Download When the population distribution is approximately normal and σ2 is

When the population distribution is approximately normal and σ 2 is
unknown, the confidence interval of µ is constructed by using the T
statistic:
X̄ − µ
√ .
T =
S/ n
P (−tα/2 < T < tα/2 ) = 1 − α
S
S
P (X̄ − √ tα/2 < µ < X̄ + √ tα/2 ) = 1 − α
n
n
A random sample of size n is selected from a population whose variance
σ 2 is unknown. The (1 − α)100% confidence interval is:
s
s
x̄ − √ tα/2 < µ < x̄ + √ tα/2 .
n
n
When n ≥ 30, tα/2 ≈ zα/2 .
Example 3. The contents of seven similar containers of sulfuric acid
are 9.8, 10.2, 10.4, 9.8, 10.0, 10.2 and 9.6 liters. Find a 95% confidence
interval for the mean of all such containers, assuming an approximate
normal distribution.
4.4 Variance σ 2
Any value s2 of S 2 is an point estimator of the population variance σ 2 .
Hence, the statistic S 2 is an estimator of σ 2 . S 2 is an unbiased estimator
of the population variance σ 2 .
(n−1)S 2
σ2
has a chi-square distribution with n − 1 degrees of freedom. We
can use it to construct the confidence interval of σ 2 .
(n − 1)S 2
<
< χ2α/2 ) = 1 − α
2
σ
(n − 1)S 2
(n − 1)S 2
2
P(
<
σ
<
)=1−α
χ2α/2
χ21−α/2
2
P (χ1−α/2
For our particular sample of size n, the sample variance s2 is computed,
and the (1 − α)100% confidence interval is given by
(n − 1)s2
(n − 1)s2
2
<
σ
<
.
χ2α/2
χ21−α/2
Example 4. The following are the weights, in decigrams, of 10 packages
of grass seek distributed by a certain company: 46.4, 46.1, 45.8, 47.0,
46.1, 45.9, 45.8, 46.9, 45.2 and 46.0. Find a 95% confidence interval
for the variance of all such packages of grass seed distributed by this
company.
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4.6. Difference between two means
If we have two populations with means µ1 and µ2 and variances σ12 and
σ22 , respectively,
Any value x̄1 −x̄2 of X̄1 − X̄2 is a point estimator of the difference between
µ1 and µ2 .
X̄1 − X̄2 is an unbiased estimator of the difference between µ1 and µ2 .
If our independent samples are selected from normal populations, we can
establish a confidence interval for µ1 − µ2 by considering the sampling
distribution of X̄1 − X̄2 .
We can expect the sampling distribution of X̄1 − X̄2 to be approximately
normally distributed
with mean µ1 − µ2 (denoted by µX̄1 −X̄2 ), and stanq
dard deviation σ12 /n1 + σ 2 /n2 (denoted by σX̄1 −X̄2 ). Thus,
(X̄1 − X̄2 ) − (µ1 − µ2 )
q
σ12 /n1 + σ22 /n2
P (−z α2 <
(X̄1 − X̄2 ) − (µ1 − µ2 )
r
σ12
n1
s
P (X̄1 − X̄2 − zα/2
+
σ22
n2
∼ N (0, 1).
< z α2 ) = 1 − α
s
σ12 σ22
σ12 σ22
+
< µ1 − µ2 < X̄1 − X̄2 + zα/2
+ )=1−α
n1 n2
n1 n2
For any two independent random samples of sizes n1 and n2 selected from
two populations whose variances σ12 and σ22 are known, the difference of
the sample means, x̄1 − x̄2 , is computed, and the (1 − α)100% confidence
interval of µ1 − µ2 is:
s
x̄1 − x̄2 − zα/2
s
σ12 σ22
σ12 σ22
+
< µ1 − µ2 < x̄1 − x̄2 + zα/2
+ .
n1 n2
n1 n2
If σ12 and σ22 are unknown and our samples are sufficiently large, we may
replace σ12 and σ22 by s21 and s22 .
Example 5. A standardized chemistry test was given to 50 girls and 75
boys. The girls made an average grade of 76 with a standard deviation of
6, while the boys made an average grade of 82 with a standard deviation
of 8. Find a 96% confidence interval for the difference µ1 − µ2 , where µ1
is the mean score of all boys and µ2 is the mean score of all girls who
might take this test.
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