Anova presentation Essay

ANALYSIS OF

VARIANCE?

WHAT NULL SPECULATION

ISTESTED BY SIMPLY ANOVA?

ANALYSIS OF VARIANCE IS A STATISTICAL APPROACH USED TO

EVALUATION DIFFERENCES BETWEEN TWO OR MORE MEANS.

IT REALLY IS USED TO EVALUATION GENERAL RATHER THAN SPECIFIC

DISSIMILARITIES AMONG MEANS. THUS THE NULL HYPOTHESIS IS

CALLED AN OMNIBUS NULL HYPOTHESIS IT MEANS THAT AT

LEAST ONE INHABITANTS MEAN DIFFERS FROM THE OTHERS FROM FOR

LEASTONE VARIOUS OTHER MEAN.

THE ANOVA DOES NOT EXPOSE WHICH COUPLE IS SIGNIFICANT,

THUS A FOLLOW UP CHECK IS NECESSARY TO DETERMINEWHICH

MATCH IS DIFFERENT FROM EACH OTHER.

PRESUMPTIONS

 1 .

THE POPULATION HAVE SIMILAR

VARIANCE. THE ASSUMPTION IS NAMED

ASSUMPTION OF HOMOGENEITY OF

VARIANCE.

 2 .

THE PEOPLE ARE NORMALLY

GIVEN AWAY

 three or more.

EACH BENEFIT IS SAMPLED

INDEPENDENTLY

PRECISELY WHAT IS MEAN SQ . WITHIN?

Mean

square Within is the value

that approximate the population

variances of each

group( considering that the

groups will be homogenous as stated

in the supposition.

Formula:

WHAT IS INDICATE SQUARE

AMONG?

  MEAN

SQ . BETWEEN CAN BE AN

ESTIMATOR OF THE INHABITANTS

VARIANCE IN CASE THE PPULATION

MEANS ARE EQUAL. IF THE MEANS

ARE NOT EQUIVALENT IT ESTIMATIONS A

BIGGER QUANTITY.

 Formula:

= / (n-1)

FACTORS AND LEVELS

 FACTOR

– A ATTRIBUTE

UNDER CONSIDERATION, BELIEVED

TO INFLUENCE THE TESTED

OBSERVATIONS.

 LEVEL

– A VALUE FROM THE FACTOR.

EINE WAY ( FACTOR) ANOVA

ON THE WHOLE ONE WAY ANOVA TECHNIQUES CAN BE USED TO

STUDY THE RESULT OF E (> 2) LEVELS OF A SINGLE FACTOR

EXAMPLE:

THE TABLE SHOWS THE LIFETIMES UNDERNEATH CONTROLLED

CIRCUMSTANCES, IN HOURS IN EXCSS OF a thousand HRS, OF SAMPLES OF

60w ELECTRIC LIGHT BULBS OF THREE DIFFERENT BRAND

BRAND

one particular

2

three or more

16

18

26

12-15

22

thirty-one

13

twenty

24

21

16

30

14

twenty-four

24

SUMMARIZE DATA

COMPANY

1

2

3

SAMPLE SIZE

your five

5

a few

SUM

85

100

one hundred thirty five

SUM OF

SQUARES

1316

2040

3689

MEAN

sixteen

20

twenty-seven

VARIANCE

9

10

10

SINCE EACH OF THESE THREE SAMPLE VARIANCES IS AN

ESTIMATION OF THECOMMON POPULATION VARIANCE. A POOLED

ESTIMATE CAN BE CALCULATED INSIDE THE USUAL WAY.

= 10

THE VARIABILITY BETWEEN SAMPLES CAN BE COMPUTED BY TE

3 SAMPLE MEANS

BRAND

MPLE MEAN

you

2

a few

16

20

27

SUM

63

TOTAL OF

SQUARES

1385

INDICATE

21

VARIANCE

31

THIS

DIFFERENCE, DENOTED SIMPLY BY IS CALLED DIFFERENCE BETWEEN

TEST MEANS BASED UPON (3-1) a couple of DEGREES OF LIBERTY,

BUT ONLY WHEN THE NULL IS HOLDS TRUE, IF THE NULL IS PHONY,

THEN THE FUTURE " SIGNIFICANT DFFERENCES AMONG

SAMPLE MEANS WILL BE SEEN.

ANOTHER WAY OF SOLVING

ANOVA

COMPUTATIONAL FORMULA

TOTAL SUM OF PIECES SSt – total coming from all squared info

Among sample amount of potager SSb = the quantity of square-shaped total per column divided by d – rectangular of summation of all data over in

SSw = SSt – SSb

MSt = SSt / n-1; MSb sama dengan SSb as well as (k-1); MSw = SSw / and - k

Anova computation (II)

=

1316

+2040 +3689 - = 7845 – 6615 = 430

-- 6615 = 1280+2000+3645 – 6615 = 6925 – 6615 = 310

MSb = 310 / two = 155

MSw sama dengan 120 /12 = 12

F = 155/10 = 15. 5

EXAMPLE a couple of

WITHIN A COMPARISON OF THE CLENING ACTION OF 4

DETERGENTS, twenty PIECES OF WHITE CLOTH HAD BEEN

FIRST DIRTY WITH INDIA INK. THE CLOTHS

HAD BEEN THEN CLEANED UNDER CONTROLLED

CONDITIONS WITH 5 PARTS EACH LAUNDERED BY

EACH OF THE DETERGENTS. UNFORTUNATELY

THREE PIECES OF CLOTH WERE LOST IN

THECOURSE OF EXPERIMENTATION. WHITENESS

READING MANUFACTURED ON THE 17 REMAINING BITS

OF CLOTH HAPPEN TO BE SHOWN:

DETERGENT

A

N

C

M

77

seventy four

73

seventy six

81

66

78

85

61

58

57

seventy seven

76

69

64

69

63

ASSUMING MOST WHITENESS PSYCHIC READINGS TO BE NORMALLY

DISTRIBUTED WITH COMMON DIFFERENCE, TEST THE

HYPOTHESISOF SIMPLY NO DIFFERENCE BEETWEEN THE 4

BRANDS IN RELATION TO MEAN WHITENESS READINGS FOLLOWING

WASHING....

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