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Hypothesis Testing and Comparison of Two Populations. Dr. Burton. If the heights of male teenagers are normally distributed with a mean of 60 inches and standard deviation of 10, And the sample size was 25, what percentage of boy’s heights in inches would be: Between 57 and 63 Lass than 58

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## Hypothesis Testing and Comparison of Two Populations

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### Hypothesis Testing and Comparison of Two Populations

Dr. Burton

If the heights of male teenagers are normally distributed with a mean of 60 inches and standard deviation of 10, And the sample size was 25, what percentage of boy’s heights in inches would be:

Between 57 and 63

Lass than 58

61 or larger

7.2a

%

Height

Z

60

0

57

63

57 - 60

Z= -1.5 = .4332

10 / 25

.8664 = 86.8%

x - 

Z =

s / n

63 - 60

Z= 1.5 = .4332

10 / 25

7.2b

%

58

-1.0

60

0

Height

Z

x - 

58 - 60

Z =

Z = -1.0 = .5000 - .3413

s / n

10 / 25

.1587 = 16%

7.2c

%

61

60

0 0.5

Height

Z

x - 

61 - 60

Z =

Z = 0.50

s / n

10 / 25

= 0.50

- .1915 = .3085 = 30.9%

Hypothesis Testing

Hypothesis: A statement of belief…

Null Hypothesis, H0: …there is no difference between the population mean  and the hypothesized value 0.

Alternative Hypothesis, Ha: …reject the null hypothesis and accept that there is a difference between the population mean  and the hypothesized value 0.

Probabilities of Type I and Type II errors

Truth

H0 True

H0 False

a

b

Type II

Error

Correct

results

Accept H0

Test

result

1 - 

c

d

Type I

Error

Correct

results

Reject H0

1 - 

Differences

H0 True = statistically insignificant

H0 False = statistically significant

Accept H0 = statistically insignificant

Reject H0 = statistically significant

http://en.wikipedia.org/wiki/False_positive

 = 0.05

0.025

0.025

Probability Distribution for a two-tailed test

1.96 SE

-3

-2

-1

1

2

3

0

SE

SE

Magnitude of (XE – XC)

XE < XC

XE > XC

Probability Distribution for a one-tailed test

 = 0.05

1.645 SE

-3

-2

-1

0

1

2

3

SE

SE

Magnitude of (XE – XC)

XE < XC

XE > XC

Box 10 - 5t =

Distance between the means

Variation around the means

A

Box 10 - 5t =

Distance between the means

Variation around the means

A

B

Box 10 - 5t =

Distance between the means

Variation around the means

A

B

C

t-Tests
• Students t-test is used if:
• two samples come from two different groups.
• e.g. A group of students and a group of professors
• Paired t-test is used if:
• two samples from the sample group.
• e.g. a pre and post test on the same group of subjects.
One-Tailed vs. Two Tailed Tests
• The Key Question: “Am I interested in the deviation from the mean of the sample from the mean of the population in one or both directions.”
• If you want to determine whether one mean is significantly from the other, perform a two-tailed test.
• If you want to determine whether one mean is significantly larger, or significantly smaller, perform a one-tailed test.
t-Test(Two Tailed)Independent Sample means

xA - xB - 0

t =

[ ( 1/NA ) + ( 1/NB) ]

Sp

d f = N A + N B - 2

Sample A (A – Mean)2

26 34.34

24 14.90

18 4.58

17 9.86

18 4.58

20 .02

18 4.58

Mean = 20.14

A2 = 2913

N = 7

(A – Mean)2 = 72.86

Var = 12.14

s = 3.48

Sample B (B – Mean)2

38 113.85

26 1.77

24 11.09

24 11.09

30 7.13

22 28.41

Mean = 27.33

B2 = 4656

N = 6

(B – Mean)2 = 173.34

Var = 34.67

s = 5.89

Independent Sample Means
Standard error of the difference between the means (SED)

Theoretical

 A 2

 B 2

SED of  E - C =

+

Population

NB

NA

Estimate of the

s A 2

s B 2

Sample

SED of xE - xC =

+

NB

NA

Pooled estimate of the SED (SEDp)

Estimate of the

1

1

SEDp of xA - xB =

Sp

+

NB

NA

s2(nA-1)

+ s2 (nB – 1)

Sp

=

nA + n B - 2

+34.67(5)

12.14 (6)

Sp

=

= 22.38 = 4.73

7 + 6 - 2

t-Test(Two Tailed)

xA - xB - 0

t =

[ ( 1/NA ) + ( 1/NB) ]

Sp

20.14 - 27.33 - 0

=

= -2.73

( 1/7 ) + ( 1/6)

4.73

d f = N E + N C - 2 = 11

Critical Value 95% = 2.201

One-tailed and two-tailed t-tests
• A two-tailed test is generally recommended because differences in either direction need to be known.

d =  D/N

=  d 2 / N - 1

S d2

 d 2

=  D 2 – ( D) 2 / N

Pairedt-test

d - 0

d - 0

= -------------

t paired = t p =

S d2

Standard error of d

N

df = N - 1

= 3.7 / 41.5667 / 10

= 3.7 / 4. 15667

df = N – 1 = 9

0.05 > 1.833

Pre/post attitude assessment

N = 10

Student Before After Difference D squared

Total 171 208 37 511

d - 0

d - 0

= -------------

t paired = t p =

S d2

Standard error of d

N

d =  D/N

= 37/10 = 3.7

 d 2 =  D 2 – ( D) 2 / N

= 511 - 1369/10 = 374.1

= 3.7 / 2.0387

=  d 2 / N - 1

S d2

= 1.815

= 374.1 / 10 – 1 = 41.5667

Probabilities of Type I and Type II errors

Truth

H0 True

H0 False

Type II

Error

Correct

results

Accept H0

Test

result

1 - 

Type I

Error

Correct

results

Reject H0

1 - 

Differences

H0 True = statistically insignificant

H0 False = statistically significant

Accept H0 = statistically insignificant

Reject H0 = statistically significant

Disease status

Present

Absent

Total

Present

a

b

a + b

Risk

Factor

Status

Absent

c

d

c + d

a + c

b + d

a+b+c+d

Total

Standard 2 X 2 table

a = subjects with both the risk factor and the disease

b = subjects with the risk factor but not the disease

c = subjects with the disease but not the risk factor

d = subjects with neither the risk factor nor the disease

a + b = all subjects with the risk factor

c + d = all subjects without the risk factor

a + c = all subjects with the disease

b + d = all subjects without the disease

a + b + c + d = all study subjects

Disease status

Present

Absent

Total

Present

a

b

a + b

Risk

Factor

Status

Absent

c

d

c + d

a + c

b + d

a+b+c+d

Total

Standard 2 X 2 table

Sensitivity = a/a+c

Specificity = d/b+d

Diabetic Screening Program

Disease status

Diabetic

Nondiabetic

Total

>125mg/100ml

5

13

18

Risk

Factor

Status

Sensitivity = a/a+c = 100 X 5/6 = 83.3% (16.7% false neg.)

Specificity = d/b+d = 100 X 81/94 = 86.2%(13.8% false pos.)

<125mg/100ml

1

81

82

6

94

100

Total