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Lesson 9 - R
Sampling Distributions
Objectives
• Define a sampling distribution
• Contrast bias and variability
• Describe the sampling distribution of a sample
proportion (shape, center, and spread)
• Use a Normal approximation to solve probability
problems involving the sampling distribution of a
sample proportion
• Describe the sampling distribution of a sample mean
• State the central limit theorem
• Solve probability problems involving the sampling
distribution of a sample mean
Vocabulary
• None New
Effect of Sample Size on σ
• Sampling distribution of sample means have
the same mean as population, but have tighter
spreads (less variance) than the population
• Sampling distribution of sample proportions
have the same proportion as the population,
but also have less variance than the
population
• As n increases, then σ decreases
Distribution
Standard
Deviation
Example
Population
σ
σ = 10
Sampling
of x-bar
σ/√n
n=9
σx-bar = σ/√n = 10/√9 = 10/3 = 3.333
Central Limit Theorem
Regardless of the shape of the population, the
sampling distribution of x-bar becomes
approximately normal as the sample size n
increases.
Caution:
– only applies to shape and not to the mean or
standard deviation
Conditions: (requirements for x-bar)
– SRS
– Rule of Thumb: n ≥ 30 (big enough to apply)
P-hat Requirements
Sampling Distribution of p-hat
• SRS
• For a simple random sample of size n such that
n ≤ 0.1N (sample size is ≤ 10% of the population size)
[keeps binomial vs going to hypergeometric]
• The shape of the sampling distribution of p-hat is
approximately normal provided
n(1 – p) ≥ 10 and n(1-p) ≥ 10
[allows use of normal approximations of binomial]
X vs X-bar
• When problem talks about a single instance
or event, then we are dealing with X and not
with X-bar.
– If normally distributed use σ, μ, and X
– If nothing is said about the distribution of
population, then you cannot assume normality
(problem is not workable)
• When problem talks about a sample of size n,
then we are dealing with X-bar and not with X
– Use σX-bar = σ/√n and X-bar
X vs P-hat
• When problem talks about values that are not
in percentages or in numbers greater than 1,
then they are talking in terms of X
– Convert x to percentage (divide by n) before using
– Work in percentages (p-hat)
• When problem talks about percentages or
values less than one, then they are (more than
likely) talking in terms of p-hat.
– Use values directly
• Read problem carefully
Chapter 9 Summary
• The sample mean and the sample proportion can be
considered as random variables in sampling distribution
• The sampling distribution of the sample mean is
approximately normal with
– A mean equal to the population mean, μx-bar = μ
– A standard deviation equal to  x   / n
• The sampling distribution of the sample proportion is
approximately normal with
– A mean equal to the population proportion , μp-hat = p
– A standard deviation equal to  p̂  p( 1  p ) / n
• Central Limit Theorem: for large n (>30), the sampling
distribution of x̄ is approximately normal for any
population (with a finite σ)
Summary and Homework
• Summary
• Homework
– pg 607 – 609; 9.47, 49-53, 56, 58
Problem 1
Based on a simple random sample of size 100,
a researcher calculated the standard deviation
associated with a sample proportion to be 0.08.
If she increases the sample size to 400, what
will be the new standard deviation associated
0.04
with the sample proportion?______________
0.08 ~ n = 100
n = 400 is an increase of 4
Inverse Square root relationship
0.08/ √4 = 0.08/ 2 = 0.04
Problem 2
unbiased
We know that p-hat is a/an _______________
statistic because the mean of the sampling
distribution of p-hat is equal to the true
population proportion p.
Problem 3a
According to the manufacturer’s specifications, the mean
time required for a particular anesthetic drug to produce
unconsciousness is 7.5 minutes with a standard deviation
of 1.8 minutes. A random sample of 36 patients is to be
selected and the average time for the drug to work will be
computed for the sample. Find the probability that
(a) the mean time for the sample will be less than 7.0 mins
x-bar ~ N(7.5, 1.8/√36) because 36 is sufficiently large to
invoke CLT (Central Limit Thrm)
P( x-bar < 7.0) = 0.0478
normcdf(-E99, 7.0, 7.5, 1.8/√36)
Problem 3b
According to the manufacturer’s specifications, the mean
time required for a particular anesthetic drug to produce
unconsciousness is 7.5 minutes with a standard deviation
of 1.8 minutes. A random sample of 36 patients is to be
selected and the average time for the drug to work will be
computed for the sample. Find the probability that
(b) a randomly selected patient requires less than 7.0
If X ~ N(7.5, 1.8) then it can be computed, otherwise not.
No statement of normality.
If X ~ N(7.5, 1.8) then P( X < 7.0) = 0.3906
normcdf(-E99, 7.0, 7.5, 1.8)
Problem 3c
According to the manufacturer’s specifications, the mean
time required for a particular anesthetic drug to produce
unconsciousness is 7.5 minutes with a standard deviation
of 1.8 minutes. A random sample of 36 patients is to be
selected and the average time for the drug to work will be
computed for the sample. Find the probability that
(c) If more random samples of size 36 were selected, the
middle 95% of the sample means should fall between
7.0757
7.9243
_______________
minutes and ______________
minutes.
middle 95% is μ  2σ
7.5  2(1.8/√72) = 7.5  0.4243
Problem 4
As we have discussed in class, a one-pound (16 ounce)
box of sugar generally weighs more than 1 lb. According
to some state laws, producers will be fined if the mean of
5 randomly selected boxes is less than 1 lb. If the
packaging equipment delivers individual weights that are
N (μ, 0.4) ounces, what setting should be used for μ so the
probability of being fined is 0.01? Provide a sketch to
support your answer.
invnorm(0.02) = -2.054
x-bar – μ
Z = ------------σ/√n
(or from the table)
16 - μ
-2.054 = -------0.4/√5
-2.054 (0.17889) = 16 – μ
-0.367431 = 16 – μ
μ = 16.3674 ounces
Problem 5
Central Limit Theorem
According to the __________________________________,
when a simple random sample of size n is drawn from any
population with mean µ and standard deviation σ, if n is
sufficiently large the sampling distribution of the sample
mean is approximately normal.
Problem 6
Place the word “true” or “false” in the blank at the end of
each of the following sentences.
(a) If the underlying population is skewed, the distribution
False
of x-bar will be normal for n = 2. _________________
(b) If the underlying population is skewed, the distribution
True
of x-bar will be normal for n = 100. _________________
(c) If the underlying population is normal, the distribution
True
of x-bar will be normal for n = 2. _________________
(d) If the underlying population is normal, the distribution
True
of x-bar will be normal for n = 100. _________________
Problem 7a
We know that 60% of the students in a large state
university are male.
(a) Determine the mean and standard deviation of the
sampling distribution of the sample proportion of males
(p-hat) when samples of 400 students are randomly
selected from this population.
Mean = 0.60
standard deviation = √p(1-p)/n
√.6(.4)/400 = 0.0245
Problem 7b&c
We know that 60% of the students in a large state
university are male.
p ~ N(0.6, 0.0245)
(b) Verify that the formula you used for your standard
deviation computation is valid in this situation. State
the condition(s) that must be satisfied and convince me
that all necessary conditions are met.
Assume total number of male students > 4000 N > 10n
400(0.6) = 240 ≥ 10
np ≥ 10
400(0.4) = 160 ≥ 10
n(1-p) ≥ 10
(c) What is the probability that a simple random sample of
400 students will contain more than 65% males?
P(p-hat > 0.65) = 0.0206
normalcdf(0.65, E99, 0.6, 0.0245)
Problem 8
The weight of eggs produced by a certain breed of hen is
N (60, 4). What is the probability that the weight of a
dozen (12) randomly selected eggs falls between 700
grams and 725 grams.
X is weight of an egg
12X is the weight of a dozen eggs
E(X) = 60
V(X) = 4² = 16
s(X) = 48
E(12X) = 12E(X) = 720
V(12X) = 12²V(X)= 144(16) = 2304
P(700 < 12X < 725) = 0.2030
normcdf(700,725,720,48)