Hypothesis Testing for

Continuous Variables

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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:

17.0:1 H

17.0:0 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:

211 : H

210 : 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 (large )λ

),(~ 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:

50:1 H

50:0 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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