Parametric Test
Minimum Detectable Difference for Two-Sample t-Test for Means
Equation and example adapted from Zar, 1984
2 s 2p
δ≥
n
where the t parameters are derived from
the t table and the variances are pooled.
(tα ,v + t β (1),v )
Example: In a two-tailed t-test we are interested in determining the minimum
detectible difference between asthma rates Denver, Colorado—essentially how
different must the rates be before we detect a difference. Our data set contains
monthly asthma rates per 10,000 people for 1992 and 1994. We will use an alpha
level of 0.05, have a sample size of 12, and want to have a 90% chance of
detecting a difference. Remember that v is the degrees of freedom.
α = 0.05
n1 = 12
n2 = 12
v = n1 + n2 - 2 = 12+12 – 2 = 22
t0.05(2), 22 = 2.074
t0.10(1), 22 = 1.321
The first parameter (t0.05(2), 22) is taken from the t table for a 2-tailed test with an
0.05 alpha, and 22 degrees of freedom. The second parameter (t0.10(1), 22) is for 90%
confidence (1 - .09 = 0.10), one-tailed test (this parameter is always 1-tailed), and
22 degrees of freedom.
Asthma Rate per 10,000
1992
547
518
546
463
427
426
359
377
516
500
433
444
SS1 = 43286
1994
652
524
603
483
489
432
370
446
513
554
537
463
SS2 = 64599
The pooled variance is calculated as:
s 2p =
s 2p =
δ≥
SS1 + SS 2
v1 + v2
where SS1 and SS2 are the sums of squares for each group, and v1 and v2 are the group
degrees of freedom.
43286 + 64599
= 4903 .86
11 + 11
2( 4903.86)
(2.074 + 1.321) = 408.66 (3.395) = 68.63
24
Therefore the minimum detectable difference for a t-test on these asthma data would be 68.63 cases. Since the
average asthma rate for 1992 was about 463 cases, 68.63 cases represents over 15%... they would have to be very
different for the t test to detect it! Solution? The sample needs to be larger, but how large?
To create a graph showing sample size function for your data set, the minimum detectable difference equation will
need to be solved for a variety of sample sizes. These estimates can then be plotted on a graph so that you can
determined the best sample size based on your specific needs (See below). Since we have no way of knowing how
the sums of squares will change as we increase and decrease the sample size, this method gives us an estimate only.
Note that the degrees of freedom (v) will be based on each sample size used.
Parametric Test
δ≥
2( 4903.86)
( 2.828 + 1.372) = 980.77 ( 4.200) = 131.53
10
δ≥
2( 4903.86)
( 2.131 + 1.341) = 653.85 (3.472) = 88.78
15
δ≥
2( 4903.86)
( 2.086 + 1.325) = 490.39 (3.411) = 75.54
20
δ≥
2( 4903.86)
( 2.042 + 1.310) = 326.92 (3.352) = 60.61
30
δ≥
2( 4903.86)
( 2.021 + 1.303) = 245.19 (3.324) = 52.05
40
δ≥
2( 4903.86)
( 2.009 + 1.299) = 196.15 (3.308) = 46.33
50
δ≥
2( 4903.86)
(1.993 + 1.293) = 130.77 (3.286) = 37.58
75
δ≥
2( 4903.86)
(1.984 + 1.290) = 98.06 (3.274) = 32.42
100
Therefore, if you wanted your minimum detectable difference to be 30 cases (or about 5%) the sample size would
have to be close to 100. Also notice that after a sample size of about 100 there are diminishing returns, in that
collecting 200 samples would not significantly impact the minimum detectable difference.