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Chapters 10 Two Sample Problems: Inference about a Population Mean
When we have to samples (treatments) and we want to test to see if there is a statistically
significant difference between the two groups this is the procedure we use. Many of the
techniques and statistics that we have used in previous chapters will be used again. So it should
seem very familiar how we go about studying and analyzing this type of procedure.
Assumptions When Comparing Two Sample Means
a. When looking at two samples we have to assume independence. If they were not independent
it would not make sense to think they are different from one another (i.e. if dependent you are
saying they ARE similar…so why test differences?).
b. The populations must be normally distributed with either known or unknown means or
variances. This implies we will be using a z-stat & t-stat.
**although the book goes through the procedure with unknown mean and variance…we should
note that if they were known then instead of using a t-stat we would simply use our Z. In so the
method would be exactly the same but, we would use the Z. I am going to outline both.
A. Hypothesis Testing: Differences between means
Recall from before that we had essentially 3 types of hypotheses. Since we are testing to see
whether or not there is a difference between the populations we get the following types of
hypotheses:
a.
Ho: u1 ≥ u2 or u1 - u2 ≥ 0
Ha: u1< u2 or u1 - u2 < 0
**to get the other one tailed test simply switch u1 and u2 around
b.
Ho: u1 = u2 or u1 - u2 = 0
Ha: u1 ≠ u2 u1 - u2≠ 0
So what we are testing is to see if one mean is bigger than the other (i.e. the one tailed tests) or to
simply see if there is a significant difference between the two (either higher or lower  two
tailed test.
Question: So when might this be used?
Answer: What if we wanted to see if there was a significant difference between the test scores of
on-line Business Statistics students vs. In-Class Students.
If we didn’t suspect one type of result or the other we would have hypotheses of:
1) Ho: u of Test Scoresin-class = u of Test Scoreson-line
or u1 - u2 = 0
**this implies that there is no difference
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2) Ha: u of Test Scoresin-class ≠u of Test Scoreson-line
or u1 - u2≠ 0
The method of analysis is exactly the same except know when find our Z-stat or t-stat. So the
first thing we need to do (as before) is find out if we have known σ’s or unknown values and
have to use the sample values (i.e. the s’s)
Performing Hypothesis Tests on Difference Between Means: 3 Conditions
1. σ – known
In this case we simply use our z-stat as before as follows:
Z=
where SE
and Do is the difference that is hypothesized. Note that
many time we simply have Do = 0. Generically in can be thought of as the difference between
)
b. Example: If we are given the avg height of males in one class of 50 students is 67 inches with
a variance of 9 inches and the avg height of 36 males in a second class is 66 inches with a
variance of 6 inches, perform a hypothesis test to see if the mean height of males is different
between the two classes. Use α = 0.05.
Since we have that n > 30 for each of the samples we may use the Z-table to get out test statistic.
i) Hypotheses:
Ho: uclass1 = uclass2
Ha: uclass1 ≠ uclass2
ii) Test Statistic:
Z=
=
≈ 1.70
SE
=
= 0.589
iii) Analysis of Critical Region: Since our alpha is 0.05 and we have a Z-stat that is two-tailed
our critical values are +/- 1.96. So we will reject Ho only if our test statistics lies farther out than
either of these two values.
iv) Conclusion: Since our test statistics = 1.70 < 1.96, we fail to reject Ho and conclude there is
not enough evidence to suggest that the mean height of males is different between the two
classes.
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2. σ – unknown
In this case we use our t-stat as before but there are two cases we must examine here. We must
decide if the sample variances are roughly equal or not (i.e. if
Case 1: We have σ-unknown and we think that
(i.e. equal variance assumption)
In this case we use the following statistic called the pooling estimate for variances
= pooled estimate
We then note that SE in this case is
and our test stat is then simply
and in this case the df = n1+n2-2
example: Suppose we are given the following values and told that our variances can be assumed
equal. If we want to test whether or not there is a difference in the samples at the 95% level of
confidence, here is how we do it.
Variables Sample 1 Sample 2
15
16
S
4
4.6
n
24
16
Step 1: Set up Hypotheses: So we are testing to simply see if there is a difference so
i) Hypotheses:
Ho: usample 1 = usample 2
Ha: usample 1 ≠ usample 2
ii) Find the test statistic, which is t =
One of the first things we want to do is find the SE or the standard deviation since our
calculation becomes much simpler then.
So recall that
with
So
variance.
; What this gives us is an approximate value for the combined
We then find SE =
; This is the standard error
So now we find that our t-stat =
iii) Compare to Critical Value: Since we want to be 95% confident we want the t-stat that gives
us the following:
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These are your t-values that put 2.5%
in each tail with
dof =
= 3. These values
are ± 2.042
Note: There is no value for t with
dof = 38 so always go to the lower
Two Sided Test
-you need to note the level of
confidence and α to find the
appropriate t-values for your CI.
value…in this case 30.
t
-tα/2 =-2.042
μ
tα/2 =2.042
iv) Conclusion: Since our test stat is not in the critical/rejection region [i.e. -1.50 > -2.042 So
it does not lie in the tail], which is always the tail, we can conclude that there is no statistical
difference between the sample means with 95% confidence.
Case 2: We have σ-unknown and we think that
then use the following information:
t-stat =
(or unequal variances assumed) and we
where SE
This method is a little simpler in calculating the test statistic, but where this becomes a bit more
difficult is find the appropriate degrees of freedom. The exact same method is used but the
degree of freedom calculation for this instance is:
; and yes…this is a bear. So if you believe the variances are close enough
to being equal you would generally want to assume this if you had to do this by hand. This
becomes much simpler when we have a software package simply give us the dof. In any case our
rule will be to assume they are equal unless you test them and find they are statistically different
or you notice that the larger sample variance is 3X the value of the smaller sample.
For example suppose we have the following data:
Variables Sample 1
15
s2
9
n
24
Sample 2
16
2
16
In this case we see that sample 1’s variance is 9 and it is the larger sample. So we must check to
see if it s 3X the size of sample 2’s variance. Since 9/2 = 4.5 we see it is in fact 4.5 times the
size and we would use the unequal variances method.
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B. Hypothesis Testing: Paired Differences
Many times experiments are run where the variation between the measurements needs to be
controlled for to truly test whether there is a difference between measurements. In this case we
want to use the paired differences procedure. It can be thought of as a before and after test.
Here are some common time and why:
1) When performing a drug test you want to see how it affects a person or group of people…but
internal chemistry, age, weight, and other characteristics need to be controlled for.
2) When one wants to test the difference in procedure or method. For example, if we wanted to
test study methods, we would want to have the same people use the different methods and get the
difference in response to a test of some sort
3) A final example could be fertilizer or some sort of agricultural process. Since soil
composition, sunlight, water retention, etc are different among different plots of land you
generally want to run a test on the same plot of land to ensure there is truly a difference that
occurs due to what you are testing for.
1.Calculation of Mean Difference and Variance/Standard Deviation
Here we get turn our two means into one mean. We do this for all the differences as follows:
i)
or you can take measure 2 and subtract
measure 1.
ii) To get the variance and standard deviation we do the following:
with
as before. If you notice, this is just the regular standard
deviation and variance formula applied to mean difference
For example suppose we have the following data on practice test scores for the SAT at Kaplan
before and after the class for seven people (n=6)
Person Before After Difference (measure 2- measure 1)
1
1050 1210
160
2
970
1310
340
3
1160 1480
320
4
890
1080
190
5
760
1110
350
6
1260 1580
320
(160-280)2
(340-280)2
(320-280)2
(190-280)2
(350-280)2
(320-280)2
/ n = [160+340+320+190+350+320]/6 =280
= [(-120)2 +(60)2 + (40)2 + (-90)2 + (70)2 + (40)2 ] /5 = 6840
2. Performing a Hypothesis Test
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-so now we can perform a hypothesis test as before like we were simply using a t-stat. In this
instance our test stat becomes t
with dof = n-1 where n is the number of pairs
Example: Using our data before suppose we want to test whether or not the Kaplan class
actually improves scores with 90% confidence.
i) Hypotheses:
Ho: ud ≤ 0 (i.e. that it doesn’t improve anything)
Ha: ud > 0 (i.e. that the Kaplan improves scores since the difference would be positive
ii) Test Stat: t
iii) Compare to critical value:
One Sided Test
-It is only one sided since we are testing
to see if the score is increased
-you need to note the level of
confidence and α to find the
appropriate t-values for your CI.
This is your t-value that put 10% in
the upper tail with
dof =
. The value turns
out to be 1.476
t
μ
tα=0.1 =1.476
iv) Conclusion: Since our test stat is clearly bigger than 1.476 (8.29 >> 1.476 so it lies in the
critical or rejection region) we can conclude that with 90% confidence that the Kaplan study
program does indeed help to raise scores.
C. Hypothesis Testing: Comparing Two Population Proportions using Large Samples
All of the hypothesis tests we have done before have involved means and differences of means,
but hypothesis testing can be applied to any type of distribution. In this next section we will look
at how it can be used to compare differences in population proportions.
As before we want to see if there is any difference between two values, in this case
must be the case that we have random independent larger samples.
1. How to calculate the standard deviation:
; note that if we don’t have p1 and p2 and
a. We can estimate
must use sample values then we can calculate
subbing in our sample proportions.
by simply using the previous formula but
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It
b. No Difference Assumed between
One thing to note is that in hypothesis testing we are testing to see if there are differences in the
proportions. So if we assume that there is no difference (i.e. Do = 0 as in most cases) then we
can amend the calculation of the standard deviation as follows:
where we have that
So in this case our proportion is constructed using both samples.
Example: Suppose we are looking at the number of students at OCCC who wear shorts and we
take two samples. One during the day and one at night both of size 50. If we have that in the
first sample during the day that 23 wore shorts and in the evening that 28 did then our
Note:
2. Performing the Hypothesis Test
-this is done with as before with the same four steps. So suppose we want to see if there is a
difference between the two samples and we want to test this at the 95% confidence level.
i) Set up hypotheses
Ho: p1- p2 = 0
Ha: p1- p2 ≠ 0
*it is two sided because we are only seeing if there is a difference and not if one is specifically
bigger than the other.
ii) Test Stat: We use a Z since we have large samples and in this case
Z=
**if you have
=
= -1
then you would use this as the standard deviation
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iii) Find the critical value – These are just our typical z values for 95% confidence level
Two Sided Test
Z
-zα/2 =-1.96
Zα/2 =1.96
0
iv) Conclusion: When we compare our critical values and our test stat we find -1 is not in either
tail. So we fail to reject Ho and conclude there is no difference in the proportion of students who
wear shorts at night or during the day.
D. Hypothesis Testing: Comparing Two Population Variances assuming Independence
-recall from parts A and B that sometimes we assume that the variances of two samples are equal
and sometimes we assume they are unequal. Aside from simply using our rule of equality of
variances unless the larger sample variance is 3X the size of the smaller sample we can actually
test to see if they are different.
To do this we must introduce a new distribution called the F distribution. The F-distribution has
the following characteristics.
i) It is skewed to the right (i.e. tail extends to the right)
ii) It only has positive test stats. So it starts from 0.
iii) It has two degrees of freedom for the numerator (df1) and the denominator (df2).
Graphically it looks like:
Once again α is the probability in the
tail. We can have one or two sided
tails. In this case we just have one.
If there were two you would divide
up α into two parts as before.
F
Fα
For example suppose we wanted to find the F value with α =0.05 and df1 = 7 and df2 =7 would be
F= 3.50
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To get these values go to A.5-8 on 644 and 647. Make sure to use the proper table with the
appropriate α.
1. Performing a hypothesis test for equality of variances – General Set up
a. Set up the Hypotheses
So we can test if
i) Two Tailed
ii) One Tailed:
Ho:
Ha:
Or if Ho:
Ha:
**nearly all of the time we will run a two tailed test.
b. Set up the test state which is F =
or F =
It depends on whether
is larger. Put the larger on the top.
c & d. Find the critical value (based on the hypothesis test run and α) and then state your
conclusion. Note that for the degrees of freedom df1 = n1-1 and df2 = n2-1. When performing the
test if the df are not in the table always round down. For example if we have df1 = 23-1 =22
then use the value for 20. Also, when we have our alpha we must
2. Example: Suppose we want to test to see if the following variances are significantly different
from one another testing them at the 95% level.
Variables Sample 1 Sample 2
15
16
2
s
9
2
n
24
16
a. Hypotheses: It is two tailed since we are testing to see if they are different
b. Test Stat: Since sample 1 has a greater value we have F =
c. Find the critical value:
Once again α is the probability in the
tail. We have a two tailed test so
α/2=0.025 with df1 = 24-1=23 and
df2 = 15-1 =14. We use A.7 since we
want F0.025
F
Fα/2 F0.025
= 2.84
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d. Conclusion: Since our test stat of 4.5 is greater than 2.84 and lies in our tail (also known at
our critical or rejection region) then we reject Ho and conclude that with 95% confidence the
variances are NOT equal.
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