by DAN CALLOWAY
Published 4 August 2010
WEAVERVILLE, NC - In this article, I used the Standard Error and Sample Size calculator to evaluate the effects that varying sample size and standard deviation have on the standard error (Lowry, 2010). The initial values that I selected for mean, standard deviation, and sample size (see Chart No. 1) were 25, 2.3, and 50, respectively. The results of the calculation revealed a range of 22.7 and 27.3 about the population mean of 25, which was ±2.3 SD about the population mean, and with a sample size of 50, resulted in a range of 24.6757 and 25.3253 in the sample population. The resultant SE was ±0.3253. When I varied the sample size from 50 to 200 (See Chart No. 2), the range in the new sampling population revealed results from 24.8374 to 25.1626 with a SE of ±0.1626. By multiplying the sample size by a factor of 4, the SE was multiplied by a factor of ½ the previous amount. Thus, what we can determine in this exercise is that it appears the relationship between SE and n (sample population size) is expressed by the formula:
Δ SE = 1 / (Δ n)1/2 , [a]
where Δ represents the multiplicative change in the expression under consideration. Thus, using the equation above, in this example, a change in sample size, n, by a factor of 4 will result in a corresponding change in SE by a factor of ½ . Next, I returned to my original values in Chart No. 1 for hypothetical mean, SD, and sample size, n, and varied the SD by multiplying the value of SD by a factor of 2. The resultant figure for SE upon doubling SD was ±0.6505 (See Chart No. 3). Thus, in this example, the effect of doubling the SD on SE was ±0.6505 / ±0.3253 ~ 2, or a doubling of SE. Therefore, we can determine in this exercise that it appears the relationship between SE and SD can be expressed by the formula:
Δ SE = Δ SD, [b]
where Δ represents the multiplicative change in the expression under consideration. Therefore, to summarize, we can deduce from [a] and [b] above, that changing the sample population by a factor, k, changes the SE by a factor of 1 / k1/2, and that changing the SD by a factor of m, increases the SE by the same factor, m. What we can see from this deduction is that increasing the sample size in a sample population reduces the SE whereas increasing the SD increases the SE proportionally. If we assume that a normal distribution is a continuous function, then using Calculus we can convert the expression in [a] to:
d SE/dn = d (1/n1/2) / dn = d (n-1/2) / dn = 1/(2n).
Thus the rate of change of SE with respect to n (sample population) is 1/(2n). Likewise, In equation [b]:
dSE/SD = dSD/SD = 1.
Thus the rate of change of the SE with respect to the SD is 1, or, in other words, the rate of change of SE with respect to SD is proportional. Therefore, we can deduce that the greater effect on SE is in varying the sample population (n) since it decreases the value of SE by a factor of 1/(2n) rather than varying the SD of the sample population since the rate of change is 1. In other words, the rate of change of SE with respect to n is 1/(2n), which means as n increases the rate of change in SE decreases inversely. However, as one increases/decreases SD, a proportionate change in SE occurs. In light of this information, I would opt with increasing the sample population before adjusting the SD since it has the greater effect on reducing SE.
Chart No. 1
Chart No. 2
Chart No. 3
Reference:
Lowry, R. (2010). Standard error of sample means. Retrieved from http://faculty.vassar.edu/lowry/dist.html. on 4 August 2010.
Dan Calloway
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