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Transcript of UL(Inferences From Two Samples)-Salva
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INFERENCES FROM
TWO SAMPLES
SALVACION M. VINLUAN DR. IMELDA E. CUARTELUNIVERSITY OF LUZON Professor
Doctor of Philosophy in Development
Education
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Definitions
Testing Two Means, Dependent
Case: The Mean of the Differences
Testing of Two Variances
Testing of Two Means, Independent
Case: The Differences of the Means
Testing Two Proportions
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What is Statistical Inference?
This refers to the process of
drawing conclusions from data that is
subject to random variation, for
example, observational errors or
sampling variation.
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More substantially, the term
statistical inference, statistical
induction and inferential statistics
are used to describe systems of
procedures that can be used to drawconclusions from datasets arising
from systems affected by random
variation, such as observationalerrors, random sampling, or random
experimentation.
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Initial requirements of such a
system of procedures for inferenceand induction are that the system
should produce reasonable
answers when applied to well-defined situations and that it
should be general enough to be
applied across a range ofsituations.
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The outcome of statistical
inference may be an answer to thequestion what should be donenext?, where there might be a
decision about making furtherexperiments or surveys, or about
drawing conclusions before
implementing some organizationalor governmental policy.
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Statistical inference
refers to the statistical methodconcerned with making estimates
of population value. This
particular method and processwill help us determine how
accurate and acceptable our
generalizations are.
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Testing Two Means, Dependent
Case: The Mean of theDifference
There are two possible caseswhen testing population means,
the dependent case and the
independent case.
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The t-Test for Dependent
Samples
The t-Test for dependent samples is
applied to matched pairs or correlated
samples. The samples are supposedly
taken from one population. For example,in the research study on the degree of
seriousness of problems encountered
by college freshmen, data were takenbefore and after their individual
counseling sessions.
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From the 15-item problem
checklist, the corresponding degree
of seriousness of problems
encountered by the students before
and after their individual sessions
comprises the data to be
compared. This is also referred to
as repeated measures.
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To compute the t-value for dependent
samples, the formula is as follows
t = DnD (D)
n - 1
where:
t = t-value
D = differencen = number of cases
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Example:
Student Before After
1 3.00 4.00
2 3.25 3.50
3 3.00 3.50
4 2.50 3.605 2.75 3.45
6 2.50 2.75
7 2.75 4.758 3.75 2.00
9 3.50 3.51
10 3.60 4.50
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Solution:
Step 1. Ho : P1 = P2
H1 : P1 P2
Step 2: = 0.05 (two-tailed test)
Step 3: Reject the null hypothesis if the computed
t-value is greater than the critical t-value
of1.833 at =0.05 with nine degrees offreedom.
Step 4: t-test for dependent samples
t = D
n D (D)
n - 1
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Student Before (X1) After(X2) D D2
1 3.00 4.00 -1.00 1.0000
2 3.25 3.50 0.25 0.0625
3 3.00 3.50 0.50 0.2500
4 2.50 3.60 1.10 1.2100
5 2.75 3.45 0.70 0.49006 2.50 2.75 0.25 0.0625
7 2.75 4.75 -2.00 4.0000
8 3.75 4.00 0.25 0.0625
9 3.50 3.51 0.01 0.0001
10 3.60 4.50 0.90 0.8100
6.96 7.9476
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t = DnD (D)
n - 1
t = 6.96
10(7.9496) - (6.96)
10- 1
t = 3.74
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The degree of freedom:
df= 10-1df= 9
Step 5: Reject the null hypothesis since
the computed value of3.74 is
greater than the critical value of
1.833 at = 0.05 with 9 degreesof freedom.
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Conclusion:
There is a significant difference
between the degrees of
seriousness of problems
encountered by the students before
and after their exposures to
individual counseling sessions.
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Testing Two Variances
A technique commonly used under F-test
is referred the Analysis of Variance
(ANOVA), since it deals with the ratiobetween the variability occurring among
the different groups or treatments against
the variability occurring within the
members of each of the groups ortreatments.
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Moreover, it is assumed that the different groupsmust have equal variances. The following is an
example of the data layout for this test.
Groups or Treatments
1 2 3 n
1 X11 X12 X13 X1n2 X21 X22 X23 X2n
Replicates3 X31 X32 X33 X3n
.
.m Xm1 Xm2 Xm3 Xmn
Totals T1 T2 T3 Tmn
Means X1 X2 X3 Xn
Sum Squares SS1 SS2 SS3 SSn
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where:
m = is the number of replicates
n = is the number of groups or treatments
Xjk = is the observation belonging to the jth
replicate of the kth treatment or group
1 j m and 1 k n
Tk = is the total of all the kth treatments
from each of the m replicates, i.e.
Tk=Xjk
j=1
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Xk is the mean of the kth treatments
from each of the m replicate; i.e.
Xk = Xjkm
SSk is the sum of the squares of all thekth treatments from each of the m
replicates; i,e.
SSk= Xjkj=1
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To perform the test of hypothesis
comparing more than two samplemeans the following steps and needed
information must be noted.
Ho: All the means of the different groups
or treatments are the same.
Ha: At least one mean is different fromthe other means.
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The rejection criteria from this test is
stated as follows:Reject Ho is Fc > F(df1,df2). Do not reject Ho
otherwise.
where:
F(df1,df2)is the critical value, is the levelof significance, df1 = n-1, df2 = p-n, p =
mn; m = number of replicates and n =
number of groups or treatments. Fc is
the computed test statistic.
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In computing for the test statistic, we
construct the ANOVA table as follows:
Analysis of VarianceSources
of
Variation
df Sum of
Squares
Means
Squares
F Ratio
Between
Groups
Within
Groups
Total
df1 = n-1
df2 = p-n
P-1
SSB
SSW
SS
MSB
MSW
Fc
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where:
SSB - sum of squares between groups
SSW -sum of squares within groups
MSB - mean squares between groups
MSW -means squares within groupsSST - total sum of squares
Fc - test statistic
n - number of groups or treatmentsm - number of replicates
p - mn
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a) the correction factor (CF), given by
nCF= Tk
P
b) the total sum of squares (SST)
SST=SSk-CF
c) the sum of squares between groups (SSB)
SSB = Tk - CF
k=1 m
d) the sum of squares within groups (SSW)
SSW = SST - SSB
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e) the mean squares between groups (MSB)
MSB = SSB = SSBdf1 n-1
f) the mean squares within groups (MSW)
MSW = SSW = SSW
df1 n-1
g) the Fc, referred to as the test statistic is
defined by
Fc = MSB
MSW
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Example:
Five bus companies were selectedin order to determine if there is a
difference in the number of hours
travelling a 200 kilometer path from
place A to place B at 5% level of
significance. The data sets were
given as follows:
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Bus Companies
u v w x y
Travel 1 2.7 3.0 3.4 4.9 4.6
time of 2 2.8 4.2 3.6 2.6 3.8
buses 3 4.0 4.3 4.6 5.1 4.6
4 3.5 4.1 2.8 3.5 3.2
5 3.8 3.2 5.0 3.1 2.9
6 4.9 4.0 2.3 3.2 2.5
Totals 21.70 22.80 21.70 22.40 21.60Sum of
squares 81.83 88.18 83.81 88.88 81.66
Means 3.62 3.80 3.62 3.73 3.60
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Solution:
Statement of Hypothesis:
Ho: The five bus companies have the sametraveling time in hours from place A to
place B.
Ha: At least one of the five bus company has
different traveling time in hours.
Critical Region and Criteria for rejection
Level of significance: = 5
Test: F-test
df: df1= n-1= 5-1=4
df2= p-1= 30-5=25 ; p=mn
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Criteria for rejection:
Reject the Ho
is Fc
> F0.05(4.25)
Do not reject Ho otherwise
Computation:
Test Statistic: FcCF= Tk = (21.70+22.80+21.70+22.40+21.60)
30 30
= (110.2)
30
= 404.80
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SST=SSk - CF
=(81.83+88.18+83.81+88.88+81.66)- 404.80
= 19.56
SSB= (Tk) - CF
k=1 m
= 21.70+22.80+21.70+22.40+21.60 - 404.80
6 6 6 6 6
= 404.99 404.80= 0.19
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MSB = 0.19 = 0.0475
5 -1
SSW = SST SSB
= 19.56 0.19
= 19.37
MSW = 19.37 = 0.7748
30-5
.:F = 0.475 = 0.0613
0.7748
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Analysis of Variance
Sources of
variation
df Sum of
squares
Means
squares
F Ratio
Between
BusCompany
Within Bus
Company
4
25
0.19
19.37
0.0475
0.7748
0.0613
Total 29 19.56
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F0.05(4.25)= 2.76
Conclusion:
Since the test statistic does not exceeds
the critical value, the Ho is not rejected.
Thus, the traveling time in hours of thefive bus companies are the same from
place A to place B at 5% level of
significance.
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Example:
Three teachers taught statistics to three sections. The
final grades of six students per section and somedescriptive were shown below. Can we say that there
is a difference between the scores of each section?
Test at 5% level of significance.
Sections Students final grade Total SS Mean
A
B
C
87
75
90
81
84
82
86
83
89
91
86
86
87
79
80
90
83
82
522
490
509
45,476
40.096
43,265
87.00
81.67
84.83
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Solution:Statement of Hypothesis:
Ho: The three sections have the same meanfinal grades in Statistics
Ha: At least one of the three sections has
different mean final grades in Statistics
Critical Region and Criteria for rejection
Level of significance: = 5%
Test: F-test
df: df1 = n-1= 3-1= 2
df2 = p-n =18-3 =15
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Criteria for rejection:
Reject Ho is Fc > F0.05(2.15)
Do not reject Ho otherwise
Computation:
Test Statistic: Fc
CF=Tk = (522+490+509)
n 18
= (1521)18
= 128,524.50
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SST= SSk CF
k=1
= (45,476+40,096+43,265)-128,524.50
= 128,837 128,524.50
= 312.5
SSB= (Tk) - CF
k=1 m
= 522 + 490 + 509 -128,524.5
6 6 6
= (45,414 + 40,016.67 + 43,180.17) 128,524.5= 128,610.84 128,524.5
= 86.34
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MSB = 86.34 = 43.1700
3-1
SSW = SST SSB
= 312.5 86.34
= 226.16
MSW = 226.16 =15.0773
18-3
.: F = 43.1700 = 2.8632
15.0773
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Analysis of Variance
Sources ofVariation
df Sum ofSquares
MeansSquares
F Ratio
Between Sections
Within Sections
2
15
86.34
226.16
43.1700
15.0773
2.8632
Total 17 312.50
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F0.05(2.15) = 3.68
Conclusion:
Since the test statistic does not exceeds
the critical value, we have to accept Ho.Thus, the final grades of the students in
the three sections are the same at 5%
level of significance.
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Testing Two Means, Independent Case:
The Difference of the Means
The independent t-test is used when the two
sample means are taken from separate
groups of respondents or populations. For
example, the researcher is interested tocompare the research skills of assistant
professors and associate professors in a
certain university. The assistant professors
comprise one group and the associateprofessors another group.
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t = X1X2S1 + S2
n1 n2
where:
X1 = Mean of the first group
X2 = Mean of the second group
S1 = Variance of the first group
S2 = Variance of the second group
n1 = Number of cases in the first group
n2 = Number of cases in the second
group
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Example:
In a study conducted to determine theresearch skills of assistant and associate
professors in state universities and colleges,
the following data represent the mean scores of
the professors:
Assistant Prof.(X1) 3.0 4.2 2.75 3.50 4.50 2.50 2.60
Associate Prof.(X2) 3.75 4.50 3.50 4.00 4.00 3.50 3.00
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Solution:
Step 1: State the hypotheses
Ho: There is no significant difference
between the research skills of
assistant and associate
professors.
H1: There is a significant difference
between the research skills of
assistant and associate
professors.
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Step 2: Set the level of significance
= 0.05; two-tailed test
Step 3: Reject Ho if the computed
value is greater than 1.782.
Step 4: Compute the value of the test
statistics from the given data.
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Asst. Professor Assoc. Professor
(X1) (X2)
3.00 3.75
4.20 4.50
2.75 3.50
3.50 4.00
4.50 4.002.50 3.50
2.60 3.00
__________ ___________
X1 = 23.05 X2= 26.26X1 = 3.29 X2 = 3.75
S1 = 0.64 S2=0.23
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t = X1 - X2
S1 + S2
n2 n2
t = 3.29 -3.75
.64 + .23
7 7t = -.46
.12t = - 1.3143
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The degree of freedom:df = n1 + n2 n2
df = 7 + 7 2df = 14 2
df = 12
Step 5: In as much as the computed t- value of1.3143
is lower than the critical value, then there is nosufficient evidence to reject the null hypothesis.
Step 6: State the conclusion.
Since the computed t-value is lower than the critical t-
value, it means that there is no significant differencebetween the research skills of assistant and associate
professors. Thus, they manifest comparable research
skills.
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Example:A researcher wishes to compare the post-test
performance of two groups of students classified
according to their learning styles assimilators andconvergers. Ten students were assigned per group.
After their exposures to simplified instructional
materials on Basic Statistics, the assimilators mean
post-test performance was 20, with a standarddeviation of 8.92, while the convergers mean post-test performance was 26.26, with a standard
deviation of 8.67. Test the significant difference
between the two group means.
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Solution:
Step 1:
Ho: There is no significant difference in
the mean post-test performance of the
two groups of students classified
according to their learning styles.
H1: There exists significant difference in
the mean post-test performance of thetwo groups of students classified
according to their learning styles.
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Step 2: = 0.05 (two-tailed test)
Step 3: Reject the null hypothesis (Ho) if
the computed t-value is greater
than the critical t-value of1.734at = 0.05 with 18 degrees offreedom.
Step 4: t-test for independent samples
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t = X1 X2
S1 + S2 n1 n2
Convergers AssimilatorsX1 = 26.26 X2 = 20
S1 = 8.67 S2 = 8.92
n1 = 10 n2 = 10
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t = X1 X2S1 + S2
n1 n2t = 26.26 20
8.67 + 8.92
10 10t = 6.26
1.759
t = 6.26 t = 4.711.33
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The degree of freedom:
df = n1+ n2 - 2
df = 10 + 10 2df = 20 2
df = 18
Step 5: Reject the null hypothesis since the computed
t-value of4.71 is greater than 1.734.
Conclusion:
There exists a significant difference between the mean
post-test performance of the two groups of students
classified according to their learning styles after their
exposures to simplified instructional materials in Basic
Statistics.
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Testing Two Proportions
Remember that the normal distributioncan be used to approximate the binomial
distribution in certain cases. Specifically,
the approximation was considered goodwhen np and nq were both at least 5.
Now, we're talking about two
proportions, so np and nq must be atleast 5 for both samples.
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We don't have a way to specifically
test two proportions for values, what
we have is the ability to test thedifference between the proportions.
So, much like the test for two means
from independent populations, we
will be looking at the difference of
the proportions.
We will also be computing an
average proportion and calling it p-
bar. It is the total number of
successes divided by the totalnumber of trials. The definitions
which are necessary are shown to
the right.
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The test statistic used here is
similar to that for a single
population proportion, except
the difference of proportions areused instead of a single
proportion, and the value of p-
bar is used instead of p in the
standard error portion. Since we're using the normal
approximation to the binomial,
the difference of proportions has
a normal distribution. The teststatistic is given.
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Some people will be tempted to try to
simplify the denominator of this test
statistic incorrectly. It can be simplified,
but the correct simplification is not to
simply place the product of p-bar and q-
bar over the sum of the ns. Rememberthat to add fractions, you must have a
common denominator, that is why this
simplification is incorrect.
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The correct simplification would be to
factor a p-bar and q-bar out of the two
expressions. This is usually the formulagiven, because it is easier to calculate.
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Comparing Two Proportions
A commonly posed question is "Are twoproportions different?" For example, is
the end-of-course passing rate this year
significantly different from the end-of-course passing rate last year. This type of
question involves comparing two
proportions both of which are estimates,
and thus the necessary formula will be
different.
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In this case we will develop the theory
that allows us to apply the hypothesis
testing procedure to this problem. The
procedure involves the use of test
statistic that involves two proportions.But this statistic will still be a z-statistic
and we will compare it to the normal
scores, i.e. z-scores.
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Theoretical Background
Given two independent variables that areboth normal, their sum or difference will
be normal distributed. This idea can
readily be proven. The important point isthat the difference of two sample
proportions is an example of the
difference of two normally distributedvariables. Thus we can standardize them
and obtain a a-score.
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For example, If you have two estimated
proportions, 1 and 22 and you happen toknow that one has a mean of 0.6 and
the other a mean of 0.4, then you know
their difference will have a mean of12= 0.6 0.4 = 0.2. What we now need isthe standard deviation of this difference.
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Assuming that each estimate is based on 300
observations you would also be able to
determine the variance of the difference. Sincethe variance of each estimated proportion is
0.60.4 300 = 0.0008. The variance of thedifference would be the sum of the two
variances or 0.0016. Which means thestandard deviation of the difference is 0.0016= 0.04.
All this may seem complicated but the bottom
line is that we know the distribution of ( 1 2).Thus we can calculate confidence intervals
and do hypothesis tests on this variable.
f
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Distribution of 1 2Let 1 is a proportion based on a sample size n1 and
2 is based on a sample of size n2. Assume that bothsample sizes are large enough (>100) then we can
assume that the estimates of p are normally
distributes.
If we assume that the underlying proportions from thetwo samples are the same, i.e. p1 = p2, then ( 12)will have:
a normal distribution (i.e. the variable can be
standardized to a z-score.)
a mean = 0
an approximate variance = (1 )( 1 n1 + 1 n2)where is the estimate of the proportions.
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This fact will be useful to deriving a test-
statistic for proportion. In this case 95%
of the time the corresponding z-scoreshould not deviate more than 1.96 units
from 0. This z-score also called the test
statistics for comparing two proportionsis:
z =
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If we do not assume that the two
proportions are the same then ( 12) canestimate their difference. In this case thisexpression will have:
a normal distribution. (i.e. we can use z-scores to find the error margin.)
a mean, =p1p2
an approximate variance of = 1(1 1)n1 + 2(1 2)n2.
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This means if we want to estimate the
difference (or gap) between twoproportions, p1 and p2 then the 95%
confidence interval will be:
( 1 2) 1.96 p1p2 ( 1 2) + 1.96
Example: The five- step hypothesis test for
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Example: The five step hypothesis test for
two proportions.Example:
Test the assertion that the EOG Mathematics test of
female fifth grade students is no different than that of
the male students. (Data for Asheville schools, 2008.
NC DPI)
STEP I: Set up two opposing hypotheses.
Let p1 be the proportion of female students passing
andp2be the proportion of male student.
H0 : p1 = p2 VS Ha : p1p2
STEP II G t d t
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STEP II: Get data.
Even though all students in each grade are tested, you
still only have a sample since your population is all
students that could potentially be taught in the
Asheville school system. According to NC Department
of Public Instruction 89 of 126 (or 70.6%) female
students passed the test and 119 of 142 (or 83.8%)
male students passed the test.# Passed Sample Size Estimated
Proportion
Female
Students
89 n1 = 126
1 = 0.706
Male Students 119 n2 = 142
2 = 0.838
All Students 208 n = 268
3 = 0.776
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STEP III: Decide on a statistical test.
We will use the two proportion z test given in this
lesson.STEP IV: Calculate your test statistic.
Every test has a formula that standardizes the
estimator, in this case that estimator
is . The statistic is:Z statistic =
Z- stat = (0.706 0.838)0.7760.224(1126 + 1142)
Z-stat0.1320.0026 2.587.
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STEP V: Arrive at a conclusion and state it in
clear English.
Remember that the two critical values for the differentlevels of certainty are: z0.025 = 1.96 orz0.005 = 2.576.
Since |z-stat| >z*(the critical value) we reject the
Null Hypothesis.
Conclusion:
With 99% certainty we can state that there is astatistically significant larger proportion of boys
passing the fifth grade EOC in mathematics in the
Asheville school district.
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Thank you