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Analyze Statistic by Using SPSS

2nd Day

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اإلعجاز اإلحصائي للقرآناإلعجاز اإلحصائي للقرآن يع2ني أننا عاجزون عن تأليف كتاب فيه نظام رقمي

.دقيق لتكرار الكلمات مثل القرآن } ه سبيال خذ إلى رب { هذه اآلية الكريمة نجدها في إن هذه تذكرة فمن شاء ات

موضعين فقط من القرآن : ]19{ ] المزمل : إن هذه تذكرة فمن شاء اتخذ إلى ربه سبيال 2 } 1]29{ ] اإلنسان : إن هذه تذكرة فمن شاء اتخذ إلى ربه سبيال 2 } 2

2 لماذا تكررت هذه اآلية مرتين في القرآن ؟1 ؟29 2 19 2 لماذا كان رقم اآليتين : 2

2 لماذا كان تسلسل سورة المزمل قبل سورة اإلنسان ؟3

سوف نجيب على هذه األسئلة وغيرها بلغة األرقام ، فتكرار هذه اآلية في القرآن وذلك لحكمة 29 و 19مرتين له حكمة ، وقد اختار الله تعالى لهذه اآلية رقمين

. والفكرة 7أيضا ، ويمكن استنتاج جزء من هذه الحكمة باالعتماد على الرقم األساسية في هذا البحث تعتمد على صف أرقام اآليات بجانب بعضها وذلك

حسب تسلسل هذه اآليات في القرآن )وليس جمع األرقام (

هذا العدد يقرأ : 2919 ينتج عدد جديد هو 229 19وعندما نصف أرقام اآليتين ، أي يقبل القسمة 7ألفان وتسعمئة وتسعة عشر ، وهو من مضاعفات الرقم

من دون باق :7تماما على 417 × 7 = 2919

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Probability Distributionsبعض التوزيعات االحتمالية

1. The normal distribution2. The Standard normal distribution3. T distribution4. Chi-square distribution

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(1)Normal Distribution• The normal distribution is a probability distribution that

associates the normal random variable X with a cumulative probability . The normal distribution is defined by the following equation: Y = [ 1/σ * sqrt(2π) ] * e -(x - μ)2/2σ2

• the sampling distribution of a statistic will follow normal distribution, as long as : 1- the sample size is sufficiently large. 2-we know the standard deviation of the population.

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The Curve of Normal Distribution:

• Find P(Z > a)?P(Z > a) = 1 - P(Z < a)

• Find P(a < Z < b)? P(Z < b) - P(Z < a)

= P(Z < z)

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Difference between these curves:

The curve on the left is shorter and wider than the curve on the right, because the curve on the left has a bigger standard deviation.

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(2)The standard normal distribution

• The standard normal distribution is a special case of the normal distribution.

• the following equation: z = (X - μ) / σ

• Z ~ N ( μ , σ ) in standard normal distribution: Z ~ N ( 0 , 1 )

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Standard Normal Distribution Table• Find the cumulative probability of a z-score equal to -1.31?

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08

-3.0 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010

... ... ... ... ... ... ... ... ... ...

-1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0722 0.0708 0.0694

-1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838

-1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003

... ... ... ... ... ... ... ... ... ...

3.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990

The table shows that the probability that a standard normal random variable will be less than -1.31, that is, P(Z < -1.31) = 0.0951.

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قال تعالى :ذي خلقكم من ضعف ثم جعل من بعد ه ال )الل

ضعف قوة ثم جعل من بعد قوة ضعفا وشيبة [54]الروم : يخلق ما يشاء وهو العليم القدير(

ضعف

قـوة

ضعف

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(3) t Distribution• sample sizes are sometimes small, and often

we do not know the standard deviation of the population. When either of these problems occur, statisticians rely on the distribution of the t statistic.

• The equation : t = [ x - μ ] / [ s / sqrt( n ) ]

• Degrees of Freedom:the sample size minus one (n-1,α).

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Curve of t Distribution

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(4)Chi-Square Distribution• using the following equation:

Χ2 = [ ( n - 1 ) * s2 ] / σ2

• degrees of freedom Χ2 :n - 1

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Curve of Chi-SquareAs the degrees of freedom increase, the chi-square

curve approaches a normal distribution.

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Inferential Statisticsاالحصاء االستداللي

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The First Topic inInferential Statistics

Estimation

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Estimation التقدير• When a parameter is being estimated, the estimate can be either a

single number or it can be a range of scores.

• When the estimate is a single number, the estimate is called a "point estimate"

• When the estimate is a range of scores, the estimate is called an interval estimate. Confidence interval are used for interval estimates.

)(x

the population ..… estimate of Sample..… Mean µ mean M or

variance σ² variance s²

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Estimation by SPSSAnalyze Descriptive Statistics Explore

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Statistics

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Output:

105.42 < µ < 118.98

µ = 112.2

σ² = 89.73

All the estimations at Confidence interval 95%

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The Second Topic inInferential Statistics

Testing Hypotheses

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Inferential Statisticsاالحصاء االستداللي

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Tests Concerning a Single Mean اختبارات الفروض حول متوسط المجتمع

Single meanµ = µₒ

n>30 Z Distribution

n≤30T

Distribution

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Summary of Computational Steps Specify the null hypothesis and an alternative

hypothesis. Compute M = ΣX/N. Compute , if σ unknown compute

Compute , if n<30 compute where M is the sample mean and µ is the hypothesized

value of the population. Use a z table to determine p from z , or Use a t table to compute p from t and df (df=N-1).

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Testing Hypothesesاختبارات الفروض

Consider an experiment designed to test the null hypothesis that µ = 10. The test would be conducted with the following formula:

where M (the statistic) is the sample mean.

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Tests Concerning a Single Mean Example

n > 30

A random sample of 100 deaths in the Philippines last year showed an average life span of 69.3 years. Assuming a population standard deviation of 7.8 years.

• does this seem to indicate that the life span today is lesser than 70 years?

• Use a 0.01 level of significance?

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n

90.078.0

703.69/

n

xz o

We will solve this testing problem in 5 steps:

1. H0: = 70 years.H1: < 70 years.

2. Use a = 0.01.3. Since we are testing the mean life span and the population standard deviation is known

( / = 0.78). Using normal distribution ,

4. At 0.01 level of significance, we reject if and only if Z < z 0.01= -2.33

5. Based from the rejection region (and critical value –2.33), we see that the z value is outside the rejection region (or –0.90 is greater than –2.33). Thus, we do not reject the null hypothesis and conclude that the life

expectancy of Filipinos is 70 years.

Solution:

Z = -0.90

-2.33

Rejection Region

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Note

The previous Example We cannot use SPSS because:

1- the standard deviation is known .2- and we haven’t data .

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Tests Concerning a Single Mean Example

n ≤ 30

A teacher of Arabic language was assumed that the mean of students' degree is 68, drew a sample from the students' degree as follows (n=30):

80 85 75 65 55 52 44 3330 25 45 80 95 50 3095 88 90 77 72 75 60 4057 55 52 48 84 87 78

Is teacher's claim is true at 0.01 level of significance?

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H0: µ = 68H1: µ 68

SPSS Solution:

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Analyze compare mean One-sample test

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Option

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Output

(P-value) 0.240 > 0.01 (α)don't reject Hₒ

the teacher's claim is true.

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Tests of Differences between Means اختبارات الفروض حول متوسطي مجتمعين

Two meansµ1 = µ2

TwoIndependent

Samples

Independent samples t-test

TwoDependent

Samples

Paired samples t-test

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Tests of Differences between Means,Independent Samples

Example: If we have the following marks for students (male & female):

Is there a difference between the mean of marks at level of significant 0.05?

75 77 30 50 84 80 52 25 30 33 44 52 55 65 75 85 80 F

emale

73 78 87 95 48 45 55 57 40 60 75 72 77 90 88 9570 85 45 84 38 79 75 77 44 65 40 44 30 25 30 77 75 98

Male85 95 61 60 48 36 95 98 65 66 77 42 44 62 76 93 80

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SPSS Solution

H0: µ1 = µ2H1: µ1 µ2

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Analyze Compare means Independent Samples T Test

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Press Define Groups

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Press Option

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Output:

0.771 > 0.05Equal Variance

Assumed

Inside the Box:T-Test for Equality of means

t= .045, df =(33+35)-2 =68-2= 66

Sig.=0.964> α=0.05Don’t reject Hₒ

Confidence Interval of the difference(-9.949,10.412)

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What is learned:• Test of Hypotheses : T-Test of two Independent

Samples.

• Estimate : Confidence Interval of the difference between means.

• Mean difference :• µ1-µ2 = 64.69 – 64.45 = 0.24

(from Group Statistics Table)• Or mean difference = 0.23 1

(from Independent samples test Table )

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Make a decision :

1) From P-value(Sig.)

P-value > αDon’t reject Hₒ

P-value < αReject Hₒ

2) From Confidence Interval

0 inside the Interval Don’t reject Hₒ

0 outside the Interval Reject Hₒ

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Tests of Differences between Means,dependent Samples

Example : We have the data of 20 students in two courses (Arabic & English):

Examine if there is difference between the mean of students’ marks in two courses, at level of significant 0.10?

10 9 8 7 6 5 4 3 2 1 Num

60 60 80 90 80 40 85 50 80 85 Arabic

30 40 84 95 90 50 65 50 50 80 English

20 19 18 17 16 15 14 13 12 11 Num

85 80 55 60 70 75 90 85 80 87 Arabic

75 90 70 40 60 75 70 65 75 80 English

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SPSS Solution

H0: µ1 = µ2H1: µ1 µ2

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Analyze Compare means Paired Samples T Test

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Press Option

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Output: t= 2.309, df=20-1=19

Sig.=0.032 < α=0.10Reject Hₒ

Confidence Interval of the difference

(1.796,12.504)Note: the interval doesn’t contain

ZERO

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اخطاء شائعة

:tاستخدام اختبارات • غير مسحوبة من مجتمع صغيرةلبيانات طبيعي

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Statistical HumorA boy asked his statistician father, "Why is my

body not well proportioned just like my brother's?" His father's response, "Because, when your mother had your pregnancy, its distribution was skewed!!"