Parametric Stat Excel MS2007 Prez

8/10/2019 Parametric Stat Excel MS2007 Prez

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STATISTICS(Using

Microsoft Office Excel 2007

Program )An Adaptation from Antonio S. Broto 2008)

Microsoft

Office Excel

You should see

Analysis

on the far right

If not, Click

Office

Button

Office Button

Dialogue Box appears

Options

Excel Options Dialogue

Box appears Click

Add-Ins

Add-Ins Dialogue Box

appears Click Analysis

Tool Pack-VBA

Click Go…

Click OK.This

installs

Analysis

Add-Ins Dialogue Box

appears

Click Boxof

Analysis

Tool Pack

and VBA

Click YES.You’ll be prompted toinstall the tool.

NOW, you are readyto use Microsoft Excel

2007 Program to solvedifferent statistical

tests needed in yourresearch problems

REMEMBER…

If the nature of test is test

of difference, the

parametric tests are:1. t-test

o Independent sampleso correlated samples

REMEMBER…

of difference, the

parametric tests are:2. z-test

o Independent sampleso correlated samples

REMEMBER…

of difference, the

parametric tests are:3. F-test Analysis of

Variance (ANOVA)o One-Way

REMEMBER…

of relationship, the

parametric test is: Pearson Product

Moment of Coefficientof Correlation, r

REMEMBER…

of association, the

parametric tests are:1. Simple Linear

Regression Analysis2. Multiple Regression

Analysis

When do we know if

the test is directionalor non-directional?

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

H1: µ < k, the test isleft-tailed, so the

REJECTION region is inthe left tail: Directional

H1: µ > k, the test isright-tailed, so the

REJECTION region is inthe right tail: Directional

Solving by the

Stepwise Method

I. ProblemII. Hypotheses

III. Level of Significance

Critical or tabular

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

Solving by the

Stepwise Method

IV. StatisticV. Decision Rule

2. If the computed

value is < or within

the critical value,

ACCEPT H0

Solving by the

Stepwise MethodVI. Conclusion

S-1: DecisionS-2: What your decision

means: NOT significantor significantS-3: Implication

Let’s get started!!!

How do we use the t-test

for independent samples?

Use the formula:

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.

Given Data

Male (X1): 14, 18, 17, 16, 414, 12, 10, 9, 17

Female (X2): 12, 9, 11, 5, 103, 7, 2, 6, 13

Let your Excel do the job!

Enter the scores as shown:

Click Data Analysis

A dialogue box appears

Scroll down to t-test: TwoSample Assuming Equal

Variances Click OK

This box appears Click this!

This box appears

Highlight from 14 to 17

Then Click

Click this!

Highlight from 12 to 13under Female

Then Clickthis!

This box appears

Click this! Then Click

Then Click

This box appears

Click D5cell, or any

cell you

Then Clickthis!

This box appears Click OK!

Voila! You now have the answer!

t-computed

t-critical

value!

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.

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

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.

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

Enter data for X1 and X2 as

Click Data

Analysis

Click t-Test: Two-Sample

Assuming Unequal Variances

Click Variable 1 Range

Highlight from 9.8 to 9.7

Click Variable 2 Range

Highlight from 12 to 13.5

Click O of Output Range

t-computed

You should have this result!

t-critical

Your NDC looks like this since

w/c is non-directional!

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

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.

How do we use the t-

test for correlatedsamples?

Use this formula

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

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

Sample Problem

…the pretest and posttest, or

the use of the programmed

materials did not affect thestudents’ performance in

English. Hence,

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

Enter the

data asshown:

Click Data

Click Data Analysis

Click t-Test: Paired TwoSample for Means

The NDC looks like this since

w/c is directional:

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.

Conclusion

It can be implied that the useof the programmed materials

in English is effective because

the post-test result is higherthan the pretest.

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

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,

Enter the Data

For , type 90 hundred

times in column A

For , type 85 hundred

times in column B

Enter the Data

Y h ld h hi

You should have this

result

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

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

Conclusion

…freshmen of the College of

Nursing are better than theincoming freshmen of the

College of Veterinary Medicine.

How do we use the F-test?

Use the following

computations:

1. Compute the correctionfactor, CF

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

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

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

A B C D

Cli k D t d D t A l i

Click Data and Data Analysis

Fi d A Si l F t

Find Anova: Single Factor

Cli k i d fi ld

Click required fields

Y h ld h thi lt

You should have this result

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

…difference in the averagesales of the four brands of

shampoo.

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

hours among the 3 types ofpocket scientific calculators

before a recharge is required.

Do the analysis of variance at.05 level of significance

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

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.

F-TEST

TWO-WAY ANOVA

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

Problem 1

…the term. Use the two-

way ANOVA with

interaction effect at .05level of significance to test

the following hypothesis:

Hypotheses

H01: There is no significant

difference in the

performance of thethree groups of students

under three different

instructors

Y h ld h thi lt

You should have this result

F-Test (Two-Way ANOVA)

H01: There is no significant difference

Conclusion-1

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.

in the performance of the three

groups of students under threedifferent instructors

Conclusion-1

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

Conclusion-1

…better than instructor A or C

Conclusion-1

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

groups of students under thethree different methods of

teaching

Conclusion-1

This means that there is nosignificant difference in the

performance of the students

under the three different

methods of teaching.

teaching

Conclusion-1

It can be implied that eachmethod of teaching is as

effective as the other methods.

teaching

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

±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

±1.00 –

perfect correlation, perfectrelationship

Source: Basic Probability andStatistics by Winston S. Sirug

(2011)

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

Parametric Stat Excel MS2007 Prez

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