Analyze Statistic by Using SPSS

Analyze Statistic by Using SPSS

5

^th

Day

Fadwa Flemban 2

ميركلا نآرقلا يف يددعلا طبارتلا

بسانتلا هيف نا ذا .. ظافللاا رركت يف يواستلا ىلع رصتقيلا يددعلا زاجعلاا نا دادعلاا يف

ميركلا نارقلا يف نمحرلا ظفل رركت لاثمف 57

ميحرلا ظفل رركتو .. هرم 114

نمحرلا فعض رركت ميحرلا نا يا .. هرم

ءازجلا رركتو 117

هرفغملا ترركتو .. هرم 234

فعض هرفغملا نا يا .. هرم

ءازجلا

تركذ تارم تس ىلولاا تركذ امنيبف .. راجفلا ترركتام فعض راربلاا ترركتو تارم ثلاث هيناثلا

رسيلا ترركت دقف .. رسعلا رركتام فاعضا ثلاث رسيلا ترركتو 36

يف هرم

رسعلا نيح 12

هرم

يددعلا طبارتلا هيف كلذكو

رركت ىلاعتو هناحبس هللا نم رملاا وهو.. لق ظفل نا دجنف 332

نارقلا يف هرم

..

هرخلااو ايندلا ةايحلا يف نجلاو سنلاا نم اولاق ظافلا عومجمو 332

هرم

رركت دق رهشلا نا دجن اضياو 12

هنس ردقب يا .. هرم

Non Parametric Tests ةيملعملالا تارابتخلاا

بجاولا طورشلا دحأ يف للخ دوجو دنع • تارابتخا مادختسلإ اهرفوت t

تارابتخلال أجلن

.ةيملعملالا تارابتخلا ةرظانملا ةيملعملالا تارابتخلاا • t

. ةوق لقأ نوكت

Non Parametric Tests

Non Parametric Tests One-Sample

The Binomial Test

• Example: random sample was withdrawn for a number of children's weights and were as follows: 35 , 25 , 15 , 20 , 30 , 24 , 35 , 15 , 25 , 30 , 32 , 33 , 34 , 16 , 19 , 20 , 31, 10 . Required test the hypothesis average weight of the child over than 25 kilos at the confidence interval

99%?

Mathematically solution

H₀:= 25 H_1: > 25

Use a = 0.01

Compute for the z – statistic. = = 0

from observations the sign= + 0 - - + - + - 0 + + + + - - - + - (It means s = 8 , n = 18-2 = 16 ).

Compute for the critical region. At a = 0.01, we reject H₀ if z>z 0.01 (z>+2.33).

Thus, we don't reject the null hypothesis and conclude that the average weight of the child equal 25 kilos.

n n

z 

²S^

Analyze  Nonparametric Tests  Binomial

Solution by SPSS

The result: p-value

(1-tailed) = 0.815/2= 0.41 a = 0.01

0.41>0.01 don't reject H₀

Non Parametric Tests

Two-Independent Samples The Man-Whitney test

Example: To compare the ratios of intelligence between the two schools A and B, selected a random sample of students from these schools, intelligence test was conducted and the results were:

A 22 19 23 21 24 20 18

24 22 25 24 20

B 18 20 17 22 19 23 21

16 18 20

test the hypothesis that the students in the school A are smarter than the students in the school B at the level 0.05?

Mathematically solution

rank school Obs.

15 A 22

6.5 A 19

17.5 A 23

12.5 A 21

20 A 24

9.5 A 20

4 A 18

20 A 24

15 A 22

22 A 25

20 A 24

9.5 A 20

rank school Obs.

4 B 18

9.5 B 20

2 B 17

15 B 22

6.5 B 19

17.5 B 23

12.5 B 21

1 B 16

4 B 18

9.5 B 20

Rank all

observations in

increasing order from smallest to largest.

Assign the smallest observation the rank 1.

If some observations have tied values, assign the rank that is the average of the ranks they would have been assigned if there

were no ties.

Mathematically solution

1. H₀ : µ1=µ2 , H1 : µ1>µ2

2. W=(15+6.5+17.5+12.5+20+9.5+4+20+15+22+20+9.5)=171.5 5

. 93 )

2 / ) 13 ( 12 ( 5 . 2 171

) 1 1 (

1    



 n n W

U

) 1 , 0 ( ...

, 8 2 ,

1 U N

Z then

n n

u

u 

 

 



2 60 ) 10 ( 12 2

2

1  

n n

u

, 12 230

) 23 )(

10 ( 12 12

) 1 2 1

( 2

2 n1n n n   

u _u  230 15.17

21 . 17 2

. 15

60 5

.

93  

 Z

Compute for the critical region. At a = 0.05, we reject H₀ if z>z 0.05 (z>+1.65), from above 2.21 > 1.65

Thus, we reject the null hypothesis and conclude that the

students in the school A are smarter than the students in the school B at the level 0.05

Solution by SPSS

Analyze  Nonparametric Tests 2 Independent Samples

Output

:

p-value: (1-tailed) = 0.014

a = 0.05, 0.014<0.05

we reject H₀

Students of school A are smarter

Non Parametric Tests Two-Related Samples

The Wilcoxon test

Example : Two people decided correct the answer sheet of 20 students in statistics, were them degrees as follows:

A 68 75 79 87 80 69 72

90 75 67 23 56 65 90 88

51 68 57 48 45

B 63 75 72 85 83 69 68

87 72 70 21 56 67 85 84

53 66 52 48 43

Test the hypothesis that the person A gives the degrees different than the person B at the confidence interval 95%?

Mathematically solution

H₀: µ1 = µ2 H_1: µ1 > µ2 Use a = 0.05

Compute for the z – statistic.

(T =(14+16+3.5+11.5+8.5+8.5+3.5+14+11.5+3.5+14+3.5) =112 ).

Compute for the critical region. At a = 0.05, we reject H₀ if z>z 0.05 (z>+1.65).

Thus, we reject the null hypothesis and conclude that the person A gives the degrees larger than the person B.

275 . 34 2

. 19

68 112  





 

 



z T

4 68 ) 17 ( 16 4

) 1

(   

n n

T ³⁷⁴

24 ) 33 )(

17 ( 16 24

) 1 2

)(

1

2 n(n  n  

T

34 . 19 374 

T 



Solution by SPSS

Analyze  Nonparametric Tests 2 Related Samples

Output

:

The result: p-value= 0.01

0.01<0.05 we reject H₀

the teacher A gives the degrees different than the teacher B.

Summary

Non Parametric Tests

1-Sample The Binomial

Test

2-Independent Samples

The Man-Whitney

test 2-Related

Samples The Wilcoxon

test

Non Parametric Tests K-Independent Samples

The Kruskal-Wallis test

• Example : To study if the marks of statistics course are dependent on the specialization of students :

Are there difference between the marks of students in the three specializations at level of significance 0.05?

Development Education Computer

50 90 80 40 40 31 42 85 95 98 75 65 85 90 75 77 40 85 95 90

70 65 70 80 75 40 35 55 70 65

55 45 30 74 78 82 88

65 55 75 45 30 35 65 80 90 65

60 62 68

Solution by SPSS

H₀ : there is no difference between the mean of samples.

H1 : there is difference between the

mean of samples.

Analyze  Nonparametric Tests  k

Independent Samples

Define Range

Exact ..

Output

:

P-value (0.124)>0.05

Don’t reject Hₒ

There is no difference between the mean of samples.

Non Parametric Tests K-Related Samples The Freidman test

Assume we have this data:

Examine if there is difference in the methods of treatment

at the confidence interval 99%?

T3 T2 T1 Num

7 18 10 1

8 19 12 2

16 17 15 3

12 14 13 4

17 20 15 5

10 15 12 6

6 7 11 7

11 18 13 8

11 19 15 9

12 13 7 10

18 13 12 11

Solution by SPSS

H₀ : there is no difference between the methods of treatments.

H1 : there is difference between the

methods of treatments.

Solution by SPSS

Analyze  Nonparametric Tests k Related Samples

By default : the confidence interval is 99%

Output :

p-value = 0.004 a = 0.01,

0.004<0.01 we reject H₀

there is difference between the methods of treatments

Summary

Non Parametric Tests

K-Independent Samples

Kruskal-Wallis test

k-Related Samples

The Freidman

test

يئاصحلاا ريرقتلا

ريخ

.. ملاكلا لّق ام

.. لّدو

؟ kايئاصحا kاريرقت يبتكت فيك

لمشي نا دبلا يئاصحلاا ريرقتلا • :ةيلاتلا رصانعلا

1 ديـهمت -

2 ةـمدقم -

3 ليلحتلا قرط -

4 جـئاتنلا -

5 ةـصلاخلا -

6 عجارملا -

7

قـحلاملا -

ةيئاصحلاا ريراقتلا لضفأ : ـب مستت يتلا يه

ديق ةلئسلأا عيمج نع ةباجلاا يف ةلوهسلاو ةطاسبلا •

.ةساردلا

ةينايب موسرب امإ اهصيخلتو اهعمج قرطو تانايبلا فصو •

.اهريغ وأ لوادج وأ

نع ةيفاك تامولعمب ئراقلا ديوزت وه فدهلا لعج •

.تانايبلا

.ريرقتلا ءانثأ ينايبلا لكشلا ةشقانمو فصو •

.ةيوغللا ةراهملا ةاعارم •

ريغ يشاوحلاو ةدقعملا ةيوغللا بيكارتلا نع داعتبلاا •

.ةيرورضلا

عم قفرم ينايب لكش وأ لودج لك ىلع قيلعتلا نم دبلا •

.ريرقتلا

بسح م|وقي فوس ريرقتلا نا يملعا : ةيلاتلا صئاصخلا

: يئاصحلاا قسنلا • ةحصو ليلحتلا يف ةراتخملا ةيئاصحلاا قرطلا ةبسانم

.اهقيبطت : يملعلا قسنلا • .اهل تلصوت يتلا جئاتنلاو ثحبلل ةيملعلا ةميقلا ثيح نم

: يوغللا قسنلا •

.حوضولاو بيترتلاو قطنملاو ةباتكلا ثيح نم

Statistical Humor

• What do you call a tea party with more than 30 people?

A Z party!!!

) هللاب لاإ يقيفوت امو (

: عجارملا لّلاج.د / ةيئاصحلاا قرطلا يف ةمدقم •

بيبح دمحم.د و دايصلا مادختساب يئاصحلاا لّيلحتلا • SPSS

زع.د /

حاتفلادبع نسح

دمحأ قيتع.د / =ايئاصحا =اريرقت بتكت فيك •

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