UL(Inferences From Two Samples)-Salva

7/28/2019 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
http://www.ehow.com/how_7213406_calculate-t_statistic.html


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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
http://www.ehow.com/how_7213406_calculate-t_statistic.html


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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.
http://www.ehow.com/how_5145417_calculate-ttest.html


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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

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