Last time, we talked about the types of samples
and introduced two
types of samples. One is called non-probability
samples, and the other one is probability samples.
And also, we have discussed two types of non
-probability, which are judgment and convenience.
For the product samples, we also produced four
types, random sample, systematic, stratified, and
clustered sampling. That was last Sunday. Let's
see the comparison between these sampling data. A
simple, random sample, systematic random sample,
first, for these two techniques. First of all,
they are simple to use because we just use the
random tables, random number tables, or by using
any statistical software. But the disadvantage of
this technique
So it might be this sample is not representative
of the entire population. So this is the mainly
disadvantage of this sampling technique. So it can
be used unless the population is not symmetric or
the population is not heterogeneous. I mean if the
population has the same characteristics, then we
can use simple or systematic sample. But if there
are big differences or big disturbances between
the items of the population, I mean between or
among the individuals. In this case, stratified
sampling is better than using a simple random
sample. Stratified samples ensure representation
of individuals across the entire population. If
you remember last time we said a IUG population
can be splitted according to gender, either males
or females, or can be splitted according to
students' levels. First level, second level, and
fourth level, and so on. The last type of sampling
was clusters. Cluster sampling is more cost
effective. Because in this case, you have to split
the population into many clusters, then you can
choose a random of these clusters. Also, it's less
efficient unless you use a large sample. For this
reason, it's more cost effective than using the
other sampling techniques. So, these techniques
are used based on the study you have. Sometimes
simple random sampling is fine, and you can go
ahead and use it. Most of the time, stratified
random sampling is much better. So, it depends on
the population you have underlying your study.
That was what we talked about last Sunday.
Now, suppose we design a questionnaire or survey.
You have to know, number one, what's the purpose
of the survey. In this case, you can determine the
frame of the population. Next,
survey
Is the survey based on a probability sample? If
the answer is yes, then go ahead and use one of
the non-probability sampling techniques, either
similar than some certified cluster or systematic.
Next, we have to distinguish between four types of
errors, at least now. One is called coverage
error. You have to ask yourself, is the frame
appropriate? I mean, frame appropriate means that
you have all the individual list, then you can
choose one of these. For example, suppose we
divide Gaza Strip into four governorates. North
Gaza, Gaza Middle Area, Khanon, and Rafah. So we
have five sections of five governments. In this
case, if you, so that's your frame. Now, if you
exclude one, for example, and that one is
important for you, but you exclude it for some
reasons, in this case, you will have coverage as
well, because you excluded one group out of five
and that group may be important for your study.
Next is called non-response error. Suppose I
attributed my questionnaire for 100 students and I
gave each one 30 minutes to answer the
questionnaire or to fill up the questionnaire, but
I didn't follow up. The response in this case, it
might be you will get something error, and that
error refers to non-responsive, so you have to
follow up, follow up. It means maybe sometimes you
need to clarify the question you have in your
questionnaire so that the respondent understand
what do you mean exactly by that question
otherwise if you don't follow up it means it may
be there is an error, and that error is called non
-response. The other type of error is called
measurement error, which is one of the most
important errors, and we have to avoid.
It's called measurement error. Good questions
elicit good responses. It means, suppose, for
example, my question is, I feel this candidate is
good for us. What do you think? It's my question.
I feel this candidate, candidate A, whatever he
is, is good for us. What do you think? For sure
there's abundant answer will be yes. I agree with
you. So that means you design the question in the
way that you will know they respond directly, that
he will answer yes or no depends on your design of
the question. So it means leading question. So
measurement error. So but if we have good
questions, just ask any question for the
respondent, and let him or let his answer based on
what exactly he thinks about it. So don't force
the respondent to answer the question in the
direction you want to be. Otherwise you will get
something called Measurement Error. Do you think?
Give me an example of Measurement Error.
Give me an example of Measurement Error. Just ask
a question in a way that the respondent will
answer, I mean, his answer will be the same as you
think about it.
Maybe I like coffee, do you like coffee or tea? So
maybe he will go with your answer. In this case
it's measurement. Another example.
Exactly.
So it means that if you design a question in the
way that you will get the same answer you think
about it, it means that you will have something
called measurement error. The last type is
sampling error. Sampling error always happens,
always exists. For example, suppose you are around
50 students in this class. Suppose I select
randomly 20 of you, and I am interested suppose in
your age. Maybe for this sample.
I will get an average of your age of 19 years
someone
select another sample from the same population
with the same size maybe
the average of your age is not equal to 19 years
maybe 19 years, 3 months
Someone else maybe also select the same number of
students, but the average of the class might be 20
years. So, the first one, second tier, each of them
has different sample statistics. I mean different
sample means. This difference or this error
actually is called sampling error and always
happens. So, now we have five types of errors. One
is called coverage error. In this case, you have
a problem with the frame. The other type is called
non-response error. It means you have a problem with
following up. Measurement error. It means you have
bad questionnaire design. The last type is called
sampling error, and this one always happens and
actually we would like to have this error. I mean
this sampling error as small as possible. So, these are
the steps you have to follow up when you design
the questionnaire.
So again, for these types of errors, the first one
coverage error, or selection bias. This type of
error exists if some groups are excluded from the
frame, and have no chance of being selected. That's
the first type of error, coverage error. So, it
means there is a problem on the population frame.
Non-response error bias. It means people who don't
respond may be different from those who do
respond. For example, suppose I have a sample of
tennis students.
And I got responses from number two, number five,
and number 10. So, I have these points of view for
these three students. Now, the other seven students
might be they have different opinions. So, the only
thing you have, the opinions of just the three,
and maybe the rest have different opinions, it
means in this case you will have something called
non-responsiveness. Or the same as we said before,
if your question is designed in a correct way. The
other type, sample error, variations from sample
to sample will always exist. As I mentioned here,
we select six samples, each one has different
sample mean. The other type, Measurement Error,
due to weakness in question design. So that's the
type of survey errors. So, one more time, average
error, it means you exclude a group or groups from
the frame. So, in this case, suppose I excluded
these from my frame. So I just select the sample
from all of these, except this portion, or these
two groups. Non-response error means you don't
have a follow-up on non-responses. Sampling error,
random sample gives different sample statistics.
So it means random differences from sample to
sample. Finally, measurement error, bad or leading
questions. This is one of the most important ones
that you have to avoid. So, that's the first part
of this chapter, assembling techniques. Do you
have any questions? Next, we'll talk about
assembling distributions. So far, up to this
point. I mean, at the end of chapter 6, we
discussed the probability, for example, of
computing X greater than, for example, 7. For
example, suppose X represents your score in
business statistics course.
And suppose we know that X is normally distributed
with a mean of 80, standard deviation of 10. My
question was, in chapter 6, what's the probability
that the student scores more than 70? Suppose we
select randomly one student, and the question is,
what's the probability that his score, so just for 197 00
score of the student. Now we have to use something
other called x bar. I'm interested in the average
of this. So x bar minus the mean of not x, x bar,
then divided by sigma x bar. So this is my new,
the score.
Here, there are three questions. Number one,
what's the shape of the distribution of X bar? So,
we are asking about the shape of the distribution.
It might be normal. If the entire population that
we select a sample from is normal, I mean if the
population is normally distributed, then you
select a random sample of that population, it
makes sense that the sample is also normal, so any
statistic is computed from that sample is also
normally distributed, so it makes sense. If the
population is normal, then the shape is also
normal. But if the population is unknown, you
don't have any information about the underlying
population, then you cannot say it's normal unless
you have certain condition that we'll talk about
maybe after 30 minutes. So, exactly, if the
population is normal, then the shape is also
normal, but otherwise, we have to think about it.
This is the first question. Now, there are two
unknowns in this equation. We have to know the
mean, Or x bar, so the mean of x bar is not given,
the mean means the center. So the second question,
what's the center of the distribution? In this
case, the mean of x bar. So we are looking at
what's the mean of x bar. The third question is
sigma x bar is also unknown, spread. Now shape,
center, spread, these are characteristics, these
characteristics in this case sampling
distribution, exactly which is called sampling
distribution. So by sampling distribution we mean
that, by sampling distribution, we mean that you
have to know the center of distribution, I mean
the mean of the statistic you are interested in.
Second, the spread or the variability of the
sample statistic also you are interested in. In
addition to that, you have to know the shape of
the statistic. So three things we have to know,
center, spread and shape. So that's what we'll
talk about now. So now sampling distribution is a
distribution of all of the possible values of a
sample statistic. This sample statistic could be
sample mean, could be sample variance, could be
sample proportion, because any population has
mainly three characteristics, mean, standard
deviation, and proportion.
So again, a sampling distribution is a
distribution of all of the possible values of a
sample statistic or a given size sample selected
from a population. For example, suppose you sample
50 students from your college regarding their mean
GPA. GPA means Graduate Point Average. Now, if you
obtain many different samples of size 50, you will
compute a different mean for each sample. As I
mentioned here, I select a sample the same sizes,
but we obtain different sample statistics, I mean
different sample means. We are interested in the
distribution of all potential mean GBA We might
calculate for any given sample of 50 students. So
let's focus into these values. So we have again a
random sample of 50 sample means. So we have 1, 2,
3, 4, 5, maybe 50, 6, whatever we have. So select
a random sample of size 20. Maybe we repeat this
sample 10 times. So we end with 10. different
values of the simple means. Now we have new ten
means. Now the question is, what's the center of
these values, I mean for the means? What's the
spread of the means? And what's the shape of the
means? So these are the mainly three questions.
For example, let's get just simple example and
that we have only population of size 4. In the
real life, the population size is much bigger than
4, but just for illustration.
Because size 4, I mean if the population is 4,
it's a small population. So we can take all the
values and find the mean and standard deviation.
But in reality, we have more than that. So this
one just for as example. So let's suppose that we
have a population of size 4. So n equals 4.
And we are interested in the ages. And suppose the
values of X, X again represents H,
and the values we have.
So these are the four values we have.
Now simple calculation will
give you the mean, the population mean.
Just add these values and divide by the operation
size, we'll get 21 years. And sigma, as we
mentioned in chapter three, square root of this
quantity will give 2.236
years. So simple calculation will give these
results. Now if you look at distribution of these
values, Because as I mentioned, we are looking for
center, spread, and shape. The center is 21 of the
exact population. The variation is around 2.2.
Now, the shape of distribution. Now, 18 represents
once.
I mean, we have only one 18, so 18 divided one
time over 425. 20% represent also 25%, the same as
for 22 or 24. In this case, we have something
called uniform distribution. In this case, the
proportions are the same. So, the mean, not
normal, it's uniform distribution. The mean is 21,
standard deviation is 2.236, and the distribution
is uniform. Okay, so that's center, spread and
shape of the true population we have. Now suppose
for example,
we select a random sample of size 2 from this
population.
So we select a sample of size 2. We have 18, 20,
22, 24 years. We have four students, for example.
And we select a sample of size two. So the first
one could be 18 and 18, 18 and 20, 18 and 22. So
we have 16 different samples. So number of samples
in this case is 16. Imagine that we have five. I
mean the population size is 5 and so on. So the
rule is number
of samples in this case and the volume is million.
Because we have four, four squared is sixteen,
that's all. 5 squared, 25, and so on. Now, we have
16 different samples. For sure, we will have
different sample means. Now, for the first sample,
18, 18, the average is also 18. The next one, 18,
20, the average is 19.
20, 18, 24, the average is 21, and so on. So now
we have 16 sample means. Now this is my new
values. It's my sample. This sample has different
sample means. Now let's take these values and
compute average, sigma, and the shape of the
distribution. So again, we have a population of
size 4, we select a random cell bone. of size 2
from that population, we end with 16 random
samples, and they have different sample means.
Might be two of them are the same. I mean, we have
18 just repeated once, but 19 repeated twice, 23
times, 24 times, and so on. 22 three times, 23
twice, 24 once. So it depends on The sample means
you have. So we have actually different samples.
For example, let's look at 24 and 22. What's the
average of these two values? N divided by 2 will
give 22.
So again, we have 16 sample means.
Now look first at the shape of the distribution.
18, as I mentioned, repeated once. So 1 over 16.
19 twice. 23 times. 1 four times. 22 three times.
then twice then once now the distribution was
uniform remember now it becomes normal
distribution so the first one x1 is normal
distribution so it has normal distribution so
again the shape of x1 looks like normal
distribution we need to compute the center of X
bar, the mean of X bar. We have to add the values
of X bar, the sample mean, then divide by the
total number of size, which is 16. So in this
case, we got 21, which is similar to the one for
the entire population. So this is the first
unknown parameter. The mu of x bar is the same as
the population mean mu. The second one, the split
sigma of x bar by using the same equation
we have, sum of x bar in this case minus the mean
of x bar squared, then divide this quantity by the
capital I which is 16 in this case. So we will end
with 1.58.
Now let's compare population standard deviation
and the sample standard deviation. First of all,
you see that these two values are not the same.
The population standard deviation was 2.2, around
2.2. But for the sample, for the sample mean, it's
1.58, so that means sigma of X bar is smaller than
sigma of X.
It means exactly, the variation of X bar is always
smaller than the variation of X, always.
So here is the comparison. The distribution was
uniform. It's no longer uniform. It looks like a
bell shape. The mean of X is 21, which is the same
as the mean of X bar. But the standard deviation
of the population is larger than the standard
deviation of the sample mean or the average.
Different samples of the same sample size from the
same population will yield different sample means.
We know that. If we have a population and from
that population, so we have this big population,
from this population suppose we selected 10
samples, sample 1 with size 50.
Another sample, sample 2 with the same size. All
the way, suppose we select 10 samples, sample 10, 412 00:33:
smaller than sigma of the standard deviation of
normalization. Now if you look at the relationship
between the standard error of X bar and the sample
size, we'll see that as the sample size increases,
sigma of X bar decreases. So if we have large
sample size, I mean instead of selecting a random
sample of size 2, if you select a random sample of
size 3 for example, you will get sigma of X bar
less than 1.58. So note that standard error of the
mean decreases as the sample size goes up. So as n
increases, sigma of x bar goes down. So there is
an inverse relationship between the standard error of
the mean and the sample size. So now we answered
the three questions. The shape looks like a bell
shape. If we select our sample from a normal
population with a mean equal to the population mean
and standard deviation of the standard error equals
sigma over the square root of n. So now, let's talk
about the sampling distribution of the sample mean if
the population is normal. So now, my population is
normally distributed, and we are interested in the
sampling distribution of the sample mean of X bar.
If the population is normally distributed with
mean mu and standard deviation sigma, in this
case, the sampling distribution of X bar is also
normally distributed, so this is the shape. With
the mean of X bar equals mu and sigma of X bar equals
sigma over the square root of n. So again, if we sample from a normal
population, I mean my sampling technique, I select
a random sample from a normal population. Then if
we are interested in the standard distribution of
X bar, then that distribution is normally
distributed with a mean equal to mu and standard
deviation sigma over mu. So that's the shape. It's
normal. The mean is the same as the population
mean, and the standard deviation of x bar equals
sigma over the square root of n. So now let's go back to the z
-score we discussed before. If you remember, I
mentioned before
that the z-score, generally speaking, is X minus the mean
of X divided by sigma X.
And we know that Z has a standard normal
distribution with a mean of zero and a variance of one. In
this case, we are looking for the sampling
-distribution of X bar. So Z equals X bar. minus
the mean of x bar divided by sigma of x bar. So
the same equation, but different statistics. In the
first one, we have x, for example, which represents the
score. Here, my sample statistic is the sample
mean, which represents the average of the scores.
So x bar, minus its mean, I mean the mean of x
bar, divided by its standard error. So x bar minus
the mean of x bar divided by sigma of x bar. By
using that mu of x bar equals mu, and sigma of x
bar equals sigma over the square root of n, we will end with
this equation z square.
So this equation will be used instead of using the
previous one. So z square equals sigma, I'm sorry,
z equals x bar minus the mean divided by sigma
bar, where x bar is the sample mean, mu is the
population mean, sigma is the population standard
deviation, and n is the sample size. So that's the
difference between chapter six,
and that one we have only x minus y by sigma. Here
we are interested in x bar minus the mean of x bar
which is mu. And sigma of x bar equals sigma over the square root of n.
Now when we are saying that mu of x bar equals mu,
That means the expected value of
the sample mean equals the population mean. When
we are saying mean of X bar equals mu, it means
the expected value of X bar equals mu. In other
words, the expectation of X bar equals mu. If this
happens, we say that X bar is an unbiased
estimator
of
mu. So this is a new definition, an unbiased
estimator X bar is called an unbiased estimator if
this condition is satisfied. I mean, if the mean
of X bar or if the expected value of X bar equals
the population mean, in this case, we say that X
bar is a good estimator of Mu. Because on average,
The expected value of X bar equals the population
mean, so in this case, X bar is a good estimator of
Mu. Now if you compare the two distributions,
a normal distribution here with the population mean Mu
and a standard deviation for example sigma.
That's for the scores, the scores. Now instead of
the scores above, we have x bar, the sample mean.
Again, the mean of x bar is the same as the
population mean. Both means are the same, mu of x
bar equals mu. But if you look at the spread of
the second distribution, it is more than the
other one. So that's the comparison between the
two populations.
So again, to compare or to figure out the
relationship between sigma of x bar and the sample
size. Suppose we have this blue normal
distribution with a sample size of say 10 or 30, for
example.
As n gets bigger and bigger, sigma of x bar
becomes smaller and smaller. If you look at the
red one, maybe if the red one has n equal to 100,
we'll get this spread. But for the other one, we
have a larger spread. So as n increases, sigma of x
bar decreases. So this, the blue one for a smaller
sample size. The red one for a larger sample size.
So again, as n increases, sigma of x bar goes down
four degrees.
Next, let's use this fact to
figure out an interval for the sample mean with 90
% confidence and suppose the population we have is
normal with a mean of 368 and sigma of 15 and suppose
we select a random sample of a size of 25 and the question
is find symmetrically distributed interval around
the mean that will include 95% of the sample means
when mu equals 368, sigma is 15, and your sample
size is 25. So in this case, we are looking for
the
estimation of the sample mean.
And we have this information,
Sigma is 15 and N is 25.
The problem mentioned there, we have a symmetric
distribution and this area is 95% bisymmetric and
we have only 5% out. So that means half to the
right and half to the left.
And let's see how we can compute these two values.
The problem says that the average is 368
for this data and the standard deviation sigma of
15. He asked about what are the values of x bar. I
mean, we have to find the interval of x bar. Let's
see. If you remember last time, z score was x
minus mu divided by sigma. But now we have x bar.
So your z score should be x bar minus mu divided by
sigma over the square root of n. Now cross multiplication, you
will get x bar minus mu equals z sigma over the square root
of n. That means x bar equals mu plus z sigma over
the square root of n. Exactly the same equation we got in
chapter six, but there, in that one, we have x
equals mu plus z sigma. Now we have x bar equals
mu plus z sigma over the square root of n, because we have
different statistics. It's x bar instead of x. Now
we are looking for these two values. Now let's
compute z-score.
The z-score for this point, which has an area of 2.5%
below it, is the same as the z-score, but in the
opposite direction. If you remember, we got this
value, 1.96.
So my z-score is negative 1.96 to the left. and 1
.9621 so now my x bar in the lower limit in this
side on the left side equals mu which is 368 minus
1.96 times sigma which is 15 divide by the square root of 25.
So that's the value of the sample mean in the
lower limit, or lower bound. On the other hand,
expand our limit to the other hand equals 316 plus 1.96
sigma over the square root of n. Simple calculation will give this
result.
The first X bar for the lower limit is 362.12, the
other is 373.1.
So again for this data, for this example, the mean
was, the population mean was 368, the population
has a standard deviation of 15, we select a random
sample of size 25, Then we end with this result
that 95% of all sample means of sample size 25 are
between these two values. It means that we have
this big population and this population is
symmetric, it's normal. And we know that The mean of
this population is 368 with a sigma of 15.
We select from this population many samples. Each
one has a size of 25.
Suppose, for example, we select 100 samples, 100
random samples. So we end with different sample
means.
So we have 100 new sample means. In this case, you
can say that 95 out of these, 95 out of 100, it
means 95, one of these sample means. have values
between 362.12 and 373.5. And what's remaining?
Just five of these sample means would be out of
this interval either below 362 or above the upper
limit. So you are 95% sure that
the sample mean lies between these two points.
So, 5% of the sample means will be out. Make
sense? Imagine that I have selected 200 samples.
Now, how many X bar will be between these two
values? 95% of these 200. So how many 95%? How
many means in this case?
95% out of 200 is 190.
190. Just multiply. 95 multiplies by 200. It will
give you 190.
values between 362
larger and larger, or gets larger and larger, then
the standard distribution of X bar is
approximately normal in this. Again, look at the
blue curve. Now, this one looks like skewed
distribution to the right.
Now, as the sample gets large enough, then it
becomes normal. So, the sample distribution
becomes almost normal regardless of the shape of
the population. I mean if you sample from unknown
population, and that one has either right skewed
or left skewed, if the sample size is large, then
the sampling distribution of X bar becomes almost
normal distribution regardless of the… so that’s
the central limit theorem. So again, if the
population is not normal, The condition is only
you have to select a large sample. In this case,
the central tendency mu of X bar is same as mu.
The variation is also sigma over root N.
So again, standard distribution of X bar becomes
normal as N. The theorem again says If we select a
random sample from unknown population, then the
standard distribution of X part is approximately
normal as long as N gets large enough. Now the
question is how large is large enough?
There are two cases, or actually three cases. For
most distributions, if you don’t know the exact
shape, n above 30 is enough to use or to apply
that theorem. So if n is greater than 30, it will
give a standard distribution that is nearly
normal. So if my n is large, it means above 30, or
30 and above this. For fairly symmetric
distribution, I mean for nearly symmetric
distribution, the distribution is not exactly
normal, but approximately normal. In this case, N
to be large enough if it is above 15. So, N
greater than 15 will usually have same
distribution as almost normal.
For normal population, as we mentioned, of
distributions, the semantic distribution of the
mean is always.
Okay, so again, there are three cases. For most
distributions, N to be large, above 30. In this
case, the distribution is nearly normal. For
fairly symmetric distributions, N above 15 gives
almost normal distribution. But if the population
by itself is normally distributed, always the
sample mean is normally distributed. So that’s the
three cases.
Now for this example, suppose we have a
population. It means we don’t know the
distribution of that population. And that
population has mean of 8. Standard deviation of 3.
And suppose a random sample of size 36 is
selected. In this case, the population is not
normal. It says A population, so you don’t know
the exact distribution. But N is large. It’s above
30, so you can apply the central limit theorem.
Now we ask about what’s the probability that a
sample means. is between what’s the probability
that the same element is between these two values.
Now, the difference between this lecture and the
previous ones was, here we are interested in the
exponent of X. Now, even if the population is not
normally distributed, the central limit theorem
can be abused because N is large enough. So now,
the mean of X bar equals mu, which is eight, and
sigma of X bar equals sigma over root N, which is
three over square root of 36, which is one-half.
So now, the probability of X bar greater than 7.8,
smaller than 8.2, Subtracting U, then divide by
sigma over root N from both sides, so 7.8 minus 8
divided by sigma over root N. Here we have 8.2
minus 8 divided by sigma over root N. I will end
with Z between minus 0.4 and 0.4. Now, up to this
step, it’s in U, for chapter 7. Now, Z between
minus 0.4 up to 0.4, you have to go back. And use
the table in chapter 6, you will end with this
result. So the only difference here, you have to
use sigma over root N.