01.11.2022, 12:01
Lesson activity: Sampling distributions: Attempt review
Started on Thursday, 27 October 2022, 1:10 PM
State Finished
Completed on Tuesday, 1 November 2022, 12:01 PM
Time taken 4 days 22 hours
Marks 64.00/64.00
Grade 10.00 out of 10.00 (100%)
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Lesson activity: Sampling distributions: Attempt review
Question 1
Correct
Mark 5.00 out of 5.00
Sampling or sample-to-sample variability
You know that a good sample is
representative of the population
can be used to draw conclusions about the population as a whole.

and therefore it

It is important to note that sampling variability occurs and that it is an expected phenomenon.
Small differences between samples are to be expected. Since each subject is an individual,
responses will vary. Watch a video to answer a few questions about the concept of sampling
variability:
A statistic is a characteristic of the population:
TRUE
FALSE You are right! A statistic is any quantity computed from a sample whereas a
quantity computed from entire population is referred to as a population parameter or population
characteristic!
Mark 1.00 out of 1.00
In the video example, each of the sample means of turtle weights represents a
statistic
statistic!
You are right! Each of the sample means in the video example represents a
population parameter
Mark 1.00 out of 1.00
In the video example, all the sample means of turtle weights
are identical
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Lesson activity: Sampling distributions: Attempt review
vary a little You are right! Indeed, the sample means in the video example vary a little,
which is expected
vary a lot
Mark 1.00 out of 1.00
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Lesson activity: Sampling distributions: Attempt review
Question 2
Correct
Mark 18.00 out of 18.00
The following exercise will help you to
start investigating the
properties of the sampling distribution
of sample means:
The applet below uses a uniform probability distribution of X where the random variable X has
an equal probability of taking a value between 10 and 30. It's like sampling with replacement
from set of integers {10,11,12,13,..,28,29,30} when simulating a random sample, after each
successive number is selected for this sample, the number is “replaced” back into the
population {10,11,12,13,..,28,29,30} , so the same number may be selected again.
The actual mean, μ, of this population distribution of values {10,11,12,13,..,28,29,30} is equal
to
20 You are right! It is equal to [10+11+..+30]/21=20 or just [10+30]/2=20 for discrete
uniform probability distribution
15
21
Mark 1.00 out of 1.00
What you're observing in the applet below:
100 random samples are simulated from this distribution. The means of each such random
sample, , is computed and frequency distribution of all 100 values of sample means is
constructed.
First, check the box at n=1 to display the frequency distributions of sample means for
samples of size n=1. When each sample consists of just 1 value, the mean of each of the
100 random samples drawn from {10,11,12,13,..,28,29,30} is equal to this same single value
in this sample. That is why the red
 -colored frequency histogram of one hundred
1- sample means is identical to the
I don't know
uniform probability distribution of X where random variable X is equally likely to have any
value between 10 and 30. Right!
Mark 1.00 out of 1.00
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Lesson activity: Sampling distributions: Attempt review
Next, check the box n=5 and you will have simulated 100 random samples of size 5 from the
population distribution of {10,11,12,13,..,28,29,30}. The values of sample means
automatically calculated for each of the 100 samples each consisting of
are
5
values randomly drawn from {10,11,12,13,..,28,29,30}, and these 100 sample means, , are
then used to construct the yellow
 -colored frequency histogram displayed in the

applet above.
You would expect the sample means
of each of these 100 samples to be
not necessarily equal, but rather close, to the actual population mean you have computed
above You are right! The resulting yellow histogram shows there is a lot of sample-tosample variability in the value of sample means. For some samples, sample mean is around
15, and for other samples it is around 24. A sample of size 5 from the population won’t
always provide precise information about the true mean in the population, μ, which
actually is equal to 20.
exactly equal to the actual population mean you have computed above
Mark 1.00 out of 1.00
What if a larger sample is selected?
To investigate the effect of sample size on the behavior of sample means, select 100 samples of size
10 by checking n=10 box. The
mean values
green

-colored frequency histogram of the resulting sample
is
centered around 20, the actual mean of population distribution
Correct!
not centered around 20, the actual mean of population distribution
Mark 1.00 out of 1.00
and
spreads
more
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less
Lesson activity: Sampling distributions: Attempt review
Correct!
Mark 1.00 out of 1.00
around the population mean value of μ compared to red and yellow distributions of sample means
for smaller sample sizes n.
Check the remaining boxed of sample size n=20, n =30, n=60.
You notice that the resulting histograms tend to be
centered around 20, the actual mean of population distribution
Correct!
not centered around 20, the actual mean of population distribution
Mark 1.00 out of 1.00
which tells you that sample mean
an unbiased
indeed is
You are right!
an unprejudiced
a biased
Mark 1.00 out of 1.00
estimator of a population mean, μ,
and that as sample size n increases, the sampling distribution of sample means spreads
more
less
Correct!
Mark 1.00 out of 1.00
around the population mean value, μ. Which means that as sample size n increases, sample mean
becomes a more
 accurate estimate of population mean, μ.
For the largest sample size n=60, the frequency histogram of the resulting sample mean values
 -colored and most
values in this histogram are relatively
ocean-wave
close to
is
Correct!
far from
Mark 1.00 out of 1.00
the actual or true population mean value of μ equal to
20

whereas for smaller sample sizes most
farther away from Correct!
values are relatively
closer to
Mark 1.00 out of 1.00
the actual/true population mean value of μ.
This lesson activity also introduces/reminds you the Central Limit Theorem which tells that when sample
size n contains enough observations observations randomly drawn from a population, no
matter what the shape of the population distribution is, then the sampling distribution of
sample means is well approximated by a normal curve.
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Lesson activity: Sampling distributions: Attempt review
Take a look again at the distributions of sample means
from 100 samples of size n=30
and n=60. Do they look normal (=bell-shaped and symmetric) to you?
Kind of..
Right!
They look absolutely abnormal!
Mark 1.00 out of 1.00
Do not confuse the sampling distribution with the sample distribution. The sampling distribution considers the
distribution of sample statistics (e.g. sample mean), whereas the sample distribution is basically the distribution
of a particular sample taken from the population.
Does Introduction to the lesson activity make sense? There is no wrong answer here.
Yes! Got it!
No
Not quite
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Lesson activity: Sampling distributions: Attempt review
Question 3
Correct
Mark 17.00 out of 17.00
In a random sample of 100 people, we find that 20 are left-handed. What can we say about the proportion of
the population who are left handed?
In the previous lesson activities, we discussed the idea of a random variable - a single data-point drawn from
a probability distribution described by parameters (like parameters p and n for binomial  or μ and σ for
normal

). But we are seldom interested in just one data-point - we generally have a mass of data
which we summarize by determining means, medians and other statistics. The fundamental step we take
in inferential statistics is to consider those statistics as themselves being random variables, drawn from their
own distributions.
This is a big advance, the one that has challenged generations of statisticians who have tried to work out
what distributions we should assume these statistics are drawn from. Maybe in the end of the course I will have
time to introduce you into simulation-based bootstrap methods. For example, the question posed at the
beginning of this part could be answered by taking our observed data of 20 left-handed and 80 righthanded individuals, and repeatedly resampling one hundred observations from this data set, with replacement,
and looking at the distribution of the observed proportion of the left-handed people.
But these simulations are clumsy and time consuming, especially with large data sets, and in
more complex circumstances it is not straightforward to work out what should be simulated. In
contrast, formulae derived from probability theory provide both insight and convenience, and
always lead to the same answer since they don't depend on a particular simulation. But the flip
side is that this theory relies on assumption, and we should be careful not to be deluded by the
impressive algebra into accepting unjustified conclusions.
Suppose we draw samples of different sizes from a population containing exactly 20% left- and 80% righthanded people (so, true population probability of success of drawing a left-handed person, p, is equal to
0.2

.
Calculate the probability of observing different possible proportions of left-handers (=possible values of
sample proportions
of left-handers resulting from different random samples).
Note: Of course, this is the wrong way round - we want to use the
unknown
known
 sample to learn about the
 population - but we can only get to this conclusion by first exploring how a known population gives
rise to different samples.
The simplest case is a sample of 1 (n =
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Lesson activity: Sampling distributions: Attempt review
1
) when the observed proportion must be either 0 (right-hander) or 1 (left-hander) and these events occur
with probability of 0.8 and

0.2
, respectively. Use the applet below set "The number of trials: n =1"; set "the probability of success at each
trial: p =0.2". You will be able to see the resulting sampling distribution for the sample proportions of lefthanded people in all possible random samples of size n=1. Do not confuse the resulting probability
histogram with sample distribution: Each particular sample of n =1 people taken from the population will
have either x=0 (~a single person in your sample is right-handed) or x=1 (~a single person in your sample
is left-handed)!

Then take two individuals at random (n =
2

). Then the probability of 0 left-handers will be
0.64

, the probability of 1 left-hander will be
0.32

handers), and the probability of 2 left-handers will be
0.04
. (Hint: to display corresponding probabilities, set relevant value of x corresponding to the number/count
of left-handers equal to 0, then 1, and then 2, correspondingly)
Similarly, you can use the probability theory to work out the probability distribution for the observed
numbers of left-handers in the 5-, 10-, 50- and 100-person samples.

Consider a population of 1000 people, 200 of whom are left-handed and the rest are right-handed. If you select
100 people at random but with replacement, then
px(1-p)n-x=
the probability that 20 are left-handed and 80 are right-handed is p(x=20)=
0.2
 20(10.2
 )100-20
≈
0.0993

. (Go through the previous lesson activity to refresh the binomial distribution formula)
If you select without replacement, then the probability is *
)/
≈
0.1047

This distributions are already familiar to you. They are based on what is known as
binomial

distribution,
and can tell you the probability, for example, of getting at least 30%left-handed people if you sample 100.
The mean of the random variable is also known as its expectation, and in all these samples we expect a
proportion of 0.2 or 20%: all of the distributions you have simulated above will have n*p as their
corresponding mean. The standard deviation for each is equal to [n*p*(1-p)]0.5 and is usually referred to as
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Lesson activity: Sampling distributions: Attempt review
standard error to distinguish it from the standard deviation of the population distribution from which it
derives.
You may notice that as sample size n increases, the probability distributions tend to be
more

symmetric, regular, and normal in shape.
Do not confuse the sampling distribution with the sample distribution:
The sampling distribution for a proportion of left-handed people refers to the distribution of sample statistics
(e.g. sample proportion)} whereas the sample distribution is basically the distribution of a particular sample
taken from the population.
Particularly, the sampling distribution of sample proportions
refers to the distribution of sample
proportions from all possible random samples of particular size n.
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Lesson activity: Sampling distributions: Attempt review
Question 4
Correct
Mark 24.00 out of 24.00
Sampling Distributions of a Sample Proportion
In this part of the lesson activity, you will take a quick and informal look at the role that sampling distributions
play in learning about population characteristics. The two examples here show how a sampling distribution
provides important information in a estimation setting and in a hypothesis testing setting.
In an estimation situation, you need to understand sampling variability to assess how close an estimate is likely
to be to the actual value of the corresponding population characteristic. Published research reports often
include statements about a margin of error. This margin of error is based on an assessment of sampling
variability as described by a sampling distribution. This is illustrated in the following example.
Will Cash Become a Thing of the Past?
The article “Most Americans Foresee Death of Cash in Their Lifetime” used data from a random sample of 1024
adults to estimate the proportion of all adults in the United States who think it is likely that within their
lifetime, the United States will become a cashless society with all purchases being made by credit card, debit
card, or some other form of electronic payment. Of the 1024 people surveyed, 635 indicated that they thought
this was likely, resulting in a sample proportion of
=635/1024 =0.62.
The population proportion who think that a cashless society is likely probably isn’t exactly 0.62. How accurate is
this estimate likely to be?
To answer this question, you can use what you know about the sampling distribution of
for random samples
of size n=1024. Using the general rules described in the video above, you know three things about the
sampling distribution of
:
Rule 1. The sampling distribution of sample proportion
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is centered at
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Lesson activity: Sampling distributions: Attempt review
I don't know
zero
the actual population proportion of all adults in the U.S. who think it is likely that within their lifetime,
society will become cashless, p
Correct!
Mark 1.00 out of 1.00
This means that the
values from random samples cluster around
I don't know
zero
the actual value of population proportion, p which is not known
Correct!
Mark 1.00 out of 1.00
and this is only true for
convenience samples
random samples
Correct!
Mark 1.00 out of 1.00
. The study above says that this sample of 1024 adults was a
convenience sample
random sample
Correct!
Mark 1.00 out of 1.00
Rule 2. The standard deviation of sample proportion
describes how tightly the
values from different
possible random samples of size 1024 spread out around p, the actual population proportion of all adults in
the U.S. who think it is likely that within their lifetime, society will become cashless, and the
standard deviation( )=[ *(1- )/n]0.5= [0.62*(10.62

)/
1024

]0.5 =
0.015

.
Rule 3. Because: n*
=1024*
0.62

=
634.88

expected successes in the sample of 1024 adults and which is
more than OK
less than
equal to
Mark 1.00 out of 1.00
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Lesson activity: Sampling distributions: Attempt review
10
AND
n*(1-
)=1024*[1-
0.62
]=

389.12
expected failures in the sample of 1024 adults and which is

less than
equal to
more than
OK
Mark 1.00 out of 1.00
10
THEN the sampling distribution of
may be
OK
may not be
Mark 1.00 out of 1.00
approximated by normal distribution.
By using the information and what you know about normal distributions, you can now get a sense of the
accuracy of the estimate
= 0.62:
For any variable described by a normal distribution, about
95
% of the values are within 2 standard deviations of the center (Hint: you may use the applet at the bottom
of the page to refresh your memory).
Since the sampling distribution of
is approximately

normal
Correct!
abnormal
paranormal
Mark 1.00 out of 1.00
and is centered at
zero
the actual population proportion of all adults in the U.S. who think it is likely that within
their lifetime, society will become cashless, p Correct!
Mark 1.00 out of 1.00
you now know that about
95
% of the values of all possible random samples will produce a sample proportion that is
within 2 standard deviations( )= 2*

0.01516

=
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Lesson activity: Sampling distributions: Attempt review
0.03032
of the actual value of the population proportion p.
So, a margin of error =2 standard deviations( ) =

0.03032
can be reported. This tells you that the sample estimate

= 0.62 is likely to be within
0.03032
of the actual proportion of U.S. adults who think that the United States will become a cashless society in
their lifetime.

Alternatively, you could say that plausible values for the actual proportion of U.S. adults, p, who think so are
those between:
0.62-margin of error =
0.59
and

0.62+margin of error =
0.65

.
This example shows that the sampling distribution of the sample proportion is the key in helping you assess the
accuracy of the estimate of true population proportion, p.
Use the applet by dragging the values of lower and upper boundaries of corresponding intervals along the
lines:
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