8.19

16 25

177 - 170

16 5

7=

3.2

7= = 2.188

Critical Value = 1.64

X = 177= 170S = 16N = 25

Z =

8.25

21.2 100

130.1 - 120

21.2 10

10.1=

2.12

10.1= = 4.764

Critical Value = 1.662

X = 130.1= 120S = 21.2N = 100

t =

9.10 Pre/post

Before After Difference D squaredTotal 108.2 108.4 3 105

t paired = t p = d - 0

Standard error of d

= -------------d - 0

S d 2

N

d = D/N

N = 15

d 2 = D 2 – ( D) 2 / N

S d2 = d 2 / N - 1

= 3/15 =.2

= 105 - 9/15 = 175

= 175 / 15 – 1 = 12.5

= 0.2 / 0.694

= 0.288

= 0.2 / 12.5 / 15

= 0.2 / 0.833

df = N – 1 = 140.05 > 1.7613

9.11 Pre/post

Before After Difference D squaredTotal 114.2 114.7 15 279

t paired = t p = d - 0

Standard error of d

= -------------d - 0

S d 2

N

d = D/N

N = 30

d 2 = D 2 – ( D) 2 / N

S d2 = d 2 / N - 1

= 15/30 =.5

= 279 - 225/30 = 271.5

= 271.5 / 30 – 1 = 9.362

= 0.5 / 0.559

= 0.894

= 0.5 / 9.362 / 30

= 0.5 / 0.312

df = N – 1 = 290.05 > 1.6991

Pooled estimate of the SED (SEDp)

1Estimate of the 1

N sN ns

+SEDp of x E - x C = Sp2

Sp2 = Pooled estimate of the variance

(x s - x s)2 + (x ns - x ns )2 Sp2 =

N s + N ns - 2

9.22

Pooled estimate of the SED (SEDp)

1Estimate of the 1

14 s18 ns

+SEDp of x E - x C = Sp2

1399.5 + 1814.5 Sp2 =

14 + 18 - 2

3214Sp2 =

30= 107.133

9.22

t-Test(Two Tailed)

x s - x ns - 0

t =

Sp2 [ ( 1/Ns ) + ( 1/Nns) ]

d f = N s + N ns - 2

9.22

t-Test(Two Tailed)

83.5 - 72.5

t =

107.133 [ ( 1/14 ) + ( 1/18) ]

d f = 14 + 18 - 2

9.22

t-Test (Two Tailed)83.5 - 72.5

t =107.133[ ( 1/14 ) + ( 1/18) ]

d f = 14 + 18 - 2 = 30

9.22

83.5 - 72.5

107.133[ ( 1/14 ) + ( 1/18) ]

t =11

107.133 (0.071 + 0.055)

=11

107.133 (0.126)=

13.498

11=

3.674

11= 2.994

Critical value = 2.0423

ANOVA

Analysis of Variance

• Allows the statistician to analyze multiple data sets.

• Number of combinations to be made take two groups at a time– N(N-1)/2

• If individual z tests were performed on each combination of a large number of groups the number of calculations would be prohibitive.

Assumptions underlying the use of ANOVA

1. The individuals in the various subgroups should be selected on the basis of random sampling from normally distributed populations.

2. The variance of the subgroups should be homogeneous. (H0: s1 = s2 = … = sn)

3. The samples comprising the groups should be independent.

Single classification ANOVAGroup A

X

Group B

X

Group C

X

Group A

X2

Group B

X2

Group C

X2

12 18 6 144 324 36

18 17 4 324 289 16

16 16 14 256 256 196

8 18 4 64 324 16

6 12 6 36 144 36

12 17 12 144 289 144

10 10 14 100 100 196

X =82 108 60 X2 = 1068

1726 640

X = 11.71 15.43 8.57Xt = 11.90

Values needed for ANOVA

• The Total Sum of the Squaresx2

t = X2 – (X)2 / N

• The “Between” Sum of Squaresx2

b = (X – XT)2 n

• The “Within” Sum of Squaresx2 = X2 – (X)2 / n for each group orx2

w = X2t - x2

b

• The Degrees of FreedomN between groups –1 plus N within groups -1

Values needed for ANOVA

• The Total Sum of the Squaresx2

t = X2 – (X)2 / N = 1068+1726+640-[(82+108+60)2/21] = 457.8

• The “Between” Sum of Squaresx2

b = [(X )2/ n] - x2t/N

=[(82)2/7 + (108)2/7 +(60)2/7] – (250)2/21 =165.0

• The “Within” Sum of Squaresx2 = X2 – (X)2 / n for each group orx2

w = X2t - x2

b = 457.8 - 165.0 = 292.8

• The Degrees of FreedomN between groups –1 plus N within groups –13 – 1 + (7 – 1 + 7 – 1 + 7 – 1) = 2 + 18 = 20

ANOVA TableSource of variation df

Sum of

Squares

Mean

Square

“Between”

Groups2 165.0 82.5

“Within”

Groups18 292.8 16.3

Total 20 457.8

The F-Test

F = mean square for “between”groups mean square for “within” groups

= 82.5 16.3

= 5.06

“Between” df = 2 “Within” df = 18

Value of F needed of significance at the 5% level = 3.55

Page 325

Tests after the F test

• F = (X1– X2)2/s 2w (N1+ N2)/ N1 N2

• A vs. B

F = (11.71– 15.43)2/16.3 (14)/49 = (3.72)2/4.66 = 2.97

• A vs. C F = (11.71– 8.57)2/16.3 (14)/49 = (3.14)2/4.66 = 2.12

• B vs. CF = (15.43– 8.57)2/16.3 (14)/49 = (6.86)2/4.66 = 10.1

Page 181

X = 11.71

A B C D

1 7 9 8

1 7 6 6

1 7 5 4

1 7 3 1

1 7 2 1

5 35 25 20

Page 181

X

A X2 B X2 C X2 D X2

1 1 7 49 9 81 8 64

1 1 7 49 6 36 6 36

1 1 7 49 5 25 4 16

1 1 7 49 3 9 1 1

1 1 7 49 2 4 1 1

5 35 25 20

5 245 155 118X2

Page 181

X

A X2 B X2 C X2 D X2

1 1 7 49 9 81 8 64

1 1 7 49 6 36 6 36

1 1 7 49 5 25 4 16

1 1 7 49 3 9 1 1

1 1 7 49 2 4 1 1

5 35 25 20

5 245 155 118X2

Xa = 1 Xb = 7 Xc = 5 Xd = 4

Xt = 4.25

=85

=523

Values needed for ANOVA

• The Total Sum of the Squaresx2

t = X2 – (X)2 / N = [5+245+155+118] -[(5+35+25+20)2/20] = 161.75

• The “Between” Sum of Squaresx2

b = [(X )2/ n] - x2t/N

=[(5)2/5 + (35)2/5 +(25)2/5+(20)2/5 ] – (85)2/20 =93.75

• The “Within” Sum of Squaresx2 = X2 – (X)2 / n for each group orx2

w = X2t - x2

b = 161.75 – 93.75 = 68

• The Degrees of FreedomN between groups –1 plus N within groups –14 – 1 + (5 – 1 + 5 – 1 + 5 – 1+ 5 - 1) = 3 + 16

= 19

ANOVA TableSource of variation df

Sum of

Squares

Mean

Square

“Between”

Groups3 93.75 31.25

“Within”

Groups16 68 4.25

Total 20 161.75

F = 31/25/4.25 = 7.35

HSD = 4.05 4.25

5= 4.05(9.22) = 3.73

Tukey’s HSD test = 0.5k = 4n – k = 16Appendix C: q = 4.05

Pair Mean DifferenceA-B 6A-C 4A-D 3B-C 2B-D 3C-D 1

CHAPTER 11

Inferences Regarding Proportions

OUTLINE  

11.1            INFERENCES WITH QUALITATIVE DATA

Discusses the problem of inference in qualitative data

 

11.2            MEAN AND STANDARD DEVIATION OF THE BINOMIAL DISTRIBUTION

Explains how to compute a mean and a standard deviation for the binomial distribution

 

11.3            APPROXIMATION OF THE NORMAL TO THE BINOMIAL DISTRIBUTION

Shows that, using the normal approximation it is possible to compute a Z score for a number of successes

 

11.4            TEST OF SIGNIFICANCE OF A BINOMIAL PROPORTION

Gives instructions on how to test hypothesis regarding proportions if the distribution of the proportion of successes is known

 

11.5            TEST OF SIGNIFICANCE OF THE DIFFERENCE BETWEEN

Illustrates that, because the difference between two proportions is approximately normally distributed, a hypothesis test for the difference may be easily set up

 

11.6            CONFIDENCE INTERVALS

Discusses and illustrates confidence intervals for

 

LEARNING OBJECTIVES

•  

• 1.                    Compute the mean and the standard deviation of a binomial distribution

• 2.                    Compute Z scores for specific points on a binomial distribution

• 3.                    Perform significance tests of a binomial proportion and of the difference between two binomial proportions

• 4.                    Calculate confidence intervals for a binomial proportion and for the difference between two proportions

•  

INFERENCES WITH QUALITATIVE DATA

•   • A.      Qualitative data – data for which individual quantitative measurements are not

available but that relate to the presence or absence of some characteristic

• B.       p the estimate of the true proportion, , of individuals who possess a certain characteristic

• C.       To best understand the difference between the distribution of binomial events (x) and the distribution of binomial proportion (p)

– 1.                    Compare these distributions with those in the approximate analogous quantitative situation

– 2.                    The x’s of a binomial distribution with a mean and a standard error

•  

MEAN AND STANDARD DEVIATION OF THE BINOMIAL DISTRIBUTION

A. Probability of x successful outcomes in n independent trials is given by:•  

– 1.

•  • where P is the probability of a success in one individual trial• will be used to designate the probability of x successful

outcomes

• B. In a binomial distribution the mean for the number of successes, x, is•  •

•  • and the standard deviation is

xnxn

x

pp

1

n

)1( n

APPROXIMATION OF THE NORMAL TO THE BINOMIAL DISTRIBUTION

• A. The normal distribution is a reasonable approximation to the binomial distribution when n is large

• B.      We can find a point on the Z distribution that corresponds to a point x on the binomial distribution by using

)1(

n

nxZ

APPROXIMATION OF THE NORMAL TO THE BINOMIAL DISTRIBUTION

• C.      Because we are using a normal (continuous) distribution to approximate a discrete one, we may apply the continuity correction to achieve an adjustment. The correction is made by subtracting ½ from the absolute value of the numerator, that is,

•  • •  

• D.      When n is very large and is very small, another important distribution, the Poisson distribution, is a good approximation to the binomial

)1(2

1

n

nxZ

75.12

5.3

)8)(.2(.252

159

TEST OF SIGNIFICANCE OF A BINOMIAL PROPORTION

•  • A.      The mean of the distribution of a binomial proportion p is given by

the population parameter•  •  

• and the standard error of p is given by•  •  

• B.       When p appears to be normally distributed, providing n is reasonably large, we can find the Z score corresponding to a particular p and perform a test of significance

populationincasesofnumber

populationinsuccessesofnumber

n

x

npSE

)1()(

TEST OF SIGNIFICANCE OF THE DIFFERENCE BETWEEN

•  • A.      In order to compare proportions from two different samples we must:

– 1.                    assume that the proportions are equal, that is, in estimating

– 2.                    learn if , the proportion with the given characteristic in one sample differs significantly from , the proportion with the same characteristic in the second

sample

• B.       Three thing that must be know to determine if the proportions are significantly different

– 1.                    the distribution of the differences -

– 2.                    the mean -

– 3.                    the standard error of this distribution – (SE)

• C.       Statisticians have shown that follows a nearly normal distribution

21 )( 21 ppSE

1p2p

21 pp

21 pp

TEST OF SIGNIFICANCE OF THE DIFFERENCE BETWEEN

• D.      The standard error is estimated by

•  

•  

• where

•  

• and

•  

•  

• and

21

21 n

qp

n

qpppSE

tttt

ttt pqandnn

xxp

121

21

1

11 n

xp

2

22 n

xp

TEST OF SIGNIFICANCE OF THE DIFFERENCE BETWEEN

•  

• Knowing the mean and the standard error of the distribution differences, we can calculate a Z score:

•  

•  

• If , the formula for is

•  

21

2121

ppSE

ppZ

21 21 ppSE

2

22

1

11 11

nn

CONFIDENCE INTERVALS

•   

• A.      Although hypothesis testing is useful, we often go a step further to learn:

– 1.                    the true proportion

– 2.                  the true difference in proportion between the baseline data and the revised data

• B.       To answer these questions we compute confidence intervals for and for by employing a method to the one used for computing confidence intervals for and

21 21

CONFIDENCE INTERVALS

•  • C.      Confidence interval for •  • Chapter 8 version:•  

• Similar version•  • This expression presents a dilemma: it requires that we know , which is an

unknown. Solution is to have a sufficiently large sample size, permitting the use of p as an estimate of

•  • The expression then becomes•  

nZx

n

Zp

1

n

ppZp

1

CONFIDENCE INTERVALS

•   

• A.      Confidence interval for

•  

• The confidence interval for the difference of two means is:

•  

•

•  

• The confidence interval for the difference of two proportions is similar:

•  

21

212121 xxSEZxxforCI

2

22

1

112121

11

n

pp

n

ppZppforCI

CONCLUSION

The normal approximation to the binomial distribution is a useful statistical tool. It helps answer questions regarding qualitative data involving proportions where individuals are classified into two categories. With an understanding of the distribution of the binomial proportion p and of the distribution of the difference between two proportions we can perform tests of significance and calculate confidence intervals.