~Calculating the minimum sample size under certain restrictions is one of the most missed problems by students of beginning statistics. Hopefully, this discussion will help in your understanding.

~The problem is as follows: Compute the smallest sample size required  for a given margin of error and confidence level.

~With the aid of our TI-83, this problem is relatively easy to do.  Just read and follow my directions carefully.

~We need to know the formula for the margin of error.  If your teacher is kind, he/she will not have you memorize it. Most instructors let students, in beginning statistics, use some sort of reference sheet or insert for quizzes and exams.

~When this formula is used, it's important to use it with the square root (for proportions) split over the top & bottom of the fraction so that the square root of the sample size is visible. Here they are:

     For proportions:  E = Z_alpha/2 {√[(p cap)(q cap)]} divided by √(n)

     For means: E=Z_alpha/2(s) divided  by √(n)

                    or, E=t_alpha/2(S_x) divided by √(n)

~Now, all you need to do is to use a property of mathematical proportions. That is, for any proportion, a/b =c/d, you may switch the "means" (b & c) or the "extremes" (a & d) or both and not change the value of the proportion.

~In the cases where the margin of error formula is used, you simply switch the E with the √(n). Then compute the right side. The result will be the value of the √(n). Do not round off this number.

~Then, square the result to get n (round up, if necessary).

~Example:  Find the minimum sample size required to estimate a population proportion with margin of error of .05, confidence level of 95%, where the Sample Proportion has been estimated (prior study) to be .19.

~Solution:  Use the formula for the margin of error with the switch (described earlier) performed.

                         sqr(n) = Z_alpha/2[(.19)(.81)] divided by .05

~The 95% confidence level enables us to find Z_alpha/2. Do this by menu 3 under 2nd Vars. Enter the following: invnorm(1-.05/2), enter. Multiply that result by 2nd square root symbol, enter .19 times .81, enter. Now, divide that result by .05. This result is equal to the √(n). Finally, square  it by using x², enter. You should get 236.4802054. Round up to 237.

~Finding minimum sample sizes for means is done in a very similar way.

~However, the critical values, using t-curves, must be found by table, since menu 3, under 2nd Vars, gives critical values for the standard normal curve only.

~Also, calculating sample size when the finite population correction is factored in is quite involved and requires a good knowledge of basic mathematical techniques. 

See  Finding Sample Size with The Finite Population Correction