Bonferroni confidence intervals

Suppose we want to find confidence intervals I1, I2, . . . , Im for parameters 1, 2, . . . , m. In the analysis ofvariance problem 1 = 1 2, 2 = 1 3, . . . , m = k1 k. We want

P (j Ij , j = 1, 2, . . . ,m) 1 ,

i.e. we want a simultaneous confidence level 1 .

Aside note:De Morgans law states that:

P (A1 A2 . . . Am) = 1 P (A1 A2 . . . Am)

Bonferroni inequality:

P (A1 A2 . . . Am) mi=1

P (Ai)

Therefore,

P (A1 A2 . . . Am) 1mi=1

P (Ai) (1)

Suppose P (j Ij) = 1j . Lets denote with Aj the event that j Ij . Then from Bonferroni inequality(see expression (1) above) we get:

P (1 I1, 2 I2, . . . m Im) 1mi=1

P (j / Ij)

1mi=1

j .

If all j , j = 1, 2, . . . ,m are chosen equal to we observe that the simultaneous confidence level is only 1 m which is smaller than 1 because m 1. In order to have a simultaneous confidence interval1 we will need to use confidence level 1 m for each confidence interval for 12, 13, . . . , k1k.

Bonferroni confidence interval for i j :

xi xj t 2m ;Nksp

1ni

+1nj,

where sp is the estimate of the common but unknown standard deviation:

sp =

(n1 1)s21 + (n2 1)s22 + . . .+ (nk 1)s2k

N k

Example (see ANOVA handout):Construct a Bonferroni confidence interval for 1 4 using 95% simultaneous confidence level. The dataare x1 = 4.062, x4 = 3.920, n1 = 10, n2 = 10, sp = 0.06061. We also need t 0.05

2m ;63= t 0.05

221 ;63= t0.00119;63 =

3.1663.

1 4 4.062 3.920 3.1663(0.06061)

110

+110,

or

1 4 0.142 0.086,

or

0.056 1 4 0.228.