Transcript + The Sampling Distribution of

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Chapter 7: Sampling Distributions
Section 7.1
What is a Sampling Distribution?
The Practice of Statistics, 4th edition – For AP*
STARNES, YATES, MOORE
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Chapter 7
Sampling Distributions
 7.1
What is a Sampling Distribution?
 7.2
Sample Proportions
 7.3
Sample Means
+ Section 7.1
What Is a Sampling Distribution?
Learning Objectives
After this section, you should be able to…

DISTINGUISH between a parameter and a statistic

DEFINE sampling distribution

DISTINGUISH between population distribution, sampling distribution,
and the distribution of sample data

DETERMINE whether a statistic is an unbiased estimator of a
population parameter

DESCRIBE the relationship between sample size and the variability
of an estimator
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Introduction
Different random samples yield different statistics. We need to be able
to describe the sampling distribution of possible statistic values in
order to perform statistical inference.
We can think of a statistic as a random variable because it takes
numerical values that describe the outcomes of the random sampling
process. Therefore, we can examine its probability distribution using
what we learned in Chapter 6.
Population
Sample
Collect data from a
representative Sample...
Make an Inference
about the Population.
What Is a Sampling Distribution?
The process of statistical inference involves using information from a
sample to draw conclusions about a wider population.
Definition:
A parameter is a number that describes some characteristic of the
population. In statistical practice, the value of a parameter is usually
not known because we cannot examine the entire population.
A statistic is a number that describes some characteristic of a sample.
The value of a statistic can be computed directly from the sample data.
We often use a statistic to estimate an unknown parameter.
Remember s and p: statistics come from samples and
parameters come from populations
We write µ (the Greek letter mu) for the population mean and x ("x bar") for the sample mean. We use p to represent a population
proportion. The sample proportion pˆ ("p - hat") is used to estimate the
unknown parameter p.
What Is a Sampling Distribution?
As we begin to use sample data to draw conclusions about a
wider population, we must be clear about whether a number
describes a sample or a population.
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Parameters and Statistics
Problem: Identify the population, the parameter, the sample, and the
statistic in each of the following settings.
(a)A pediatrician wants to know the 75th percentile for the distribution of
heights of 10-year-old boys so she takes a sample of 50 patients and
calculates Q3 = 56 inches.
(b) A Pew Research Center poll asked 1102 12- to 17-year-olds in the
United States if they have a cell phone. Of the respondents, 71% said
yes. http://www.pewinternet.org/Reports/2009/14--Teens-and-Mobile-Phones-Data-Memo.aspx
Solution:
(a) The population is all 10-year-old boys; the parameter of interest is the
75th percentile, or Q3. The sample is the 50 10-year-old boys included in
the sample; the statistic is the sample Q3 = 56 inches.
(b) The population is all 12- to 17-year-olds in the US; the parameter is the
proportion with cell phones. The sample is the 1102 12- to 17-year-olds in
the sample; the statistic is the sample proportion with a cell phone, = 0.71.
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Example: Heights and Cell Phones
This basic fact is called sampling variability: the value of a
statistic varies in repeated random sampling.
To make sense of sampling variability, we ask, “What would
happen if we took many samples?”
Sample
Population
Sample
Sample
Sample
Sample
Sample
Sample
Sample
?
What Is a Sampling Distribution?
How can x (sample) be an accurate estimate of µ (population)? After all,
different random samples would produce different values of x .
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Sampling Variability
Sampling Distribution Applet
Definition:
The sampling distribution of a statistic is the distribution of values
taken by the statistic in all possible samples of the same size from the
same population.
In practice, it’s difficult to take all possible samples of size n to
obtain the actual sampling distribution of a statistic. Instead, we can
use simulation to imitate the process of taking many, many samples.
One of the uses of probability theory in statistics is to obtain
sampling distributions without simulation. We’ll get to the theory
later.
What Is a Sampling Distribution?
In the penny activity, we took several samples of 20 pennies. There
are many, many possible SRSs of size 20 from a population of size
100. If we took every one of those possible samples, calculated the
sample proportion for each, and graphed all of those values, we’d
have a sampling distribution.
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Sampling Distribution
1) The
population distribution gives the values of the
variable for all the individuals in the population.
2) The
distribution of sample data shows the values of
the variable for all the individuals in the sample.
3) The
sampling distribution shows the statistic values
from ALL the possible samples of the same size from
the population.
What Is a Sampling Distribution?
There are actually three distinct distributions involved
when we sample repeatedly and measure a variable of
interest.
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Population Distributions vs. Sampling Distributions
p = proportion obtained in each
p = actual proportion of in
entire population
sample
The proportion of red chips (parameter p) is 1/3
Choose 12 chips and record the proportion of RED chips you have.
Mark using
p̂ on the graph.
Population Distribution:
What Is a Sampling Distribution?
Activity: Counting Chips
Given: 147 colored chips (1/3 red, 1/3 blue, 1/3 white)
Distribution of Sample (n=12)
Sampling Distribution
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Example: Choosing Cards
Imagine a deck of cards with aces and face cards removed so that only the cards
2 through 10 remain, shuffle the deck, randomly select 5 cards, and note the median
value of the cards. For example, if the selected cards were 2, 2, 4, 5, and 9, the
median would be 4. Record the value of the sample median on a dotplot going from
2 to 10.
We used Fathom to simulate choosing 500 SRSs of size 5 from the deck of
cards described above. The graph below shows the distribution of the sample
median for these 500 samples.
Problem:
(a)Is this the sampling distribution of the sample median?
Justify your answer.
No, since the distribution didn’t use all possible samples of
size 5, this isn’t the exact sampling distribution.
(b) Describe the distribution. Are there any obvious outliers?
Shape: The graph is roughly symmetric with a single peak at 6.
Center: The mean of the distribution is about 6. Spread: The values fall mostly
between 4 and 8, although there are values as low as 2 and as high as 10.
Outliers: There don’t seem to be any outliers.
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Example: Choosing Cards
Imagine a deck of cards with aces and face cards removed so that only the cards
2 through 10 remain, shuffle the deck, randomly select 5 cards, and note the median
value of the cards. For example, if the selected cards were 2, 2, 4, 5, and 9, the
median would be 4. Record the value of the sample median on a dotplot going from
2 to 10.
We used Fathom to simulate choosing 500 SRSs of size 5 from the deck of
cards described above. The graph below shows the distribution of the sample
median for these 500 samples.
(c) Suppose that another student prepared a different deck of
cards and claimed that it was exactly the same as the one used
in the activity. However, when you took an SRS of size 5, the
median was 4. Does this provide convincing evidence that the
student’s deck is different?
(c) Getting a sample median of 4 does not provide convincing evidence that the
student’s deck is different. Getting a sample median of 4 or lower occurs fairly
often just by chance when taking random samples of size 5 from a deck of cards
with the aces and face cards removed.
Center: Biased and unbiased estimators
In the penny example, we collected many samples of size 20 and calculated
the sample proportion of 21st century pennies. How well does the sample
proportion estimate the true proportion of 21st century pennies, p = 0.47?
Note that the center of the approximate sampling
distribution is close to 0.47. In fact, if we took ALL
possible samples of size 20 and found the mean of
those sample proportions, we’d get exactly 0.47.
Definition:
A statistic used to estimate a parameter is an unbiased
estimator if the mean of its sampling distribution is equal
to the true value of the parameter being estimated.
What Is a Sampling Distribution?
The fact that statistics from random samples have definite sampling
distributions allows us to answer the question, “How trustworthy is a
statistic as an estimator of the parameter?” To get a complete
answer, we consider the center, spread, and shape.
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Describing Sampling Distributions
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Describing Sampling Distributions
To get a trustworthy estimate of an unknown population parameter, start by using a
statistic that’s an unbiased estimator. This ensures that you won’t tend to
overestimate or underestimate. Unfortunately, using an unbiased estimator doesn’t
guarantee that the value of your statistic will be close to the actual parameter value.
n=100
n=1000
Larger samples have a clear advantage over smaller samples. They
are much more likely to produce an estimate close to the true value of the
parameter.
What Is a Sampling Distribution?
Spread: Low variability is better!
Variability of a Statistic
The variability of a statistic is described by the spread of its sampling distribution. This
spread is determined primarily by the size of the random sample. Larger samples give
smaller spread. The spread of the sampling distribution does not depend on the size of
the population, as long as the population is at least 10 times larger than the sample.
We can think of the true value of the population parameter as the bull’s- eye on a
target and of the sample statistic as an arrow fired at the target. Both bias and
variability describe what happens when we take many shots at the target.
Bias means that our aim is off and we
consistently miss the bull’s-eye in the
same
. Our sample values do not
center on the population value.
High variability means that repeated
shots are widely scattered on the target.
Repeated samples do not give very
similar results.
The lesson about center and spread is
clear: given a choice of statistics to
estimate an unknown parameter,
choose one with no or low bias and
minimum variability.
What Is a Sampling Distribution?
Bias, variability, and shape
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Describing Sampling Distributions
Sampling distributions can take on many shapes. The same statistic can have
sampling distributions with different shapes depending on the population distribution
and the sample size. Be sure to consider the shape of the sampling distribution
before doing inference.
Sampling distributions for different
statistics used to estimate the
number of tanks in the German
Tank problem. The blue line
represents the true number of
tanks.
Note the different shapes. Which
statistic gives the best estimator?
Why?
What Is a Sampling Distribution?
Bias, variability, and shape
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Describing Sampling Distributions
+ Section 7.1
What Is a Sampling Distribution?
Summary
In this section, we learned that…

A parameter is a number that describes a population. To estimate an unknown
parameter, use a statistic calculated from a sample.

The population distribution of a variable describes the values of the variable
for all individuals in a population. The sampling distribution of a statistic
describes the values of the statistic in all possible samples of the same size from
the same population.

A statistic can be an unbiased estimator or a biased estimator of a parameter.
Bias means that the center (mean) of the sampling distribution is not equal to the
true value of the parameter.

The variability of a statistic is described by the spread of its sampling
distribution. Larger samples give smaller spread.

When trying to estimate a parameter, choose a statistic with low or no bias and
minimum variability. Don’t forget to consider the shape of the sampling
distribution before doing inference.
+
Chapter 7
Sampling Distributions
 7.1
What is a Sampling Distribution?
 7.2
Sample Proportions
 7.3
Sample Means
+ Section 7.2
Sample Proportions
Learning Objectives
After this section, you should be able to…

FIND the mean and standard deviation of the sampling distribution of
a sample proportion

DETERMINE whether or not it is appropriate to use the Normal
approximation to calculate probabilities involving the sample
proportion

CALCULATE probabilities involving the sample proportion

EVALUATE a claim about a population proportion using the sampling
distribution of the sample proportion
Sampling Distribution of
pˆ
Consider the approximate sampling distributions generated by a
simulation in which SRSs of Reese’s Pieces are drawn from a
population whose proportion of orange candies is either 0.45 or 0.15.
What do you notice about the shape, center, and spread of each?
Sample Proportions
How good is the statistic pˆ as an estimate of the parameter p? The
sampling distribution of pˆ answers this question.
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 The
Sampling Distribution of pˆ
Shape : In some cases, the sampling distribution of pˆ can be
approximated by a Normal curve. This seems to depend on both the
sample size n and the population proportion p.
Center : The mean of the distribution is m pˆ = p. This makes sense
because the sample proportion pˆ is an unbiased estimator of p.
Spread : For a specific value of p , the standard deviation s pˆ gets
smaller as n gets larger. The value of s pˆ depends on both n and p.
pˆ =
count of successes in sample X
=
size of sample
n
Sample Proportions
What did you notice about the shape, center, and spread of each
sampling distribution?
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 The
Sampling Distribution of pˆ
mX = np
sX = np(1 - p)
Since pˆ = X /n = (1/n) × X, we are just multiplying the random variable X
by a constant (1/n) to get the random variable pˆ . Therefore,
1
m pˆ = (np) = p
n
1
s pˆ =
np(1- p) =
n
pˆ is an unbiased estimator of p
np(1- p)
=
2
n
p(1- p)
n
As sample size increases, the spread decreases.
Sample Proportions
In Chapter 6, we learned that the mean and standard deviation of a
binomial random variable X are
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 The
Sampling Distribution of pˆ
Sampling Distribution of a Sample Proportion
Choose an SRS of size n from a population of size N with proportion p
of successes. Let pˆ be the sample proportion of successes. Then:
The mean of the sampling distribution of pˆ is m pˆ = p
The standard deviation of the sampling distribution of pˆ is
p(1- p)
s pˆ =
n
as long as the 10% condition is satisfied : n £ (1/10)N.
As n increases, the sampling distribution becomes approximately Normal. Before
you perform Normal calculations, check that the Normal condition is satisfied: np ≥
10 and n(1 – p) ≥ 10.
Sample Proportions
We can summarize the facts about the sampling distribution
of pˆ as follows :
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 The
A polling organization asks an SRS of 1500 first-year college students how far away
their home is. Suppose that 35% of all first-year students actually attend college within
50 miles of home. What is the probability that the random sample of 1500 students will
give a result within 2 percentage points of this true value?
STATE: We want to find the probability that the sample proportion falls between 0.33
and 0.37 (within 2 percentage points, or 0.02, of 0.35).
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Sample Proportions
ˆ
 Using the Normal Approximation for p
Inference about a population proportion p is based on the sampling distribution
of pˆ . When the sample size is large enough for np and n(1- p) to both be at
least 10 (the Normal condition), the sampling distribution of pˆ is
approximately Normal.
PLAN: We have an SRS of size n = 1500 drawn from a population in which the
proportion p = 0.35 attend college within 50 miles of home.
m pˆ = 0.35
s pˆ =
(0.35)(0.65)
= 0.0123
1500
DO: Since np = 1500(0.35) = 525 and n(1 – p) =
1500(0.65)=975 are both greater than 10, we’ll standardize
and then use Table A to find the desired probability.
z=
0.33 - 0.35
= -1.63
0.0123
z=
0.37 - 0.35
= 1.63
0.0123
P(0.33 £ pˆ £ 0.37) = P(-1.63 £ Z £1.63) = 0.9484 - 0.0516 = 0.8968
CONCLUDE: About 90% of all SRSs of size 1500 will give a result
within 2 percentage points of the truth about the population.
State: We want to find the probability that the proportion of middle school
students who plan to attend a four-year college or university falls between
ˆ ≤ 0.87).
73% and 87%. That is, P(0.73 ≤ p
Sample Proportions
The superintendent of a large school district wants to know what proportion of
middle school students in her district are planning to attend a four-year college
or university. Suppose that 80% of all middle school students in her district
are planning to attend a four-year college or university. What is the probability
that a SRS of size 125 will give a result within 7 percentage points of the true
value?
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Planning for College
Plan: To calculate standard deviation of “p-hat”, we must satisfy the n ≤ (1/10)N
rule. Because the school district is large, we can assume that there are
more than 10(125) = 1250 middle school students so:
0.73 - 0.80 within
Conclude:
of (0.80)(0.20)
all SRSs of size 125 will give a samplez proportion
m pˆ = 0.80 About 95%
=
= -1.94
s pˆ =
= 0.036
0.036
7 percentage points of the true
proportion
of
middle
school
students
who
are
planning
125
to attend a four-year college or university.
We can consider the distribution of to be approximately
0.87 - 0.80
z=
= 1.94
Normal since np = 125(0.80) = 100 ≥ 10 and
0.036
n(1 – p) = 125(0.20) = 25 ≥ 10.
P(0.73 £ pˆ £ 0.87) = P(-1.94 £ Z £ 1.94) = 0.9738 - 0.0262 = 0.9476
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Assignment: Pg. 430
21 – 25
Pg. 439 28, 29, 33, 35, 37, 41
+ Section 7.2
Sample Proportions
Summary
In this section, we learned that…

When we want information about the population proportion p of successes, we
often take an SRS and use the sample proportion pˆ to estimate the unknown
parameter p. The sampling distribution of pˆ describes how the statistic varies
in all possible samples from the population.
The mean of the sampling distribution of pˆ is equal to the population proportion

p. That is, pˆ is an unbiased estimator of p.
p(1- p)
The standard deviation of the sampling distribution of pˆ is s pˆ =
for
n
an SRS of size n. This formula can be used if the population is at least 10 times
as large as the sample (the 10% condition). The standard deviation of pˆ gets
smaller as the sample size n gets larger.

When the sample size n is larger, the sampling distribution of pˆ is close to a
p(1- p)
Normal distribution with mean p and standard deviation s pˆ =
.
n

In practice, use this Normal approximation when both np ≥ 10 and n(1 - p) ≥ 10 (the
Normal condition).
+
Chapter 7
Sampling Distributions
 7.1
What is a Sampling Distribution?
 7.2
Sample Proportions
 7.3
Sample Means
+ Section 7.3
Sample Means
Learning Objectives
After this section, you should be able to…

FIND the mean and standard deviation of the sampling distribution of
a sample mean

CALCULATE probabilities involving a sample mean when the
population distribution is Normal

EXPLAIN how the shape of the sampling distribution of sample
means is related to the shape of the population distribution

APPLY the central limit theorem to help find probabilities involving a
sample mean
Means
Consider the mean household earnings for samples of size 100.
Compare the population distribution on the left with the sampling
distribution on the right. What do you notice about the shape, center,
and spread of each?
Sample Means
Sample proportions arise most often when we are interested in
categorical variables. When we record quantitative variables we are
interested in other statistics such as the median or mean or
standard deviation of the variable. Sample means are among the
most common statistics.
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 Sample
Sampling Distribution of
x
Mean and Standard Deviation of the Sampling Distribution of Sample Means
Suppose that x is the mean of an SRS of size n drawn from a large population
with mean m and standard deviation s . Then :
The mean of the sampling distribution of x is mx = m
The standard deviation of the sampling distribution of x is
sx =
s
n
as long as the 10% condition is satisfied: n ≤ (1/10)N.
Note : These facts about the mean and standard deviation of x are true
no matter what shape the population distribution has.
Sample Means
When we choose many SRSs from a population, the sampling distribution
of the sample mean is centered at the population mean µ and is less
spread out than the population distribution. Here are the facts.
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 The
from a Normal Population
In one important case, there is a simple relationship between the two
distributions. If the population distribution is Normal, then so is the
sampling distribution of x. This is true no matter what the sample size is.
Sampling Distribution of a Sample Mean from a Normal Population
Suppose that a population is Normally distributed with mean m and standard deviation
s . Then the sampling distribution of x has the Normal distribution with mean m and
standard deviation s / n, provided that the 10% condition is met.
Sampling Applet
Sample Means
We have described the mean and standard deviation of the sampling
distribution of the sample mean x but not its shape. That's because the
shape of the distribution of x depends on the shape of the population
distribution.
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 Sampling
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Movie going students
Suppose that the number of movies viewed in the last year by high school
students has an average of 19.3 with a standard deviation of 15.8. Suppose
we take an SRS of 100 high school students and calculate the mean number
of movies viewed by the members of the sample.
Problem:
(a)What is the mean of the sampling distribution of
x?
(b) What is the standard deviation of the sampling distribution of
Check whether the 10% condition is satisfied.
s
15.8
sX =
=
= 1.58
n
100
x?
100 is less than 10% of the population of high
school student.
Example: Young Women’s Heights
Find the probability that a randomly selected young woman is
taller than 66.5 inches.
Sample Means
The height of young women follows a Normal distribution with mean
µ = 64.5 inches and standard deviation σ = 2.5 inches.
Let X = the height of a randomly selected young woman. X is N(64.5, 2.5)
66.5 - 64.5
P(X > 66.5) = P(Z > 0.80) =1- 0.7881 = 0.2119
z=
= 0.80
2.5
The probability of choosing a young woman at random whose height exceeds 66.5 inches
is about 0.21.
Find the probability that the mean height of an SRS of 10 young women
exceeds 66.5 inches.
For an SRS of 10 young women, the
sampling distribution of their sample
mean height will have a mean and
standard deviation
s
2.5
mx = m = 64.5 sx =
=
= 0.79
n
10
Since the population distribution is Normal,
the sampling distribution will follow an N(64.5,
0.79) distribution.
P(x > 66.5) = P(Z > 2.53)
66.5 - 64.5
z=
= 2.53
= 1- 0.9943 = 0.0057
0.79
It is very unlikely (less than a 1% chance) that
we would choose an SRS of 10 young women
whose average height exceeds 66.5 inches.
Central Limit Theorem
It is a remarkable fact that as the sample size increases, the distribution of
sample means changes its shape: it looks less like that of the population
and more like a Normal distribution! When the sample is large enough, the
distribution of sample means is very close to Normal, no matter what shape
the population distribution has, as long as the population has a finite
standard deviation.
Definition:
Draw an SRS of size n from any population with mean m and finite
standard deviation s . The central limit theorem (CLT) says that when n
is large, the sampling distribution of the sample mean x is approximately
Normal.
Note: How large a sample size n is needed for the sampling distribution to be
close to Normal depends on the shape of the population distribution. More
observations are required if the population distribution is far from Normal.
Sample Means
Most population distributions are not Normal. What is the shape of the
sampling distribution of sample means when the population distribution isn’t
Normal?
+
 The
Central Limit Theorem
Describe the shape of the sampling
distributions as n increases. What do
you notice?
Sample Means
Consider the strange population distribution from the Rice University
sampling distribution applet.
+
 The
Normal Condition for Sample Means
If the population distribution is Normal, then so is the
sampling distribution of x. This is true no matter what
the sample size n is.
If the population distribution is not Normal, the central
limit theorem tells us that the sampling distribution
of x will be approximately Normal in most cases if
n ³ 30.
Example: Servicing Air Conditioners
Your company will service an SRS of 70 air conditioners. You have budgeted
1.1 hours per unit. Will this be enough?
Sample Means
Based on service records from the past year, the time (in hours) that a
technician requires to complete preventative maintenance on an air
conditioner follows the distribution that is strongly right-skewed, and
whose most likely outcomes are close to 0. The mean time is µ = 1
hour and the standard deviation is σ = 1
Since the 10% condition is met (there are more than 10(70)=700 air conditioners in
the population), the sampling distribution of the mean time spent working on the 70
units has
s
1
sx =
=
= 0.12
mx = m =1
n
70
The sampling distribution of the mean time spent working is approximately N(1, 0.12)
since n = 70 ≥ 30.
We need to find P(mean time > 1.1 hours)
z=
1.1 -1
= 0.83
0.12
P(x > 1.1) = P(Z > 0.83)
= 1- 0.7967 = 0.2033
If you budget 1.1 hours per unit, there is a 20%
chance the technicians will not complete the
work within the budgeted time.
Buy Me Some Peanuts and Sample Means
Problem: At the P. Nutty Peanut Company, dry roasted, shelled peanuts are
placed in jars by a machine. The distribution of weights in the bottles is
approximately Normal, with a mean of 16.1 ounces and a standard deviation
of 0.15 ounces.
(a)Without doing any calculations, explain which outcome is more likely,
randomly selecting a single jar and finding the contents to weigh less than
16 ounces or randomly selecting 10 jars and finding the average contents to
weigh less than 16 ounces.
Averages are less variable than individual measurements, thus, it is
more likely that a single jar would weigh less than 16 ounces than the
average of 10 jars to be less than 16 ounces.
(b) Find the probability of each event described above.
Let X = weight of the contents of a randomly selected jar of peanuts.
X is N(16.1, 0.15). P(X < 16)
OR
z=
16 -16.1
= -.67
0.15
P(X<16)=0.2514
normalcdf(–100, 16, 16.1, 0.15) = 0.2525
(b) Find the probability of each event described above.
Let X = weight of the contents of a randomly selected jar of peanuts.
X is N(16.1, 0.15). P(X < 16)
OR
z=
16 -16.1
= -.67
0.15
P(X<16)=0.2514
normalcdf(–100, 16, 16.1, 0.15) = 0.2525
Let X = average weight of the contents of a random sample of 10 jars.
X is N(16.1, 0.15 ). Find P( X < 16)
10
z=
16 -16.1
= -2.13
0.047
P(z < 16) = 0.0166
OR
normalcdf(–100, 16, 16.1, 0.047) = 0.0166
Therefore, there is a better chance of choosing single jar containing less than 16 oz.
then there is of choosing an SRS of 10 jars with a mean less than 16 oz.
Mean Texts
Suppose that the number of texts sent during a typical day by a randomly selected
high school student follows a right-skewed distribution with a mean of 15 and a
standard deviation of 35. Assuming that students at your school are typical texters,
how likely is it that a random sample of 50 students will have sent more than a total
of 1000 texts in the last 24 hours?
State: What is the probability that the total number of texts in the last 24 hours is
greater than 1000 for a random sample of 50 high school students?
Plan: A total of 1000 texts among 50 students is the same as an average number of
texts of 1000/50 = 20
We want to find P( X > 20), where
n is large (50 > 30),
X = sample mean number of texts.
X is approximately N(15, 35/√50 ).
Do: P( X > 20) normalcdf (20, 9999,15, 4.95) = 0.1562.
Conclude: There is about a 16% chance that a random sample of 50 high school
students will send more than 1000 texts in a day.
+ Section 7.3
Sample Means
Summary
In this section, we learned that…


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When we want information about the population mean m for some variable,
we often take an SRS and use the sample mean x to estimate the unknown
parameter m. The sampling distribution of x describes how the statistic
varies in all possible samples of the same size from the population.
The mean of the sampling distribution is m, so that x is an
unbiased estimator of m.
The standard deviation of the sampling distribution of x is s / n for an SRS
of size n if the population has standard deviation s . This formula can be used
if the population is at least 10 times as large as the sample (10% condition).
+ Section 7.3
Sample Means
Summary
In this section, we learned that…
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Choose an SRS of size n from a population with mean m and standard
deviation s . If the population is Normal, then so is the sampling
distribution of the sample mean x. If the population distribtution is not Normal,
the central limit theorem (CLT) states that when n is large, the sampling
distribution of x is approximately Normal.
We can use a Normal distribution to calculate approximate probabilities for
events involving x whenever the Normal condition is met :
If the population distribution is Normal, so is the sampling distribution of x .
If n ³ 30, the CLT tells us that the sampling distribution of x will be
approximately Normal in most cases.