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Unformatted text preview: e coeﬃcient (1 − α) (or conﬁdence level 100(1 − α)%) for normal
random variable we deﬁne as
σ
σ
(x − zα/2 σx , x + zα/2 σx ) = x − zα/2 √ , x + zα/2 √ ,
n
n
where quantity zα/2 is founded from the requirement
P (x − zα/2 σx ≤ x ≤ x + zα/2 σx ) = 1 − α.
6 A. Zhensykbaev Sampling distributions Using z score we have
P (x − zα/2 σx ≤ x ≤ x + zα/2 σx ) = P (−zα/2 ≤ z ≤ zα/2 ) =
= 2P (0 ≤ z ≤ zα/2 ).
Thus,
P (x − zα/2 σx ≤ x ≤ x + zα/2 σx ) = 2P (0 ≤ z ≤ zα/2 ). (1) On the Pic. 2 the value of z = zα/2 such that the area α/2 lies to its
right.
T α/2
−zα/2 0 α/2
zα/2 E z Pic. 2
The z value we can ﬁnd using Table IV (McClave and Benson).
For example, if α = 0.1 in the Table IV we should ﬁnd that z0.05 will be
the z value corresponding to an area is equal to 0.50.05=0.45: z0.05 = 1.645.
Example. In a large airline company 225 ﬂights are randomly selected,
and the number of unoccupied seats is noted for each sampled ﬂights. The
sample mean and standard deviation are x = 11.6, s = 4.1. Estimate the
mean number of unoccupied seats per ﬂight during the past year, using a
90% conﬁdence interval.
Solution. Find α/2: due to (1 − α)% = 90% we have
α
1 − α = 0.9,
= 0.05,
z0.05 = 1.645.
2
Since n = 225, s = 4.1 and it is the best approximation of the standard
deviation σ , we ﬁnd
2σ
4.1
µ ≈ x ± √ = 11.6 ± 1.645 √
= 11.6 ± 0.45
n
225
7 A. Zhensykbaev Sampling distributions (it means that µ ∈ [11.15, 12.05] with the probability 90%).
Sample size determination.
Assume that we are given a 100(1 − α)% conﬁdence interval for mean
µ. We want to estimate µ to within a given boundary for sampling error
SE. How much will be the sample size n to accomplish this?
From the equality
σ
SE = zα/2 σx = zα/2 √ .
n
we ﬁnd: zα/2...
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This note was uploaded on 02/11/2014 for the course MATH 1390 taught by Professor Christopherstocker during the Spring '13 term at Marquette.
 Spring '13
 ChristopherStocker
 Math

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