Post on 02-Jun-2018
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STATISTICS(Using
Microsoft Office Excel 2007
Program )An Adaptation from Antonio S. Broto 2008)
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Open
Microsoft
Office Excel
2007
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Click
Data
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You should see
D
ata
Analysis
on the far right
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If not, Click
Office
Button
icon
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The
Office Button
Dialogue Box appears
Click
Excel
Options
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Excel Options Dialogue
Box appears Click
Add-Ins
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Add-Ins Dialogue Box
appears Click Analysis
Tool Pack-VBA
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Click Go…
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Click OK.This
installs
Data
Analysis
Tool
Add-Ins Dialogue Box
appears
Click Boxof
Analysis
Tool Pack
and VBA
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Click YES.You’ll be prompted toinstall the tool.
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NOW, you are readyto use Microsoft Excel
2007 Program to solvedifferent statistical
tests needed in yourresearch problems
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REMEMBER…
If the nature of test is test
of difference, the
parametric tests are:1. t-test
o Independent sampleso correlated samples
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REMEMBER…
If the nature of test is test
of difference, the
parametric tests are:2. z-test
o Independent sampleso correlated samples
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REMEMBER…
If the nature of test is test
of difference, the
parametric tests are:3. F-test Analysis of
Variance (ANOVA)o One-Way
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REMEMBER…
If the nature of test is test
of relationship, the
parametric test is: Pearson Product
Moment of Coefficientof Correlation, r
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REMEMBER…
If the nature of test is test
of association, the
parametric tests are:1. Simple Linear
Regression Analysis2. Multiple Regression
Analysis
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When do we know if
the test is directionalor non-directional?
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1. If the sign in the H1 is:
H1: µ ≠ k, the test istwo-tailed, so the
REJECTION region is in
both tails: Non-
directional
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2. If the sign in the H1 is:
H1: µ < k, the test isleft-tailed, so the
REJECTION region is inthe left tail: Directional
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3. If the sign in the H1 is:
H1: µ > k, the test isright-tailed, so the
REJECTION region is inthe right tail: Directional
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Solving by the
Stepwise Method
I. ProblemII. Hypotheses
III. Level of Significance
α
df
Critical or tabular
value
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Solving by the
Stepwise Method
IV. StatisticV. Decision Rule
1. If the test-computed
value is ≥ or beyond the
critical value, DO NOT
ACCEPT H0
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Solving by the
Stepwise Method
IV. StatisticV. Decision Rule
2. If the computed
value is < or within
the critical value,
ACCEPT H0
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Solving by the
Stepwise MethodVI. Conclusion
S-1: DecisionS-2: What your decision
means: NOT significantor significantS-3: Implication
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Let’s get started!!!
How do we use the t-test
for independent samples?
Use the formula:
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Sample Problem 1
The following are scores inspelling of 10 male and female
Grade 8 students. Test the null
hypothesis, H0, that there is
no significant difference
between the performance ofmale and female Grade 8
students in spelling. Use the t-
test at .05 level of significance.
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Given Data
Male (X1): 14, 18, 17, 16, 414, 12, 10, 9, 17
Female (X2): 12, 9, 11, 5, 103, 7, 2, 6, 13
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Let your Excel do the job!
Enter the scores as shown:
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Let your Excel do the job!
Click Data Analysis
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Let your Excel do the job!
A dialogue box appears
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Let your Excel do the job!
Scroll down to t-test: TwoSample Assuming Equal
Variances Click OK
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Let your Excel do the job!
This box appears Click this!
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Let your Excel do the job!
This box appears
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Let your Excel do the job!
Highlight from 14 to 17
Then Click
this!
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Let your Excel do the job!
Click this!
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Let your Excel do the job!
Highlight from 12 to 13under Female
Then Clickthis!
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Let your Excel do the job!
This box appears
Click this! Then Click
this!
Then Click
this!
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Let your Excel do the job!
This box appears
Click D5cell, or any
cell you
like!
Then Clickthis!
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Let your Excel do the job!
This box appears Click OK!
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Let your Excel do the job!
Voila! You now have the answer!
t-computed
t-critical
value!
df
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Conclusion Since the t-computed value of 2.881
is greater than the t-tabular (orcritical) value of 2.101 at .05 level of
significance with 18 degrees of
freedom, DO NOT ACCEPT the nullhypothesis. This means that there IS
A significant difference between the
performance of the male and female
Grade 8 students in spelling. It can
be implied that the male students
performed better than the female
students.
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Problem 2
Two groups of experimentalrats were injected with a
tranquilizer at 1.0 mg and 1.5mg dose, respectively. The
time given in seconds that took
them to fall asleep is hereby
given. Use the t-test for
independent samples at .01
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Problem 2
…level of significance to testthe null hypothesis that the
difference in dosage has noeffect on the length of time it
took them to fall asleep.
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Given Data
Time (in seconds)1.0 mg: 9.8, 13.2, 11.2, 9.5
13.0, 12.1, 9.8, 12.3
7.9, 10.2, 9.7
1.5 mg: 12.0, 7.4, 9.8, 11.5
13.0, 12.5, 12.5, 9.8
10.5, 13.5
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Enter data for X1 and X2 as
shown
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Click Data
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Click Data
Analysis
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Click t-Test: Two-Sample
Assuming Unequal Variances
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Click Variable 1 Range
Highlight from 9.8 to 9.7
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Click Variable 2 Range
Highlight from 12 to 13.5
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Click O of Output Range
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t-computed
You should have this result!
df
t-critical
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Your NDC looks like this since
w/c is non-directional!
C l i
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Conclusion
Since the t-computed value of-.40 is within the critical value
of -2.88 at .01 level of
significance with 18 degreesof freedom, ACCEPT the null
hypothesis. This means thatthere is NO significant
difference between 1.0 mg
C l i
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Conclusion
…and 1.5 mg dosages on thelength of time it took for the
rats to fall asleep. It can be
implied that both dosageshave the same tranquilizing
effect on the rats.
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How do we use the t-
test for correlatedsamples?
Use this formula
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Sample Problem
An experimental study was
conducted on the effect of
programmed materials inEnglish on the performance of
20 selected Senior high school
students. Before the programwas implemented the pretest
was administered and after
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Sample Problem
5 months the same instrumentwas used to get the posttest
result. Use the t-test forcorrelated samples at .05 level
of significance to test the null
hypothesis that there is nosignificant difference between
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Sample Problem
…the pretest and posttest, or
the use of the programmed
materials did not affect thestudents’ performance in
English. Hence,
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The following is the result ofthe experiment:
Pretest (X1): 20, 30, 10, 15, 20
10, 18, 14, 15, 20, 18, 15, 15,
20, 18, 40, 10, 10, 12, 20
Posttest (X2): 25, 35, 25, 25, 2020, 22, 20, 20, 15, 30, 10, 16,
25, 10, 45, 15, 10, 18, 25
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Enter the
data asshown:
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Click Data
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Click Data Analysis
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Click t-Test: Paired TwoSample for Means
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Click
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You should have this result!
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The NDC looks like this since
w/c is directional:
C l i
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Conclusion
Since the t-computed value of-3.17 is beyond the t-critical
value of -1.73 at .05 level of
significance with 19 degrees offreedom, DO NOT ACCEPT the
null hypothesis. This means
that there IS A significant
difference between the
pretest and posttest results.
C l i
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Conclusion
It can be implied that the useof the programmed materials
in English is effective because
the post-test result is higherthan the pretest.
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Sample Problem
An admission test was
administered to incoming
freshmen in the Colleges ofNursing and Veterinary
Medicine with 100 students
each college randomly
selected. The mean scores of
the given samples were
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Sample Problem
= 90 and = 85 and the
variances of the test scores
were 40 and 35, respectively.Is there a significant difference
between the two groups? Use
.01 level of significance. Hence,
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Enter the Data
For , type 90 hundred
times in column A
For , type 85 hundred
times in column B
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Enter the Data
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Y h ld h hi
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You should have this
result
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Level of significance
for the z-test for twosample means
Test .01 .05
One-Tailed +/- 2.33 +/- 1.645
Two-Tailed +/- 2.575 +/- 1.96
C l i
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Conclusion
Since the z-computed value of5.774 is greater than the z-
tabular value of 2.575 at .01
level of significance, DO NOTaccept the null hypothesis. This
means that there IS A
significant difference between
the two groups. It can be
implied that the incoming
C l i
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Conclusion
…freshmen of the College of
Nursing are better than theincoming freshmen of the
College of Veterinary Medicine.
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How do we use the F-test?
Use the following
computations:
1. Compute the correctionfactor, CF
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2. TSS is the total sum of
squares minus the CF
3. BSS is the between sum
of squares minus the CF
or correction factor
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4. WSS is the within sum of
squares or it is the
difference between the
TSS minus the BSS5. After getting the TSS, BSS
and WSS, the ANOVA
table should beconstructed
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Example
A sari-sari store is selling 4
brands of shampoo. The
owner is interested if there isa significant difference in the
average sales of the four
brands of shampoo for one
week. The following data are
recorded
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Data
Brand
A B C D
7
3
5
6
9
4
3
9
8
8
7
6
9
10
2
3
4
5
6
4
2
4
5
7
8
3
4
5
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Cli k D t d D t A l i
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Click Data and Data Analysis
Fi d A Si l F t
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Find Anova: Single Factor
Cli k i d fi ld
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Click required fields
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Y h ld h thi lt
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You should have this result
Conclusion
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Conclusion
Since the F-computed value of7.98 is greater than the F-
tabular value of 3.01 at .05
level of significance with 3and 24 degrees of freedom,
DO NOT ACCEPT the nullhypothesis. This means that
there IS A significant
Conclusion
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Conclusion
…difference in the averagesales of the four brands of
shampoo.
Example 2
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Example 2
The following data representthe operating time in hours of
the 3 types of scientific
pocket calculators before arecharge is required. Test the
null hypothesis that there isno significant difference in
the average operating time in
Example 2
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Example 2
hours among the 3 types ofpocket scientific calculators
before a recharge is required.
Do the analysis of variance at.05 level of significance
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You should have this result!
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Conclusion
Since the F-computed value
of 0.73 is lesser than the F-
tabular value of 3.52 at .05level of significance with 2
and 19 degrees of freedom,
ACCEPT the null hypothesis.This means that there is NO
significant difference in the
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Conclusion
…average operating time in
hours among the 3 types of
pocket calculators before arecharge is required. It can
be implied that all types of
pocket calculator have thesame operating time and
recharging time.
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F-TEST
TWO-WAY ANOVA
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Problem 1
Forty-five language
students were randomly
assigned to one of threeinstructors and to one of
the three methods of
teaching. Achievement wasmeasured on a test
administered at the end of
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Problem 1
…the term. Use the two-
way ANOVA with
interaction effect at .05level of significance to test
the following hypothesis:
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Hypotheses
H01: There is no significant
difference in the
performance of thethree groups of students
under three different
instructors
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Enter
Data
Click
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Click
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Click
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Click
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Y h ld h thi lt
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You should have this result
F-Test (Two-Way ANOVA)
H01: There is no significant difference
Conclusion-1
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Since the F-computed value of11.382 is greater than the F-
tabular value of 3.40 at .05 level
of significance with 14 and 2degrees of freedom, DO NOT
ACCEPT the null hypothesis.
H01: There is no significant difference
in the performance of the three
groups of students under threedifferent instructors
H01: There is no significant difference
Conclusion-1
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This means that there is A significant difference in the
performance of the three groups
of students under the threedifferent instructors. It can be
implied that instructor B is
H01: There is no significant difference
in the performance of the three
groups of students under threedifferent instructors
H01: There is no significant difference
Conclusion-1
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…better than instructor A or C
H01: There is no significant difference
in the performance of the three
groups of students under threedifferent instructors
H02: There is no significant difference
Conclusion-1
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Since the F-computed value of1.015 is less than the F-tabular
value of 2.063 at .05 level of
significance with 14 and 2
degrees of freedom, ACCEPT the
null hypothesis
H02: There is no significant difference
in the performance of the three
groups of students under thethree different methods of
teaching
H02: There is no significant difference
Conclusion-1
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This means that there is nosignificant difference in the
performance of the students
under the three different
methods of teaching.
H02: There is no significant difference
in the performance of the three
groups of students under thethree different methods of
teaching
H02: There is no significant difference
Conclusion-1
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It can be implied that eachmethod of teaching is as
effective as the other methods.
H02: There is no significant difference
in the performance of the three
groups of students under thethree different methods of
teaching
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0.00 - no correlation, no relationship±0.01 - ±0.20 – very low correlation,
almost negligible relationship
±21 - ±0.40 –
slight correlation,definite but small relationship
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±0.41 - ±0.70 - moderate correlation,substantial relationship
±0.71 - ±0.90 – high correlation,
marked relationship±0.91 - ±0.99 – very high correlation,
very dependable relationship
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±1.00 –
perfect correlation, perfectrelationship
Source: Basic Probability andStatistics by Winston S. Sirug
(2011)
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Example:
The owner of a chain of fruit shake
stores would like to study the
correlation between atmospheric
temperature and sales during thesummer season. A random sample
of 12 days is selected with the
results
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