Inferential Statistics

INFERENTIAL STATISTICS INFERENTIAL STATISTICS

Hypothesis

Hypothesis TeTestingsting Meaning of Hypothesis

Meaning of Hypothesis  A

 A hypothehypothesis sis is is a a tentativtentative e explanaexplanation tion for for certain certain events, events, phenomephenomena na or or  be

behavhavioriors. s. In In stastatististictical al lalangunguagage, e, a a hyhypotpothehesisis s is is a a stastatetemenment t of of prprededictiction or ion or  rela

relationtionshiship p betwbetween or among variabeen or among variables. les. PlaPlainly stateinly stated, a d, a hyphypotheothesis is sis is the mostthe most specific sta

specific statement of a problem. tement of a problem. It is a requirement thaIt is a requirement that these variablet these variables are related.s are related. Furthermore, the hypothesis is testable which means that the relationship between the Furthermore, the hypothesis is testable which means that the relationship between the variables can be put into test on the data gathered about the variables.

variables can be put into test on the data gathered about the variables. Null and Alternative Hypotheses

Null and Alternative Hypotheses There are two ways of stati

There are two ways of stating a hypothesis. ng a hypothesis. A A hypothehypothesis that is intended for sis that is intended for  statistic

statistical test is generally stated in the null form. al test is generally stated in the null form. eing the startieing the starting point of the testingng point of the testing process, it serves as

process, it serves as our wor!ing hypothesis. Aour wor!ing hypothesis. A null hypothesis (H null hypothesis (H oo ) ) expresses the idea expresses the idea

of

of nonon"n"sigsigninificficancance e of of difdiffeferenrence ce or or nonon" n" sisignignificficancance e of of rerelatlatioionshnship ip bebetwtween een thethe variables under study.

variables under study. It is so It is so stated for the purpose of stated for the purpose of being accepted or re#ected.being accepted or re#ected. If the null hypothesis is re#ected, the

If the null hypothesis is re#ected, the alternative hypothesis (H alternative hypothesis (H aa ) ) is is accepted. accepted. ThisThis

is the resear

is the researchecher$s way of r$s way of statstating hising his research research hypothhypothesisesis  in an operational manner.  in an operational manner. The research hypothesis is a statement of the expectation derived from the theory The research hypothesis is a statement of the expectation derived from the theory under study

under study. . If the related literatuIf the related literature points to the findings thare points to the findings that a certain techniqut a certain technique of e of  teaching for ex

teaching for example, is effectample, is effective, we have to assume the same predive, we have to assume the same prediction. iction. This is our This is our  alternat

alternative hypothesive hypothesis. is. %e canno%e cannot do otherwise sinct do otherwise since there is no scientific base there is no scientific basis for is for  such prediction.

such prediction.

&hain of reasoning for inferential statistics &hain of reasoning for inferential statistics '.

'. (amp(ample)sle)s* mus* must be t be randrandomly omly seleselectedcted +.

+. (ample e(ample estimate istimate is compas compared to red to underlyiunderlying distrng distribution ibution of the of the same ssame sieie sampling distribution

sampling distribution -.

-. etermine etermine the prothe probability bability that a that a sample sample estimate estimate reflects reflects the pothe populationpulation parameter

parameter

The four possible outcomes in hypothesis testing The four possible outcomes in hypothesis testing

Actual

Actual Population Population &omparison&omparison /

/uulll 0l 0yypp. . TTrruuee //uulll 0l 0yypp. F. Faallssee 

11&&II((II22// ))tthheerre e iis s nnoo difference* difference* )there is a )there is a difference* difference* 3e#ected /ull 3e#ected /ull 0yp 0yp Type I error  Type I error  ((alphaalpha &orrect ecision &orrect ecision id not 3e#ect

id not 3e#ect /ull /ull

&orrect ecision

&orrect ecision TyType II pe II Error Error  )Alpha 4 probability of ma!ing a Type I )Alpha 4 probability of ma!ing a Type I

error* error*

3egardless of whether statistical tests are conducted by hand or through statistical 3egardless of whether statistical tests are conducted by hand or through statistical software, there is an implicit understanding that systematic steps are being followed to software, there is an implicit understanding that systematic steps are being followed to determine statistical significance. These general steps are described on the following determine statistical significance. These general steps are described on the following pa

computation of statistics, and 6* decision regarding the null hypothesis. The underlying computation of statistics, and 6* decision regarding the null hypothesis. The underlying logic is based on re#ecting a statement of no difference or no association, called the null logic is based on re#ecting a statement of no difference or no association, called the null hypothesis. The null hypothesis is only re#ected when we have evidence beyond a hypothesis. The null hypothesis is only re#ected when we have evidence beyond a reasonable doubt that a true difference or association exists in the population)s* from reasonable doubt that a true difference or association exists in the population)s* from which we drew our random sample)s*.

which we drew our random sample)s*.

3easonable doubt is based on probability sampling distributions and can vary at the 3easonable doubt is based on probability sampling distributions and can vary at the researcher7s discretion. Alpha .86 is a common benchmar! for reasonable doubt. At researcher7s discretion. Alpha .86 is a common benchmar! for reasonable doubt. At alpha .86 we !now from the sampling distribution that a test statistic will only occur by alpha .86 we !now from the sampling distribution that a test statistic will only occur by random chance five times out of '88 )69 probability*. (ince a test statistic that results in random chance five times out of '88 )69 probability*. (ince a test statistic that results in an alpha of .86

an alpha of .86 could only occur by random chance 69 of the time, could only occur by random chance 69 of the time, we assume that thewe assume that the te

test st ststatiatiststic ic reresulsulteted d bebecacause use thethere re are are trutrue e difdiffefererencnces es bebetwtween een ththe e popopulpulatationion parameters, not because we drew an extremely biased random sample.

parameters, not because we drew an extremely biased random sample. %he

%hen n lelearnarning ing stastatististictics s we we gegenernerallally y conconducduct t ststatiatiststicaical l testests ts by by hanhand. d. In In ththeseese situations, we establish before the test is conducted what test statistic is needed )called situations, we establish before the test is conducted what test statistic is needed )called the

the !riti!al value!riti!al value* to claim statistical significance. (o, if we !now for a given sampling* to claim statistical significance. (o, if we !now for a given sampling distribution that a test statistic of plus or minus '.:; would only occur 69 of the time distribution that a test statistic of plus or minus '.:; would only occur 69 of the time randomly, any test statistic that is '.:; or greater in absolute value would be statistically randomly, any test statistic that is '.:; or greater in absolute value would be statistically significant. In an analysis where a test statistic was exactly '.:;, you would have a 69 significant. In an analysis where a test statistic was exactly '.:;, you would have a 69 chance of being wrong if you claimed statistical significance. If the test statistic was chance of being wrong if you claimed statistical significance. If the test statistic was -.88, statistical significance could also be claimed but the probability of being wrong -.88, statistical significance could also be claimed but the probability of being wrong would be much less )about .88+ if using a +"tailed test or two"tenths of one percent< would be much less )about .88+ if using a +"tailed test or two"tenths of one percent< 8.+9*. oth .86 and .88+ are !nown as alpha< the probability of a Type I error.

8.+9*. oth .86 and .88+ are !nown as alpha< the probability of a Type I error.

%hen conducting statistical tests with computer software, the exact probability of a Type %hen conducting statistical tests with computer software, the exact probability of a Type I error is calculated. It is presented in several formats but is most commonly reported as I error is calculated. It is presented in several formats but is most commonly reported as "p #"

"p #" or or "Sig$""Sig$" or or "Signif$""Signif$" or or "Signifi!an!e$""Signifi!an!e$" =sing >p ?> as an example, if a priori you =sing >p ?> as an example, if a priori you established a threshold for statistical significance at alpha .86, any test statistic with established a threshold for statistical significance at alpha .86, any test statistic with significance at or less than .86 would be considered statistically significant and you significance at or less than .86 would be considered statistically significant and you would be required to re#ect the null hypothesis of no difference. The following table lin!s would be required to re#ect the null hypothesis of no difference. The following table lin!s p

p vvaalluuees s wwiitth h a a bbeenncchhmmaarr! ! aallpphha a oof f ..8866@@

%

% ## AAllpphhaa %%rroo&&aa&&iilliity ty oof Tf Tyyppe I e I EErrrroorr FFiinnaal l ''ee!!iissiioonn ..8866 ..8866 669 c9 chhaanncce e ddiiffffeerreenncce ie is ns noott

significant significant (tatistically (tatistically significant significant ..''88 ..8866 ''889 c9 chhaanncce de diiffffeerreenncce ie is ns noott

significant significant /ot statistically /ot statistically significant significant ..88'' ..8866 ''9 c9 chhaanncce e ddiiffffeerreenncce ie is ns noott

significant significant (tatistically (tatistically significant significant ..::;; ..8866 ::;;9 c9 chhaanncce de diiffffeerreenncce ie is ns noott

significant significant /ot statistically /ot statistically significant significant (teps to

(teps to 0ypothesis Te0ypothesis Testingsting

0ypothesis testing is used to establish whether the differences exhibited by random 0ypothesis testing is used to establish whether the differences exhibited by random samples can be inferred to the populations from which the samples originated.

samples can be inferred to the populations from which the samples originated. eneral Assu)ptions

eneral Assu)ptions

•

• Population is normally distributedPopulation is normally distributed

•

• 3andom sampling3andom sampling

•

• utually exclusive comparison samplesutually exclusive comparison samples

•

For

For intervalinterval / / ratioratio data use data use 

t"tests, Pearson correlation, A/2BA, regression t"tests, Pearson correlation, A/2BA, regression For

For nominalnominal / / ordinalordinal data use data use 

ifference of proportions, chi square and related measures of association ifference of proportions, chi square and related measures of association State the Hypothesis

State the Hypothesis

/ull 0ypothesis )0o*@ There is no difference between CCC and CCC. /ull 0ypothesis )0o*@ There is no difference between CCC and CCC.  Alternati

 Alternative 0ypve 0ypothesis )othesis )0a*@ Th0a*@ There is a dere is a differenifference betwce between CC aeen CC and CC.nd CC.

/ote@ The alternative hypothesis will indicate whether a '"tailed or a +"tailed test /ote@ The alternative hypothesis will indicate whether a '"tailed or a +"tailed test is utilied to re#ect the null hypothesis.

is utilied to re#ect the null hypothesis.

0a for '"tail tested@ The CC of CC is greater )or less* than the CC of CC. 0a for '"tail tested@ The CC of CC is greater )or less* than the CC of CC. Set the Re*e!tion Criteria

Set the Re*e!tion Criteria

This determines how different the parameters andDor statistics must be before the This determines how different the parameters andDor statistics must be before the null hy

null hypothesis pothesis can be can be re#ected. re#ected. This >This >region region of re#eof re#ection> iction> is bases based on ad on alpha )lpha ) * ""* "" the error associated with the confidence level. The point of re#ection is !nown as the error associated with the confidence level. The point of re#ection is !nown as the critical value.

the critical value. Co)pute the Test Statisti! Co)pute the Test Statisti!

The collected data are converted into standardied scores for comparison with The collected data are converted into standardied scores for comparison with the critical value.

the critical value.

'e!ide Results of Null Hypothesis 'e!ide Results of Null Hypothesis

If the test statistic equals or exceeds the region of re#ection brac!eted by the If the test statistic equals or exceeds the region of re#ection brac!eted by the critical value)s*, the null hypothesis is re#ected. In other words, the chance that critical value)s*, the null hypothesis is re#ected. In other words, the chance that the difference exhibited between the sample statistics is due to sampling error is the difference exhibited between the sample statistics is due to sampling error is remote""there is an actual difference in the population.

remote""there is an actual difference in the population.

Eet us consider an experiment involving two groups, an experimental group and a Eet us consider an experiment involving two groups, an experimental group and a control group.

control group. The experimenter li!es The experimenter li!es to test wto test whether the treatment )values hether the treatment )values clarificationclarification lessons* will improve the

lessons* will improve the self"concept of the experimental group. self"concept of the experimental group. The same treatment isThe same treatment is not give

not given to n to the contthe control grourol group. p. It is It is prepresumesumed that any differd that any differencence e betbetweeween the n the twotwo groups after the treatment can be attributed to the experimental treatment with a certain groups after the treatment can be attributed to the experimental treatment with a certain degree of confidence.

degree of confidence.

The hypothesis for this experiment can be stated in various ways@ The hypothesis for this experiment can be stated in various ways@ a*

a* /o ex/o existence istence or exior existence stence of a dof a differenifference betwce between green groupsoups 0

0oo@ @ TheThere is re is no signino significaficant diffent differencrence in e in selfself"con"concept betwcept between the groupeen the group

exposed to values clarification lessons and the group not exposed to the exposed to values clarification lessons and the group not exposed to the same.

same. 0

0aa@ @ The selfThe self"con"concepcept t of the group exposof the group exposed to ed to valvalues clariues clarificafication lesstion lessonsons

differ significantly that of the other group. differ significantly that of the other group. b*

b* /o ex/o existence istence or exisor existence otence of an ef an effect ffect of the of the treatmenttreatment 0

0oo@ @ There is no There is no significant effect of the vsignificant effect of the values clarification lessons on the self"alues clarification lessons on the self"

concept of the students. concept of the students. 0

0a@a@ Balues clarification lessons have no significant effect on the self"conceptBalues clarification lessons have no significant effect on the self"concept

of students. of students. c*

0

0oo@ @ The self"coThe self"concept of the studentncept of the students is not significants is not significantly related to the valuesly related to the values

clarification lessons conducted on them. clarification lessons conducted on them. 0

0aa@ @ The self"cThe self"concept of the studoncept of the students is not significaents is not significantly related to the vantly related to the valueslues

clarification lessons they were exposed to. clarification lessons they were exposed to.

%ara)etri! %ara)etri! TeTestst What are the parametric tests? 

What are the parametric tests?  The parame

The parametric tests tric tests are tests are tests thathat t reqrequire normal distriuire normal distributibution, the on, the levlevels els of of  mea

measursuremeement nt of of whwhich ich arare e exexprepressessed d in in an an inintertervaval l or or ratratio io dadatata. . ThThe e folfollowlowinging parametric tests are@

parametric tests are@

t"test for one population mean compared to a sample mean t"test for one population mean compared to a sample mean t" test for Independent (amples

t" test for Independent (amples t" test for &orrelated (amples t" test for &orrelated (amples " test, for Two (ample eans " test, for Two (ample eans " test for 2ne (ample roup " test for 2ne (ample roup F" test

F" test )A/2BA)A/2BA**

3 )Pearson Product oment &oefficient of &orrelation* 3 )Pearson Product oment &oefficient of &orrelation* G4aHbx )(imple Einear 3egression Analysis*

G4aHbx )(imple Einear 3egression Analysis* G4

G4 bb00 _HH bb11 x x11 _HH bb22 x x22 _{H... HH... H} bbnn x xnn

)ultiple 3egression Analysis* )ultiple 3egression Analysis*

Co)paring a %opulation Mean to a Sa)ple Mean (

Co)paring a %opulation Mean to a Sa)ple Mean ( T+testT+test 1xample '@

1xample '@ &ompare the mean age of &ompare the mean age of incoming students to the !nown mean incoming students to the !nown mean age for allage for all pre

previvious ous incincomioming ng stustudedentsnts. . A A rarandndom om sasamplmple e of of -8 -8 incincomioming ng colcolleglege e frefreshshmenmen revealed the following statistics@ mean age ':.6 years, standard deviation ' year. The revealed the following statistics@ mean age ':.6 years, standard deviation ' year. The college database shows the mean age for previous incoming students was '.

college database shows the mean age for previous incoming students was '.

Solving &y the Step,ise Method Solving &y the Step,ise Method

I$

%ro&le)-I$ %ro&le)- Is there a signifIs there a significaicant diffent differenrence betwce between the mean age of past colleen the mean age of past collegeege students and the mean age of current incoming college studentsJ

students and the mean age of current incoming college studentsJ II$ Hypothesis

II$ Hypothesis 0o@

0o@ There There is no is no significansignificant dift difference ference between between the the mean mean age age of pof past ast collegecollege students and the mean age of current incoming college students.

students and the mean age of current incoming college students. 0o@

0o@  x x

´´

11

=

=´´

 x x22

0a@ There is a significant difference between the mean age of past college 0a@ There is a significant difference between the mean age of past college

students and the mean age of current incoming college students. students and the mean age of current incoming college students.

III$ Level of Signifi!an!e

Set the Re*e!tion Criteria

(ignificance level .86 alpha, +"tailed test egrees of Freedom 4 n"' or +:

&ritical value from t"distribution 4 +.856

I.$ Statisti!s

Co)pute the Test Statisti!

´

 x

−

¿

s √ n

−

1 t 

=¿

t 

=

19.5

−

18 1 √ 30

−

1 t 

=

1.5 1 √ 29 t 

=

1.5 .186 t  _{4 .8;6}

.$ 'e!ision Rule- If the t" computed value is greater than or beyond the t"tabular D critical value, re#ect  H 0 _.

.I Con!lusion- iven that the test statistic ).8;6* exceeds the critical value )+.856*, the null hypothesis is re#ected in favor of the alternative. There is a statistically significant difference between the mean age of the current class of incoming students and the mean age of freshman students from past years. In other words, this year7s freshman class is on average older than freshmen from prior  years.

If the results had not been significant , the null hypothesis would not have been re#ected. This would be interpreted as the following@ There is insufficient evidence to conclude there is a statistically significant difference in the ages of current and past freshman students.

What is the t-test for independent samples? 

The t"test is a test of difference between two independent groups. The means are being compared  x

´

1  _against  x

´

2 _.

When do we use the t-test for independent samples? 

The t"test for independent samples is used when we compare means of two independent groups.

%hen the distribution is normally distributed, (! 4 8 and Ku 4 .+;6.

we use interval or ratio data. the sample is less than -8.

Why do we use the t-test for independent sample? 

The t"test is used for independent sample because it is more powerful test compared with other tests of difference of two independent groups.

How do we use the t-test for independent samples? 

t 4

n1+¿_n2−2 SS1

+

SS2

¿

¿

[

1 n1

+

1 n2

]

¿

√ 

¿

´

 x1

− ´

 x2

¿

%here@ t 4 the t test

´

 x1 _{4 the mean of group '}

´

 x2  _{4 the mean of group +}

SS1  _{4 the sum of squares of group '}

SS_{2 4 the sum of squares of group +}

n1  _{4 the number of observations in group '}

n2  _{4 the number of observations in group +}

or 

t 4

n1+¿n2−2

(

n1

−

1

)

(

s1

)

2

+

(

n1

−

1

)

(

s2

)

2

¿

¿

[

1 n1

+

 1 n2

]

¿

√ 

¿

´

 x₁

− ´

 x₂

¿

%here@

  Typeequationhere .

t 4 the t test

´

 x1

4 the mean of group ' or sample '

´

 x2

 _{4 the mean of group + or sample +}

S1

 4 the standard deviation of group ' or sample '

S2

_{4 the standard deviation of group + or sample +}

n_{1  4 the number of observations in group '} n2  _{4 the number of observations in group +}

E/a)ple 0$ The following are the scores of '8 male and '8 female A students in spelling. Test the null hypothesis that there is no significant difference between the performance of male and female A students in the rest. =se the t"test .86 level of significance.

ale ) x1 _{* Female )} x2 _* '5 '+ ' : 'L '' '; 6 5 '8 '5 -'+ L '8 + : ; 'L '- Solution-ale Female  x1  x  2 1  x2 x  2 2 '5 ':; '+ '55 ' -+5 : ' 'L +: '' '+' '; +6; 6 +6 5 '; '8 '88 '5 ':; - : '+ '55 L 5: '8 '88 + 5 : ' ; -; 'L +: '- ';: CCCCCC CCCCCCC CCCCCCC CCCCCCC

∑

 x1  4'-' _Σ X 1 2 4 ':'

∑

 x2 4L _Σ X 2 2  4 L- n1 _{4 '8} n2 _{4 '8}

´

 x1  _{4 '-.'}  x

´

2 _{4 L.}

t 4

n1+¿n2−2

(

n1

−

1

)

(

s1

)

2

+

(

n1

−

1

)

(

s2

)

2

¿

¿

[

1 n1

+

 1 n2

]

¿

√ 

¿

´

 x₁

− ´

 x₂

¿

 t 4

13.1

−

7.8

√

[

174.9

+

129.6 10

+

10

−

2

][

1 10

+

1 10

]

t

4

5.3

√

[

304.5 18

]

(

.1

+

.1

)

t

4

5.3

√ 

(

16.92

) (

.2

)

t 4

5.3 √ 3.384

t 4

5.3 1.8395 t 1 2$33

Solving &y the Step,ise Method I$

%ro&le)-Is there a significant difference between the performance of the male and female students in spellingJ

II$

Hypothesis- H 0 : _{There is no significant difference between the performance}

of the male and the female A students in spelling.  H 0 :  x

´

1

=´

 x2

 H _a: _{There is a significant difference between the performance of} 

the male and the female A students in spelling.  H _a:  x

´

₁

=´

 x₂

III$ Level of Signifi!an!e-α 

=

.05 df 4 n1

+

n2 _{" +} = '8 H '8 M + = ' t..86 = +.'8' t"tabular value at .86

I.$ Statisti!s- t"test for two independent (amples

.$ 'e!ision Rule-  If the t" computed value is greater than or beyond the t" tabular D critical value, re#ect  H 0 _.

.I$

Con!lusion-(ince the t"computed value of +. is greater than t"tabular value of +.'8' at .86 level of significance with ' degrees of freedom, the null hypothesis is re#ected in favor of the research hypothesis. This means that there is a significant difference between the performance of the male and female A students in spelling, implying that the male students performed better than the female students considering that the meanDaverage score of the male students of '-.' is greater compared to the average score of female students of only L..

1xample +.

Two groups of experimental rats were in#ected with tranquilier at '.8 mg. And '.6 mg dose respectively. The time given in seconds that too! them to fall asleep in hereby given. =se the t"test for independent samples at .8' to test null hypothesis that the difference in dosage has no effect on the length of time it too! them to fall asleep.

'.8 mg dose@ :., '-.+, ''.+, :.6, '-.8, '+.', :., '+.-, L.:, '8.+, :.L. '.6 mg dose@ '+.8, L.5, :., ''.6, '-.8, '+.6, :., '8.6, '-.6. (olution@ )'.8 mg dose* )'.6 mg dose* N' N'+ N+ N++ :. :;.85 '+.8 '55.88 '-.+ 'L5.+5 L.5 65.L; ''.+ '+6.55 :. :;.85 :.6 :8.+6 ''.6 '-+.+6 '-.8 ';:.88 '-.8 ';:.88 '+.' '5;.5' '+.6 '6;.+6 :. :;.85 :. :;.85 '+.- '6'.+: '8.6 ''8.+6 L.: ;+.5' '-.6 '+.+6 '8.+ '85.85 ΣN+ 4'88 ΣN++4 ''58.5 :.L :5.8: n+4 :

ΣN' 4''.L

´

 x1

=

11.11 _ΣN '+ 4'-8:.+6 /' 4 '' N'4'8.L: SS1  _{4 " Σ} 1

−¿

2  X _¿

(

∑

 x1

¿

2 n1 SS2  _{4 Σ} X 2 2  "

(

∑

 x2

¿

2 n2 4 '-8:.+6" 118.7

¿

2

¿

¿

¿

4 ''58.5 M 100

¿

2

¿

¿

¿

SS1  ₄ SS2 ₄

t4

n1+¿_n2−2 SS1

+

SS2

¿

¿

[

1 n1

+

1 n2

]

¿

√ 

¿

´

 x1

− ´

 x2

¿

4

10.79

−

11.11

√

[

28.37

+

29.73 11

+

9

−

2

][

1 11

+

1 9

]

4

−

0.32

√

[

58.10 18

]

(

.09

+

.11

)

4

−

0.32

√ 

(

3.23

)(

.2

)

4

−

0.32 √ .646

4

−

0.32 .8037

t 4 "8.58

Solving &y the Step,ise Method

29.73 28.37

I$ %ro&le) - Is there a significant difference brought about by the dosages on the length of time it too! for the rats to fall asleep

II$ Hypothesis-  H 0 :   _{There is no significant differences brought about by the}

dosages on the length of time it too! the rats to fall asleep.

 H _a: _{There is a significant difference brought about by the}

dosages on the length of time it too! for the rats to fall asleep.

III$ Level of Signifi!an!e-α 

=

.01

df 4 n1

+

n2 _{" +}

= '' H : M + = '

t.05 = "+.

I.$ Statisti!s- t"test for two independent samples

.$ 'e!ision Rule- If the t" computed value is greater than or beyond the t" tabular D critical value, re#ect  H 0 _.

.I$ Con!lusion- (ince the t"computed value of D".58D is within the critical value of D"+.D at .8' level of significance with ' degrees of freedom, the null hypothesis is not re#ected. This means that no significant difference was brought about by the dosages on the length of time it too! for the rats to fall asleep.

E/a)ple 4. To find out whether a new serum would arrest leu!emia, '; patients on the advanced stage of the disease were selected. 1ight patients received the treatment and eight did not. The survival was ta!en from the time the experiment was conducted.

/o treatment %ith Treatment

N' N'+ N+ N++ 2$0 5$50 5$2 06$75 4$2 08$25 9$0 27$80 4$8 :$88 9$8 29$88 2$3 6$35 5$7 20$07 2$0 5$50 4$: 09$20 0$2 0$55 5$4 03$5: 0$3 4$25 9$2 26$85 0$: 4$70 4$: 09$20

∑

 x1  4 '.' _Σ X 1 2 4 55.':

∑

 x2  4 -;+ _Σ X 2 2  4 ';6.L; n₁ 4  n2 _{4 }

´

 x1  _{4 +.+}  x

´

2 _{4 5.6+}

Solution- (olve for

∑

 x1, _Σ X 1 2

and  x

´

1 _{li!ewise for}

∑

 x_{2 , Σ} X ₂

2

and

´

 x2.  _{Then solve for the sum of the squares of group ', and sum of squares of group +}

SS1  _{4 " Σ} X 1 2 "

(

∑

 x1

¿

2 n1 SS2  _{4 Σ} X 2 2  "

(

∑

 x2

¿

2 n2 4 55.': " 18.7

¿

2

¿

¿

¿

4 ';6.L; M 36.2

¿

2

¿

¿

¿

4 55.': M 58.:6 4';6.L; M ';-.8 SS1

=¿

 ₄₄ SS2 _{4 0$:7}

t 4

n1_+¿n 2−2 SS1

+

SS2

¿

¿

[

1 n1

+

1 n2

]

¿

√ 

¿

´

 x1

− ´

 x2

¿

4

2.26

−

4.52

√

[

3_.24

+

1_.96 8

+

8

−

2

][

1 8

+

1 8

]

4

−

2.26

√

[

5.2 14

][

2 8

]

4

−

2.26

√ 

(

.37

) (

.25

)

4

−

2.26 √ .0925

4

−

.30412.26

t 4

-7.43 Solving &y the Step,ise Method

I$ %ro&le)- %ill the new serum arrest leu!emia for the  patients who had reached the advanced stageJ

II$

 H 0 : _{The new serum will not arrest the leu!emia of the  patients who}

had reached the advanced stage of the disease.

 H 1 : _{The new serum will arrest the leu!emia of the  patients who had}

reached an advanced stage of the disease. III$ Level of

Signifi!an!e-α 

=

.05

df 4 n1

+

n2 _{" +}

=  H  M + = '5

t .05  = "+.'56

I.$ Statisti!s- t"test for independent samples

.$ 'e!ision Rule- If the t"computed value is greater than or beyond the critical or tabular  value, re#ect  H 0.

.I$ Con!lusions- The t"computed value is D"L.5-D which is beyond the critical value of  D"+.'56D at .86 level of significance with '5 degrees of freedom. The null hypothesis therefore is disconfirmed in favor of the research hypothesis. This means that the new serum will arrest the advanced stage of leu!emia.

What is the t-test correlated samples? 

The t"test for correlated samples is another parametric test applied to one group of samples. It can be used in the evaluation of a certain program or treatment. (ince this is another parametric test, conditions must be met li!e the normal distribution and the use of interval or ratio data.

When do we use the t-test for correlated samples? 

The test for correlated samples is applied when the mean before and the mean after are being compared. The pretest )mean before* is measured, the treatment of the intervention is applied and then the posttest )mean after* is li!ewise measured. Then the two means )pretest vs. the posttest* are compared.

Why do we use the t-test for correlated samples? 

The t"test for correlated samples is used to find out if a difference exists between the before and after means. If there is a difference in favor of the posttest then the treatment or intervention is effective. 0owever, if there is no significant difference then the treatment is not effective.

This is the appropriate test for evaluation of government programs. This is used in an experimental design to test the effectiveness of a certain technique or  method or program that had been developed.

How do we use the t-test for correlated samples?  The formula is

t4

 D

¿

2

¿

¿

n n

(

n

−

1

)

¿

 D2

−¿

√ 

¿

´

 D

¿

%here@  D 4 the eman difference between the pretest

´

and the posttest.

Σ D

2

4 the sum of the squares of the difference between the pretest and the posttest.

Σ 4 the summation of the difference between the pretest and the posttest.

n 4 the sample sie

E/a)ple 0$ An experiment study was conducted on the effect of programmed materials in 1nglish on the performance of +8 selected college students. efore the program was implemented the pretest was administered and after 6 months the same instrument was used to get the posttest result. The following is the result of the experiment. =se α  at .86 level. Pretest Posttest N' N'+  D 2 +8 +6 "6 +6 -8 -6 "6 +6 '8 +6 "'6 ++6 '6 +6 "'8 '88 +8 +8 8 8 '8 +8 "'8 '88 ' ++ "5 '; '5 +8 "; -; '6 +8 "6 +6 +8 '6 6 +6 ' -8 "'+ '55 '6 '8 6 +6 +8 +6 "6 ' ' '8  ;5 58 56 "6 +6 '8 '6 "6 +6 '8 '8 8 8 '+ ' "; -; +8 +6 "6 +6 CCCCCCCCCCCCC CCCCCCCCCCCCC C  Σ 4 "' Σ D 2  4 :5

´

 D 4

−

2081

´

 D  4 " 5.86

What are the steps in using the t-test for correlated samples? 

(ubtract the difference between the two observations before and after the treatment.

Find the summation of the difference algebraically Σ )consider the H and  M sign*

&ompute  D  the mean difference,

´

Σ n

(quare the difference between the before and after observation and get the summation that is Σ D

2

etermine the number of observation that is small letter n.

If the t"compound value is greater than or beyond the critical value disconfirm the null hypothesis and confirm the research or alternative hypothesis.

t 4

 D

¿

2

¿

¿

n n

(

n

−

1

)

¿

 D2

−¿

√ 

¿

´

 D

¿

4

−

81

¿

2

¿

¿

20 20

(

20

−

1

)

¿

947

−¿

√ 

¿

−

4.05

¿

4

−

4.05

√

947

−

328.05 20

(

19

)

4

−

4.05

√

618.95 380

4

−

4.05 √ 1.6288

4

1.2762

−

4.05

t 1 +4$06

Solving &y the Step,ise Method

I$ %ro&le)- Is there a significant difference between the pretest and the posttest on the use of programmed materials in 1nglishJ

II$

Hypotheses- H _{0 :   There is no significant difference between the pretest and} the posttest, or the use of the programmed materials did not affect the students$ performance in 1nglish.

 H _{1 :  The posttest result is higher than the pretest result.}

III$ Level of Signifi!an!e-α 

=

.05

df 4 n"' = +8"' = '5 t.05 = "'.L+:

I.$ Statisti!s- t"test for correlated samples

.$ 'e!ision Rule- If the t"computed value is greater than or  beyond the critical value, re#ect  H 0.

.I$ Con!lusion@ The t"compound value of -.'L is beyond the t"critical value of D"'.L-D at .86 level of significance with ': degrees of  freedom. The null hypothesis is therefore re#ected in favor of  the research hypothesis. This means that the posttest result is higher than the pretest result. It implies that the use of the programmed materials in 1nglish is effective.

What is the z-test? 

The "test is another test under parametric statistics which requires normality of  distribution. It uses the two population parameters µ and σ.

It is used to compare two means, the sample mean, and the perceived population mean.

It is also used to compare the two sample means ta!en from the same population. It is used when the samples are equal to or greater than -8. The " test can be applied in two ways@ the 2ne"(ample ean Test and the Two"(ample ean Test.

The tabular value of the "test at .8' and .86 level of significance is shown below.

Test Eevel of (ignificance

.8' .86

2ne"tailed O +.-- O '.;56 Two"tailed O +.6L6 O '.:; What is the z-test for one sample group? 

The "test for one sample group is used to compare the perceived population mean µ against the sample mean,  X 

´

When is the z-test for a one-sample group? 

The one"sample group test is used when the sample is being compared to the perceived population mean. 0owever if the population standard deviation is not !nown the sample standard deviation can be used as a substitute.

Why is the z-test used for a one-sample group? 

The "test is used for a one"sample group because this is appropriate for  comparing the perceived population mean µ against the sample mean  X  . %e

´

are interested if significant difference exists between the population µ against the sample mean. For instance a certain tire company would claim that the life span odd its product will last +6,888 !ilometers. To chec! the claim, sample tires will be tested by getting sample mean  X  .

´

How do we use the z-test for a one-sample group? 

The formula is

 4

´

 x

−¿

¿

√ n

¿

¿

%here@

´

 X  _{4 sample mean}

µ 4 hypothesied value of the population mean

σ 4 population standard deviation / 4 sample sie

E/a)ple 0$ The A& company claims that the average lifetime of a certain tire as at least +,888 !m. To chec! the claim, a taxi company puts 58 of these tires on its taxis and gets a mean lifetime of +6,6;8 !m. %ith a standard deviation of ',-68 !m, is the claim trueJ =se the "test at .86.

What are the steps in using the z-test for a one sample group?  (olve for the mean of the sample and also the standard deviation if the population σ is not !nown.

(ubtract the population from the sample mean and multiply it by the square root of n sample.

ivide the result from step + by the population standard deviation on the sample standard deviation if the population σ is not !nown.

&ompare the result in the table of the tabular value of the "test at .8' and .86 level of significance.

Eevel of (ignificance

Test .8' .86

2ne tailed O +.-- O '.;56

Two tailed O +.6L6 O '.:;ɑɑ

If the "compound value is greater than or beyond the critical value, re#ect the null hypothesis and confirm the research hypothesis.

I$ %ro&le)- Is the claim true that the average lifetime of a certain tire is at least +,888 !mJ

II$

Hypotheses- H 0 : _{The average lifetime of a certain tire is +,888 !m.}

 H 1 : _{The average lifetime of a certain tire is not +,888 !m.}

III$ Level of Signifi!an!e-4 .86

ɑ

I.$

Statisti!s-"test for a one"tailed test Co)putation-; 

 4

´

 x

−¿

¿

√ n

¿

¿

4

(

25,560

−

1,35028,000

)

√ 40

4

(−

2,4401,350

)(

6.32

)

4

−

15,420.81,350

 4 "''.5+

.$ 'e!ision Rule- if the  computed is greater than or beyond the  tabular value, re#ect the 0o.

.I$ Con!lusion- (ince the  computed of "''.5+ is beyond the critical value of "'.;56 at .86 level of significance the research hypothesis is re#ected which means that the average lifetime of a certain tire is not +,888 !m.

What is the z-test for a two-sample mean test? 

The "test for a two"sample mean test is another parametric test used to compare the means of two independent groups of samples drawn from a normal population if  there are more than -8 samples for every group.

When do we use the z-test for two sample mean? 

The "test for two"sample mean is used when we compare the means of samples of independent groups ta!en from a normal population.

Why do we use the z-test? 

The "test is used to find out if there is a significant difference between the two populations by only comparing the sample mean of the population.

How do we use the z-test for a two-sample mean test?  The formula is

 4

´

 x1

−´

 x2

√

 s1 2 n1

+

 s2 2 n2 where@

´

 x1 _{4 the mean of sample '}

´

 x2 _{4 the mean of sample +}

s1 2

4 the variance of sample ' s2

2

4 the variance of sample + n1 _{4 sie of sample '}

n2 _{4 sie of sample +}

1xample '. An admission test was administered to incoming freshmen in the colleges of /ursing and Beterinary edicine with '88 students each college randomly selected. The mean scores of the given samples where  x

´

1 ₄ :8 and  x

´

2 _{4 6 and the variances of the test scores were 58 and -6}

respectively. Is there a significant difference between the two groupsJ =se .8' level of significance.

What are the steps in solving for the z-test for two-sample mean? 

&ompute the sample mean of group ',  x

´

1  _{and also the sample mean of group} +,  x

´

2

&ompute the standard deviation of group ' SD1   _{and standard deviation of} 

group + SD2 _.

(quare the SD1 _{of group ' to get the variance of group '} s1 2

  and also square the SD2  _{of the group + to get the variance of group +} s2

2

.

etermine the number of observation in group ' n1  _{and also the number of} 

observation in group + n2 _.

&ompare the  computed value from the " tabular value at certain level of  significance.

Eevel of (ignificance

Test .8' .86

2ne"tailed O +.-- O '.;56 Two"tailed O +.6L6 O '.:;

If the computed "value is greater than or beyond the critical value re#ect the null hypothesis and do not re#ect the research hypothesis.

I$ %ro&le)- Is there a significant difference between the two groupsJ II$

Hypotheses- H 0 :  x

´

1

=´

 x2

 H 1 :  x

´

1≠

 ´

 x2

III$ Level of Signifi!an!e α   4 .8'  4 O +.6L6

do not re#ect

 H 0

re#ect

 H 0

  re#ect

 H 0

"+.6L6

"'

8

H' 8H+.6L6

 

"'.:;

H'.:;

I.$ Statisti!s- "test for two"tailed test

 4

´

 x1

−´

 x2

√

 s₁2 n1

+

 s2 2 n2

¿

90

−

85

√

40 100

+

35 100

4

5

√

75 100

1

5 √ .75

1

5 .866

 4 6.LL5

.$ 'e!ision Rule-  If the "computed value is greater than or beyond the  4tabular value, re#ect the null hypothesis..

.I$ Con!lusion-  (ince the "computed value of 6.LL5 is greater than the " tabular value of +.6L6 at .8' level of significance, the research hypothesis is re#ected which means that there is a significant difference between the two groups. It implies that the incoming freshmen of the &ollege of /ursing are better than the incoming freshmen of the &ollege of Beterinary edicine.

What is the F-test? 

The F"test is another parametric test used to compare the means of two or more groups of independent samples. It is also !nown as the analysis of variance, )A/2BA*.

The three !inds of analysis of variance are@ one"way analysis of variance

two"way analysis of variance three"way analysis of variance

The F"test is the analysis of variance )A/2BA*. This is used in comparing the means of two or more independent groups. 2ne"way A/2BA is used when there is only one variable involved. The two"way A/2BA is used when two variables are involved@ the column and the row variables. The researcher is interested to !now if there are significant differences between and among columns and rows. This is also used in loo!ing at the interaction effect between the variables being analyed.

Ei!e the t"test, the F"test is also a parametric test which has to meet some conditions, and the data to be analyed if they are normal are expressed in interval or ratio data. This test is more efficient than other tests of difference.

Why do we use the F-test? 

The F"test is used to find out if there is a significant difference between and among the means of the two or more independent groups.

When do we use F-test? 

The F"test is used when there is normal distribution and when the level of  measurement is expressed in interval or ratio data #ust li!e the t"test and the "test.

How do we use the F-test? 

To get the F computed value, the following computations should be done. CF 

=

(¿)

2

 N 

TSS is the total sum of squares minus the &F, the correction factor. <SS is the between sum of squares minus the &F correction factor.

=SS is the sum of squares or it is the difference between the T(( minus ((.

 After getting the T((, (( and %((, the A/2BA table should be constructed.  A/2BA Table (ources of F"Balue f (( ( &omputed Tabular   etween K"' (( BSS df   SB  S! 

 =

 F  see the table at .86 or the desired level of significance %ithin )/"'*")K"'* %(( !SS df  wD df between

roup andwD group

Total /"' T((

What are the steps in solving for the F-value? 

The A/2BA table has five columns. These are@

sources of variations, degrees of freedom, sum of squares, mean squares and the F"value, both the computed and the tabular  values.

The sources of variations are between the groups, within the group itself and the total variations.

The degrees of freedom for the total is the total number of  observation minus '.

The degrees of freedom from the between group is the total number of groups minus '.

The degrees of freedom for the within group is the total df minus the between groups df.

• The ( mean squares between is equal to the ((Ddf.

• The (% mean square within is equal to the %((Ddf.

• To get the F"computed value, divide (D(%.

• If the F"computed value at a given level of significance with the corresponding df$s of (( and %((.

• If the F computed value is greater than the F"tabular value, re#ect the null hypothesis in favour of the research hypothesis.

• %hen the F"computed value is greater than the F"tabular value the null is re#ected and the research hypothesis not re#ected which means that there is a significant difference between and among the means of the different groups.

1xample '@ A sari"sari store is selling 5 brands of shampoo. The owner is interested if there is a significant difference in the average sales of the four brands of shampoo for one wee!. The following data are recorded.

  rand A  &    L : + 5   -  - 6   6  5 L   ; L 6    : ; ; -  5 : 5 5   - '8 + 6

Perform the analysis of variance and test the hypothesis at .86 level of significance that the average sales of the four brands of shampoo are equal.

Solving &y the Step,ise Method

I$ %ro&le)- Is there a significant difference in the average sales of the four  brands of shampooJ

II$

Hypotheses- H 0 :  _{There is no significant difference in the average sales of the four brands}

of shampoo

 H 1 : _{There is a significant difference in the average sales of the four brands of} 

shampoo. 37

+

57

+

26

+

36

¿

2

¿

156

¿

2

¿

¿

¿

CF 

=

❑

1

+❑

2

+❑

3

+❑

4 n₁n₂n₃n₄

=¿

T((

 4

 x1 2

+

 x2 2

+

 x3 2

+

 x4 2

−

CF 

4 ++6H5L6H''8H+85";:.'5 4'8'5 M ;:.'5  T(( 4 '55.;

(( 4

 x1

¿

2

¿

 x₂

¿

2

¿

 x3

¿

2

¿

 x4

¿

2

¿

¿

¿

¿

¿

¿

4

37

¿

2

¿

57

¿

2

¿

26

¿

2

¿

36

¿

2

¿

¿

¿

¿

¿

¿

4 ':6.6LH5;5.'5H:;.6LH'6.'5";:.'5 4 :5'.5+";:.'5 (( 4 L+.+ %(( 4 T(( M (( 4'55.; M L+.+ %(( 4 L+.6

Analysis of .arious Ta&le F"Balue (ources of egrees of (um of ean

Bariation Freedom (quares (quares CCCCCCCCCCCCCCC  &omputed Tabular 

etween roups K"' - L+.+ +5.8: L.: -.8' %ithin roup )/"'*" )K"'* +5 L+.6 -.8+

Total /"' +L '55.;

III$ 'e!ision Rule- If the F computed value is greater than the F"tabular value, re#ect  H 0 _.

I.$ Con!lusion- (ince the F"computed value of L.: is greater than the F Mtabular value of -.8' at .86 level of significance with - and +5 degrees of freedom, the null hypothesis is re#ected in favor of  the research hypothesis which means that there is a significant difference in the average sales of the 5 brands of shampoo.

What is the Scheff  e ’s Test? 

´

To find out where the differences lies, another test must be used.

The F"test tells us that there is a significant difference in the average sales of the 5 brands of shampoo but as to where the difference lies, it has to be tested further by another test, the (cheff  e $s test formula.

´

´

 x1

−´

 x2

¿

2

¿

n1

+

n2

¿

S! 2

¿

¿

¿

 F " 

=¿

%here@  F "    _{4 (cheff}  e

´

_{$s test}

´

 x_{1 4 mean of group '}

´

 x2 _{4 mean of group +}

n1 _{4 number of samples in group '}

n2 _{4 number of samples in group +}

S! 2 _{4 within mean squares}

A vs$ < 5.28

−

8.14

¿

2

¿

¿

 F " 

=¿

¿

8.1796 42.28 49

¿

8.1796 .86  F " 

 ₄

9.5

A

vs

C

A

vs

'

 F " 

=¿

5.28

−

3.71

¿

2

¿

¿

¿

 F " 

=¿

5.28

−

5.14

¿

2

¿

¿

¿

1

2.4649 .86

1

.0196 .86  F " 

=¿

 F " 

=¿

<

vs

C

<

vs

'

 F " 

=¿

8.14

−

3.71

¿

2

¿

¿

¿

 F " 

=¿

8.14

−

5.14

¿

2

¿

¿

¿

1

19.6249 .86

1

9 .86  F " 

=¿

 F " 

=¿

C vs '

 F " 

=¿

−

5.14 3.71

¿

¿

¿

2

¿

¿

1

2.0449 .86  F " 

=¿

Co)parison of the Average Sales of the Four <rands of Sha)poo

Bet#een  F " 

(

 F .05

)

rand

(

 $ 

−

1

)

  Interpretation

(

3.01

) (

3

)

A vs  :.6' :.8- significant

A vs & +.L :.8- not significant

A vs  .8+ :.8- not significant

 vs & ++.+ :.8- significant

 vs  '8.5; :.8- significant

& vs  +.- :.8- not significant

.02 2.87

10.46 22.82

The above table shows that there is a significant difference in the sales between brand A and brand , brand  and brand & and also brand  and . 0owever, brands A and &, A and  and & and  not significantly differ in their average sales.

This implies that brand  is more saleable than brands A, & and .

E/a)ple 2$ The following data represent the operating time in hours of the - types of scientific poc!et calculators before a recharge is required. etermine the difference in the operating time of the three calculators. o the analysis of variance at .86 level of significance.

rand  F  _x 1  X 1 2  _F   x2  X 2 2  _F   x3  X 3 2 5.: +5.8' ;.5 58.:; 5. +-.85 6.- +.8: ;. 5;.+5 6.5 +:.'; 5.; +'.'; 6.; -'.-; ;.L 55.: ;.' -L.+' ;.6 5+.+6 L.: ;+.5' 5.- '.5: ;.- -:.;: ;.+ -.55 ;.: 5L.;' ;.L 55.: 6.- +.8: 6.- +.8: 6.: -5.' 5.' ';.' 5.- '.5: Σ x1 _{4 -+.'} Σ x1 2 4'L;.6L Σ x2 _{4 6+ Σ} x2 2  4 -8.L Σ x3  _45+.+ Σ  x3 2  4 +;8.5 n1 _{4 ;} n2  _{4 :} n3 _{4 L}

´

 x1  _{4 6.-6}  x

´

2  _{4 6.L}  x

´

3  ₄

;.8-I$ %ro&le)- Is there a significant difference in the average operating time in hours of the - types of poc!et scientific calculators before a recharge is requiredJ

II$

Hypotheses- H 0 : _{There is no significant difference in the average operating time in}

hours among the - types of poc!et calculators before a recharge is required.

 H 1 : _{There is a significant difference in the average operating time in}

hours of the - types of poc!et scientific calculators before a recharge is required. III$ Level of

Signifi!an!e-α   4 .86

df  4 + and ':

I.$ Statisti!s@

F"test one"way"analysis of variance

Co)putation-&F 4

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=SS 4 ':.;'

AN>.A TA<LE

(ources of egrees (um of ean &omputed

Bariation of Freedom (quares (quares F"Balue Tabular 

etween roups K"' + '.68 .L6 .L- -.6+

%ithin roup )/"'*")K"'* ': ':.;'

'.8-Total /"' +' +'.''

21.11

.$ 'e!ision Rule- If the F"computed value is greater than the F"tabular value, re#ect the  H 0 _.

.I$ Con!lusion- (ince the F"computed value of 8.L- is lesser than the F"tabular  value of -.6+ at .86 level of significance, the null hypothesis is not re#ected. This means that there is no significant difference in the average operating time in hours of the - types of poc!et scientific calculators before a recharge is required. What is the F-test two-way-A!"A with interaction effect? 

The F"test, two"way"A/2BA involves two variables, the column and the row variables.

It is used to find out if there is an interaction effect between two variables.

How do you use the F-test two-way-A!"A with interaction effect?  &onsider this example

1xample '. Forty"five language students were randomly assigned to one of three instructors and to one of the three methods of teaching then achievement was measured on a test administered at the end of  the term. =se the two"way A/2BA with interaction effect at .86 level of significance to test the following hypotheses@

'. H 0 _{@ There is no significant difference in the}

performance of the three groups of students under three different instructors.

 H _{1 @ There is a significant difference in the}

performance of the three groups of students under three different instructors.

+. H 0 _{@ There is no significant difference in the performance of the} three groups of students under three different methods of  teaching.

 H 1 _{@ There is a significant difference in the performance of the}

three groups of students under three different methods of  teaching.

-. H 0 _{@ Interaction effects are not present.}  H 1 _{@ Interaction effects are present.}

T=>+FACT>R AN>.A ,ith a signifi!ant Intera!tion TEACHER FACT>R  A  & ethods of Teaching 58 68 58 5' 68 5' 58 5 58 -: 5 - - 56 -

' Total ethod of  teaching + 58 56 68 5' 5+ 5; -: 5+ 5-- 5' 5-- 58 5+ Total ethod of  Teaching -58 58 58 5- 56 5' 5' 55 5' -: 55 -: - 5- - Total Total

Solving &y the Step,ise Method

I$ %ro&le)- '$ Is there a significant difference in the performance of  students under the three different teachersJ

+. Is there a significant difference in the performance of students under the three different methods of teachingJ

-. Is there an interaction effect between teacher and method of  teaching factorsJ

II$

Hypotheses-'. H 0 _{@ There is no significant difference in the performance of}  the three groups of students under three different instructors.

 H 1 _{@ There is a significant difference in the performance of the three}

groups of students under three different instructors.

+.  H 0 _{@ There is no significant difference in the performance of the}

three groups of students under three different methods of teaching.

 H 1 _{@ There is a significant difference in the performance of the three}

groups of students under three different methods of teaching. -.  H 0 _{@ Interaction effects are not present.}

 H _{1 @ Interaction effects are present.} III$ Level of Signifi!an!e

a =

.86

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df within

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df column

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T=>+FACT>R AN>.A ,ith Signifi!ant Intera!tion TEACHER FACT>R (Colu)n

 A  & ethod of  Teaching FA&T23 ' )row* 58 68 58 5' 68 5' 58 5 58 -: 5 - - 56 - Total ': +5' ':L Σ4 ;-; ethod of  Teaching FA&T23 + )row* 58 56 68 5' 5+ 5; -: 5+ 5-- 5' 5-- 58 5+ Total ':; +'8 ++5 Σ4 ;-8 ethod of  Teaching FA&T23 -)row* 58 58 58 5- 56 5' -' 55 5' -: 55 -: - 5- - Total +8' +'; ':: Σ4;'; Total 6:6 ;;L ;+8 4 '+

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AN>.A TA<LE (ources of  Bariation (( d  f  ( F"value &omput ed Tabul ar  Interpretati on etween &olumns 'L.'  + :.: +5.+ -.+; ( 3ows '5.86 5 L.8+ '.:6 -.+; /( Interacti on 'L.' 6 5 5;.L : '-.8- +.;- ( %ithin '+:.+ 8 -; -.6: Total 68.6  5 5 F"Balue &omputed@

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'-.8-F"Balue Tabular at .86

&olumns df  4 +D-; 4 -.+;

3ow df  4 +D-; 4 -.+;

Interaction df  4 5D-; 4

+.;-.$ 'e!ision Rule@ If the computed F value is greater than the F criticalDtabular value, re#ect  H 0 _.

.I$ Con!lusion- %ith the computed F"value )column* of +5.+ compared to the F" tabular value of -.+; at .86 level of significance with + and -; degrees of  freedom, the null hypothesis is re#ected in favor of the of the research hypothesis which means that there is a significant difference in the performance of the three

groups of students under three different instructors. It implies that instructor  is better than instructor A. %ith regard to the F +value )row* of '.:6, it is lesser than the F"tabular value of -.+; at .86 level of significance with + and -; degrees of  freedom. 0ence, the null hypothesis of no significant differences in the performance of the students under the three different methods of teaching is not re#ected.

0owever, the F"value )interaction* of '-.8- is greater than the F"tabular value of  +.;- at .86 level of significance with 5 and -; degrees of freedom. Thus, the research hypothesis is re#ected which means that interaction effect is present between the instructors and their methods of teaching. (tudents under instructor   have better performance under methods of teaching ' and - while students under instructor & have better performance under method +.

What is the #earson #roduct $oment %oefficient of %orrelation r? 

The Pearson Product oment &oefficient of &orrelation r is an index of relationship between two variables. The independent variable can be

represented by x while the dependent variable can also be represented by y. The value of r is H', ero to "'. If the value of r is H' or "', there is a perfect

correlation between x and y. 0owever, if r equals ero then x and y are independent of each other.

&onsider the x and y coordinates in the graph below. y

0igh

r 4 H

x

Eow

0igh

If the trend of the line graph is going upward, the value of r is positive. This indicates that as the value of x increase the value of y also increase. Ei!ewise, if  the value of x decreases, the value of y also decreases, the x and y being positively correlated.

y

0igh

r 4 CCC 

N

Eow

0igh

If the trend of the line graph is going downward, the value of r is negative. It indicates that as the value of x increases the corresponding value of y decreases, x and y being negatively correlated.

y

0igh

r 4 8

x

 Eow

0igh

If the trend of the line graph cannot be established either upward or downward, then r 4 8, indicating that there is no correlation between the x and y variables.

Why do we use r? 

%e use r because we want to analye if a relationship exists between two variables. If there is a relationship that exists between the x and y, then we can determine the extent by which x influences using the coefficient of determination which is equal to the square of r and multiplied by '889. This can answer or explain how much the independent variable influences the dependent variables or how much y depends on x. This is now the degree of relationship between x and y which can not be seen in other statistical tests of relationship.

%e also use r because it is a more powerful test of relationship compared with other  nonparametric tests.

When do we use r& the #earson #roduct $oment %oefficient of %orrelation? 

%e use the r to determine the index of relationship between two variables, the independent and the dependent variables. If there is relationship between the independent and the dependent variables, we can say that x influences y or y depends on x. however if there is no relationship that exists between x and y then x and y are independent of each other.

The value of r ranges from H' through ero "'. There is a perfect positive correlation of r 4 H', li!ewise there is a negative perfect correlation if the value of  r 4 "'. 0owever if r 4 8 then there is no correlation between the two variables x and y. If there is a positive correlation it indicates that for every increase of y or  for every decrease of x there is a corresponding decrease of y. If there is a negative correlation it indicates that for every increase of x there is a corresponding decrease of y, li!ewise for every decrease of x there is a corresponding decrease of y, li!ewise for every decrease of x there is a corresponding increase of y. The relationship is inverse.

The Pearson Product oment &oefficient of &orrelation represented by small letter r is determined by using the formula

r 4

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2

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2

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2n y¿ n x2

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√ 

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n xy

−

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%here@

r 4 the Pearson Product oment &oefficient of &orrelation

n 4 sample sie

Σxy 4 the sum of the product of x and y

Σx Σy 4 the product of the sum of Σ x and the sum of

Σ y Σ x 2 4 sum of squares of x Σ y 2 4 sum of squares of y

1xample '. elow are the midterm )x* and final )y* grades.  x75706590858580706590

 y80756595908590757090

What are the steps in solving for the value of r?  etermine the number of observation n.

et the sum of x that is Σx, the independent variable. (quare every x observation and get the sum of  x2 x2 .

et the sum of y the dependent variable Σy.

(quare every y observation and get the sum of  y 2

, that is Σ y

2

.

ultiply the x and y. Place it in column xy and get the sum of xy, that is

Σxy.

 Apply the formula indicated above and compare the computed r with the tabular value at a certain level of significance with n"+ degrees of freedom. If the computed r value is greater than r tabular value, disconfirm the null hypothesis and confirm the research hypothesis which means that there is a significant relationship between the two variables, x and y.

I$ %ro&le)@ Is there a significant relationship between the midterm and the final examination of '8 students in athematicsJ

II$

Hypotheses- H 0 _{@ There is no significant relationship between the midterm grades and} the final examinationDgrades of '8 students in athematics.

 H 1 _{@ There is a significant relationship between the midterm grades and}

the final examination grades of '8 students in athematics. III$ Level of Signifi!an!e@

4 .86 ɑ df 4 n"+ 4 '8"+ 4  r .86 4 .8-+ I.$ (tatistics@

Pearson Product oment &oefficient of &orrelation

x y _x2  y2 xy L6 8 6;+6 ;588 ;888 L8 L6 5:88 6;+6 6+68 ;6 ;6 5++6 5++6 5++6 :8 :6 '88 :8+6 668 6 :8 L++6 '88 L;68 6 6 L++6 L++6 L++6 8 :8 ;588 '88 L+88 L8 L6 5:88 6;+6 6+68 ;6 L8 5++6 5:88 5668 :8 :8 '88 '88 '88 Σx 4 LL6 Σy 4 '6 Σ x 2 4;8, :+6 Σ y 2 4;L, -+6 Σxy ;5,888

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