the required sample sizes for proportions with the PSAMPSZE program. When prompted, first enter either 1 or 2 to specify a "large" or a finite population. Then enter the parameters to obtain the sample size.

As with the SAMPSIZE program, the z-scores are not rounded off, so results may differ by +/- 1 as compared to working the problems "by hand."



Exercises

1. Suppose we want a maximum margin of error of 20 when finding a 98% confidence interval for the average GRE of Graduate School applicants, where a minimum score of 1300 out of a possible 1600 is required. Find the necessary random sample size for such a confidence interval in the following cases:

(a) Among a huge pool of applicants nationwide.
(b) Among a select pool of 500 applicants.


2. What random sample size will guarantee a maximum margin of error of 0.03 for any 99% confidence interval of a proportion from

(a) a large population?
(b) a population of size 2000?


Answers

1. (a) 305
(b) 190


2. (a) 1844
(b) 960



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