Hypothesis Testing for

Continuous Variables

Yuantao Hao

19th,Otc., 2009

Chapter4

We have learned:

Basic logic of hypothesis testing

Main steps of hypothesis testing

Two kinds of errors

One-side & two-side test

One-sample t test

4.3

The t Test for Data under

Randomized Paired Design

4.3 The t Test for Data under Randomized Paired Design

Example 4.2 The weights (kg) of 12

volunteers were measured before and

after a course of treatment with a “new

drug” for losing weight. The data is given

in Table 4.1. Please evaluate the

effectiveness of this drug.

Table 4.1 The data observed in a study of weight losing

Weight (kg) No.

Pre-treatment (X1) Post-treatment (X2)

Difference

d=X1-X2

1 101 100 1

2 131 136 -5

3 131 126 5

4 143 150 -7

5 124 128 -4

6 137 126 11

7 126 116 10

8 95 105 -10

9 90 87 3

10 67 57 10

11 84 74 10

12 101 109 -8

∑ d=16

Solution:

0:0 dH

0:1 dH

05.0

58.01291.7

33.1

/

0

nS

dt

d

11112 P ＞0.50

4.4

The Tests for Comparing Two Means

Based on Two Groups of Data under

Completely Randomized Design

4.4 The t test for Comparing Two Means

About this design:

1. The individuals are randomly divided into two

groups which correspond to two treatments

respectively;

2. The two groups are randomly selected from

two populations respectively.

Example 4.3

Assume the red cell counts of healthy

male residents and healthy female

residents of certain city follow two

normal distributions respectively, of

which the population means are

different and population standard

deviations are equal.

There are two random samples drawn from the

two populations respectively, of which the sample

sizes, sample means and sample standard

deviations are ;

; .

It is expected to estimate the 95% confidence

interval for the difference of average red cell

counts between healthy male and female

residents.

15,20 21 nn 18.4,66.4 21 xx

45.0,47.0 21 ss

Solution I t has been known that the sample variance f or males and

f emales are: 21s =(0.47)2=0.2209,

22s =(0.45)2=0.2025.

They are f airly closed each other so that the two population variances could be regarded as equal.

2131.021520

)45.0)(115()47.0)(120( 222

cs

3321520

Given 05.0 , check up the table f or t distribution, the two-side

critical value 05.0t =2.034.

)15

1

20

1(2131.0034.2)18.466.4()

11()(

21

221

nnstxx c

16.0 and 80.0)1577.0(034.248.0

Therefore, the 95% confi dence interval f or the diff erence of average red cell counts between healthy male and f emale residents is (0.16, 0.80) (1012/ L).

Question:

Please judge whether the population means of

males and females are equal or not.

210 : H

211 : H

4.4.1 Equal Variances

2

11

21

222

2112

nn

SnSnSc

dist. ~

)11

(

)(

21

2

21 t

nnS

XXt

c

221 nn

Solution:

04.3

15/120/12131.0

18.466.4

t

3321520221 nn

P＜ 0.01

4.4.2 Unequal Variances

2221

21

21

// nSnS

XXt

21

2211'

ww

twtwt aaa

2222 / nSw

12

11 / nSw

Satterthwaite’s method

Example 4.4 n1=10 patients and n2=20 healthy

people are randomly selected and measured for a

biochemical index. The mean and standard

deviation of the group of patients are =5.05

and S1=3.21, and that of the group of healthy

people are =2.72 and S2=1.52.

Please judge whether the two population means

are equal or not.

1X

2X

Solution:

18.2

20/52.110/21.3

72.205.522

t

24.2

2052.1

1021.3

09.22052.1

26.21021.3

22

22

'05.0

t

P＞ 0.05

4.5

The F-Test for Equal Variances of

Two Groups of Data under

Completely Randomized Design

4.5 The F-Test for Equal Variances of Two Groups

22

210 : H

22

210 : H

.~/

/22

22

21

21 distFS

S

1=n1-1, 2=n2-1

.~22

21 distFS

SVR

21

12

,,

',,

1

F

F

The larger variance is always taken as the

numerator of the statistic VR for convenience.

Thus, to a two-side test, given , one may

use /2 to find the upper critical value F/2 of

the F distribution, and if the current value of

VR is greater than or equal to F/2 ， then P≤

， otherwise, P ＞ ;

Returning to Example 4.3, where

VR=1.09 ， 1=19 ， 2=14. Let =0.10 ，

the two-side critical value of F

distribution is F0.10/2=F0.05=2.40. Since VR

＜ F0.05, not reject , hence the there is

no enough evidence to say that the two

population variances are not equal.

0H

4.6.1

The Z-test for the population probability

of binomial distribution (large n)

4.6.1 The Z-test for the population probability

),(~ nBX

))1(,(~ nnNX

))1(

,(~n

Nn

XP

n5

n(1-)5

4.6.1.1 One sample

Example 4.7 150 physicians being

randomly selected from the departments

of infectious diseases in a city had

received a serological test. As a result, 35

out of 150 were positive(23%). It was

known that the positive rate in the

general population of the city was 17%.

Please judge whether the positive rate

among the physicians working for the

departments of infectious diseases was

higher than that in the general

population.

Solution:

17.0:0 H

17.0:1 H

)150

)17.01(17.0,17.0(~

NP

06.2

150/17.0117.0

17.0150/35

Z

02.0P

4.6.1.2 Two samples

Example 4.8 To evaluate the effect of

the routine therapy incorporating with

psychological therapy, the patients with

the same disease in a hospital were

randomly divided into two groups

receiving routine therapy and routine

plus psychological therapy respectively.

After a period of treatment, evaluating

with the same criterion, 48 out of 80

patients (60%) in the group with routine

therapy were effective, while 55 out of 75

(73%) in other group were effective.

Please judge whether the probability of

effective were different in terms of

population.

Solution:

210 : H

211 : H

))1(

,(~1

1111 n

NP )

)1(,(~

2

2222 n

NP

))1()1(

,0(~2

22

1

1121 nnNPP

155

103

7580

5548

21

22110

nn

PnPnP

))75

1

80

1)(

155

1031(

155

103,0(~21 NPP

75

1

80

1

155

1031

155

10321 PP

Z

76.1

751

801

155103

1155103

7555

8048

Z

P=0.08

4.6.2

The Z-test for the population mean of

Poisson distribution (largeλ)

4.6.2 The Z-test for the population mean of Poisson distribution (largeλ)

),(~ NX

4.6.2.1 Single observation

Example 4.9 The quality control

criterion of an instrument specifies that

the population mean of radioactivity

recorded in a fixed period should not be

higher than 50. Now a monitoring test

results in a record of 58. Please judge

whether this instrument is qualified in

terms of the population mean.

Solution:

50:0 H

50:1 H

)50,50(~ NX

13.150

5058

Z

P = 0.13

4.6.2.2 Two observations

Example 4.10 The radioactivity of two

specimens was measured for 1 minute

independently, resulting in X1=150 and

X2=120 respectively. Please judge

whether the two corresponding

population means in 1 minute are equal

or not.

210 : H

Solution:

211 : H

),(~1 NX ),(~2 NX

)2,0(~21 NXX

)1,0(~21

21 NXX

XXZ

83.1120150

120150

Z

4.6.2.3 Two “groups” of observations

Example 4.11 The radioactivity of two

specimens was independently measured

for 10 minutes and 15 minutes

respectively, resulting in X1=1500 and

X2=1800.

Please judge whether the two

corresponding population means in 1

minute are equal or not.

Solution:

),(~ 111 NX ),(~ 222 NX

),(~1

111 n

NX ),(~

2

222 n

NX

),(~2

2

1

12121 nn

NXX

2

2

1

1

21

nX

nX

XXZ

26.6

1512010150

120150

Z

THE END

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