Unformatted text preview: σ 2
If σ is unknown but n is large we may replace it by the sample standard
zα/2 s 2
If we know the range r of observations in the population, we can take
σ ≈ r/4.
Example. Consider previous example. Assume that the company wants
the 90% conﬁdence interval for estimate of mean with a sampling error
SE=0.2. How many ﬂights will have to be randomly selected?
Solution. Using last formula we obtain:
n= 1.645 · 4.1
0.2 2 ≈ 1, 137. Thus, the airline company needs to select randomly 1,137 ﬂights.
Small-sample estimation of a population mean.
Now we try to estimate a population mean using small random samples.
The population standard deviation σ is almost always unknown. The
sample standard deviation s may provide a poor approximation for σ when
the sample size is small. Therefore using the standard normal statistic
8 A. Zhensykbaev Sampling distributions we deﬁne the t-statistic x−µ
For normal random variable t-statistic has mound-shaped symmetric
sampling distribution with mean 0. It depends on the sample size n. This
dependence we deﬁne as: t-statistic has (n − 1) degrees of freedom (df ).
The number (n − 1) in the divisor in the formula of the sample variance.
Small-sample conﬁdence interval for a population mean µ is
x − tα/2 √ , x + tα/2 √ ,
t= where tα/2 is deﬁned similarly zα/2 from assumption
P x − tα/2 √ ≤ x ≤ x + tα/2 √ = 1 − α,
using the Table VI (McClave and Benson).
For example, for α = 0.1 and df=5 we found from the Table VI (column
t0.05 , row 5) that t0.05 = 2.015.
The sample size needed to estimate µ within a given bound B is
n= tα/2 s
B 2 . Example. Suppose we get a sample of 5 expert opinions about next
year’s earnings per share for a stock, x = 2.63, s = 0.72. What is the
conﬁdence interval of a level α = 0.05 for unknown µ?
Solution. We have df = 5 − 1 = 4, P = 1 − α = 0.95, t0.05 = 2.132 and
µ ∈ x − tα/2 √ , x + tα/2 √ = 2.63 − 2.132 √ , 2.63 + 2.132 √ .
Thus, with the probability 95%
µ ∈ [1.944, 3.316]. 9...
View Full Document