Summary Sheet for Confidence Intervals
Math 1070-1, Spring 2003
April 14, 2003
Exercises:
• You are expected to know how to compute z ∗ and t∗ . Here are some
exercises that were discussed in lecture; for the t∗ calculations, assume
that there are 6 degrees of freedom:
– For a 95%-confidence interval, z ∗ ' 1.96, t∗ ' 2.571.
– For a 88%-confidence interval, z ∗ ' 1.56, 2.015 ≤ t∗ ≤ 2.571. [Always choose the conservative estimate to maintain the the confidence
level; here, t∗ ' 2.571.]
– For a 76%-confidence interval, z ∗ ' 1.18; 1.156 ≤ t∗ ≤ 1.476, so we
opt for t∗ ' 1.476.
1
For Population Means:
• If n is large, then
σ
µ = X̄ ± z ∗ √ ,
n
s
µ = X̄ ± z ∗ √ ,
n
if σ is known,
if σ is unknown.
Here, z ∗ is computed from the normal table.
• If n is small but the underlying population is normal, then
σ
µ = X̄ ± z ∗ √ ,
n
s
µ = X̄ ± t∗ √ ,
n
if σ is known,
if σ is unknown.
Here, t∗ is computed from the t-table with n − 1 degrees of freedom.
For Comparing Two Means:
• If n1 and n2 are both large, then
s
µ1 − µ2 = X̄1 − X̄2 ± z ∗
s
µ1 − µ2 = X̄1 − X̄2 ± z ∗
σ2
σ12
+ 2,
n1
n2
if σ is known,
s2
s21
+ 2,
n1
n2
if σ is unknown.
• If one of the sample sizes is small, and if the underlying population is
normal, then
s
σ2
σ12
∗
+ 2,
if σ is known,
µ1 − µ2 = X̄1 − X̄2 ± z
n1
n2
s
s2
s21
+ 2,
if σ is unknown,
µ1 − µ2 = X̄1 − X̄2 ± t∗
n1
n2
where t∗ is computed from a t-distribution but its degrees of freedom is
now the smaller of n1 − 1 and n2 − 1.
2
For a Popultion Proportion:
Recall Wilson’s proportion:
pe =
X +2
.
n+4
Then a confidence interval for p is:
r
p = pe ± z
∗
pe (1 − pe)
.
n+4
For Comparing Two Proportions:
The sample sizes n1 and n2 both have to be large. The Wilson s proportions,
here, are:
X1 + 1
X2 + 1
, pe2 =
.
pe1 =
n1 + 2
n2 + 2
The confidence interval is
s
p1 − p2 = pe1 − pe2 ± z ∗
pe1 (1 − pe1 ) pe2 (1 − pe2 )
+
n1 + 2
n2 + 2
3