Confidence Intervals:
Let us start by examining problem 7 from problem set 10.
7. Understanding the nice symmetry of the Normal distribution can helpful when dealing with
percentiles and probabilities. Suppose Z ~N(0, 1) and assume z > 0.
(a) Let p = P(
Z ≤
−
z). Write an expression for P(
−
z
≤
Z
≤
z) in terms of p.
(b) Use (a) to find an interval [
−
z, z] such that P(
−
z
≤
Z
≤
z) = 0.95.
7a)We let p=P(Z≤
z).
(
)
(
)
(
)
(1
(
))
(
)
(
))
(
)
P
z
Z
z
P Z
z
P Z
z
P Z
z
P Z
z
P Z
z
P Z
z
=12p.
7b) 12p=.95 means that 2p=.05, so p=.025. If we look up this zvalue we get 1.96. So, P(
1.96 ≤ Z ≤ 1.96)
= .95
Similarly, if you defined p as P(Z
≤
z) when z>0, then your equation for
)
(
z
Z
z
P
would be 2p1.
For problem 7b this means you would have 2p1 =.95, so p=.975. So, we obtained .975 and .025. If you
subtract .025 from .975 you get .95. This is exactly what we want with a 95% confidence interval. It
represents the MIDDLE 95% OF THE DATA. What else do you notice about .025 and .975? Do we need to
look up both of these?
The answer to the second question is no. We only need one of these because of the answer to the first
question. They have the same numerical value; the only difference is that one is a positive zvalue
whereas the other is a negative zvalue.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 MARTIN
 Statistics, Normal Distribution, Standard Deviation

Click to edit the document details