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Math 322
Probability and Statistical Methods
Chapter 8 Fundamental Sampling Distributions
Sampling Distributions
In the process of making an inference from a sample to a population we usually calculate one or
more statistics, such as the Mean or Variance. Since, samples are randomly selected, the values
that such statistics assume may change from sample to sample. Thus sample statistics such as
X , S and S 2 , etc. are, themselves random variables and their behavior can be modeled by
probability distributions.
The probability distribution of a sample statistic is called Sampling Distribution. Here, we will
study the nature and properties of sampling distributions.
•
The sample copy of µ , namely X , has a distribution in repeated sampling that centers
about µ . We might say, then that X is a good estimator of µ .
n
•
The sample variance S 2 =
∑(X
i =1
i
− X )2
is used to estimate the population variance σ X2 .
n −1
Let X i denote the random outcome of the i th sample observation to be selected (the sampled
observations are selected independently of one another ), then
n
X =
and
E(X ) =
where µ = E ( X i ) , i = 1, 2,...., n .
∑X
i =1
i
n
1 n
1 n
nµ
=
E
X
( i ) ∑µ = = µ
∑
n i =1
n i =1
n
1
⎛ n
⎞ 1
Var ( X ) = 2 Var ⎜ ∑ X i ⎟ = 2
n
⎝ i =1 ⎠ n
n
1
Var ( X i ) = 2
∑
n
i =1
n
∑σ
i =1
2
X
nσ X2 σ X2
= 2 =
.
n
n
Sampling Distributions of Means
‰
Sampling Distribution of the mean ( σ known)
Central Limit Theorem. If X is the mean of a random sample of size n taken from a
population with mean µ and finite variance σ 2 , then the limiting form of the distribution of
X −µ
Z=
σ/ n
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Sonuc Zorlu
Lecture Notes
Math 322
Probability and Statistical Methods
as n → ∞ , is the standard normal distribution n ( z;0,1) . The approximation will generally be
good if n ≥ 30 .
Example 1. An electrical firm manufactures light bulbs that have a length of life that is
approximately normally distributed with mean equal to 800 hours and a standard deviation of 40
hours. Find the probability that a random sample of 16 bulbs will have an average life less than
775 hours?
Solution. Given µ X = µ = 800 hours , σ = 40 hours , n = 16 , σ X = σ / n = 40 / 16 = 10 ,
⎛ X − µ 775 − 800 ⎞
<
P ( X < 775 ) = P ⎜⎜
⎟⎟ = P ( Z < −2.5 ) = 1 − P ( Z < 2.5 ) = 1 − 0.9938 = 0.0062 .
σ
10
X
⎝
⎠
Example 2. If a certain machine makes electrical resistors having a mean resistance of 40 ohms
and a standard deviation of 2 ohms, what is the probability that a random sample of 36 of these
resistors will have a combined resistance of more than 1458 ohms?
Solution. Given µ X = µ = 40 ohms , σ = 2 ohms , n = 36 , σ X = σ / n = 2 / 36 = 1 / 3 ,
⎛ n
⎞
Xi
∑
⎜
n
⎛ X − µ 40.5 − 40 ⎞
1458 ⎟
⎛
⎞
⎟ = P ( X > 40.5 ) = P ⎜
>
>
P ⎜ ∑ X i < 1458 ⎟ = P ⎜ i =1
⎟
36 ⎟
1/ 3 ⎠
⎜ n
⎝ i =1
⎠
⎝ σX
⎜
⎟
⎝
⎠
= P ( Z > 1.5 ) = 1 − P ( Z < 1.5 ) = 1 − 0.9332 = 0.0668.
‰
Sampling Distribution of the mean ( σ unknown)
We are attempting to estimate the mean of a population when the variance σ 2 is
unknown. If we have a random sample chosen from a normal population, then the random
variable
T=
X −µ
S/ n
has a student’s T-distribution with degrees of freedom υ = n − 1 .
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Sonuc Zorlu
Lecture Notes
Math 322
Probability and Statistical Methods
Example 1. Find P ( −t0.025 < T < t0.05 )
Solution. P ( −t0.025 < T < t0.05 ) = 1 − 0.05 − 0.025 = 0.925 .
Example 2. Find k such that P ( k < T < −1.761) = 0.045 , for a sample of size 15 selected from a
normal population and T =
X −µ
S/ n
.
Solution. From table A.4 we note that t0.05 = 1.761 when υ = 14 . Therefore
−t0.05 = −1.761 and let k = −tα . Then, we have 0.045 = 0.05 − α or α = 0.005 . Hence from
Table A.4 with υ = 14 , k = −t0.005 = −2.977 .
Example 3. (a) Find P ( −1.356 < T < 2.179 ) , when υ = 12 .
(b) Find P (T > 1.318) when υ = 24 .
(c) Find P ( −t0.005 < T < t0.01 ) when υ = 20
Example 4. Given a random sample of size 24 from a normal population, find k such that
(a) P ( −2.069 < T < k ) = 0.965
(b) P ( k < T < 2.807 ) = 0.095
(c) P ( −k < T < k ) = 0.90
Example 5. A random sample of size n=12 from a normal population has the mean x = 27.8 and
s 2 = 3.24 . Can we say that the given information supports the claim that the mean of the
population is µ = 28.5 ? (If the computed t -value falls between −t0.05 and t0.05 , the claim is
accepted.)
Example 6. A maker of certain brand of low fat cereal bars claims that their average saturated
fat content is 0.5 gr. In a random sample of 8 cereal bars of this brand the saturated fat content
was 0.6,0.7,0.3,0.4,0.5,0.4, 0,7 and 0.2. Would you agree with the claim? (If the computed t value falls between −t0.05 and t0.05 , the maker is satisfied with his claim)
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Sonuc Zorlu
Lecture Notes
Math 322
‰
Probability and Statistical Methods
Sampling Distribution of S 2
The beauty of the central limit theorem lies in the fact that X will have approximately a normal
sampling distribution no matter what the shape of the probabilistic model for the population, so
long as n is large and σ X2 is finite. For many other statistics additional assumptions are needed
before useful sampling distributions can be derived. Under the normality assumption for the
( n − 1) S 2
population, a sampling distribution can be derived for S 2 . It turns out that χ 2 =
has a
σ2
sampling distribution that is a special case of Gamma density function.
Note. Exactly 95% of a Chi-squared distribution lies between χ 2 0.975 and χ 2 0.025 . A χ 2 - value
falling to the right of χ 2 0.025 is not likely to occur unless our assumed value of σ 2 is too small.
Similarly a χ 2 -value falling to the left of χ 2 0.975 is unlikely to occur unless our assumed value of
σ 2 is too large. In other words, it is possible to have a χ 2 -value to the right of χ 2 0.975 or to the
left of χ 2 0.025 when σ 2 is correct.
Example 1. For a chi-squared distribution find χ 2α such that
(a) P ( χ 2 > x 2α ) = 0.99 when υ = 4
Solution. P ( χ 2 > x 2α ) = 0.99 ⇒ χ 2 0.99 = 0.297, υ = 4.
(b) P ( χ 2 > x 2α ) = 0.025 when υ = 19
Solution. P ( χ 2 > x 2α ) = 0.025 ⇒ χ 2 0.025 = 32.852, υ = 19.
(c) P ( 37.652 < χ 2 < x 2α ) = 0.045 when υ = 25
Solution. χ 2 0.05 = 37.652, υ = 25 ; χ 2 0.05− 0.045 = χ 2 0.005 = 46.928, υ = 25.
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Sonuc Zorlu
Lecture Notes
Math 322
Probability and Statistical Methods
Example 2. Find the probability that a random sample of 25 observations, from a normal
population with variance σ 2 =6, will have a variance S 2 of
(a) greater than 9.1
(b) between 3.462 and 10.745
⎛ ( n − 1) S 2 24(9.1) ⎞
2
Solution. (a) P ( S 2 > 9.1) = P ⎜
>
⎟⎟ = P ( χ > 36.4 ) = 0.05,
⎜ σ2
6 ⎠
⎝
(b)
⎛ 24(3.462) ( n − 1) S 2 24(10.745) ⎞
2
<
<
P ( 3.462 < S < 10.745 ) = P ⎜
⎟⎟
⎜
6
6
σ2
⎝
⎠
with υ = 24.
= P (13.848 < χ 2 < 42.98 ) = table(13.848, 24) − table(42.98)
= 0.95 − 0.01 = 0.94.
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Sonuc Zorlu
Lecture Notes
Math 322
Probability and Statistical Methods
Exercises
Exercise 1. (a) If X is a normal random variable with
µ = 20 and σ = 4 . Suppose a sample of
40 is chosen, find P( X > 15) .
(b) Find P (T13 < 2.650) when ν = 13 .
(c) If in a random sampling problem the sample size is n = 11 and
σ = 0.8 , find
n
P (∑ ( X i − X ) 2 > 3.1136) .
i =1
Exercise 2. Find the probability that the average number of cherries in 36 cherry puffs will be
less than 5.5 when µ = 6 and σ = 1.2 .
Solution. Given n = 36, µ = 6, σ = 1.2,
⎛ X −µ
5.5 − 6 ⎞
<
P ( X < 5.5 ) = P ⎜
⎟ = P ( Z < −2.5 ) = table(−2.5) = 0.0062 .
⎝ σ / n 1.2 / 36 ⎠
Exercise 3. The scores on a placement test given to college freshmen for the past five years are
approximately normally distributed with a mean of 74 and a variance of 8. If a random sample
of 20 students taking this course is selected, what is the probability that variance of the scores of
these 20 students is more than 13.83?
Exercise 4. For a Chi-Squared distribution find
2
χ 0.01
when n = 16 .
Exercise 5. The amount of time that a drive-through bank teller spends on a customer is a
random variable with a mean µ = 3.2 min and a standard deviation σ = 16 min , if a random
sample of 64 customers is observed find the probability that their mean time at the teller’s
counter is
(a) at most 2.7 minutes.
(b) more than 3.5 minutes.
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Lecture Notes