One-sided alternative

One-sided alternative

0

0: =

H µ µ H1:µ ≠ µ0

two-sided

0 0:

H µ≥µ H₁:µ <µ₀ one-sided

Example 3

Ni content of a solution is measured: 3.25, 3.27, 3.24, 3.26, 3.24

The required minimum value is 3.25 g/cm³ Is it reached?

1

0 0:

H µ≥µ H₁:µ <µ₀

t

n s n s x n s

t₀=x−µ⁰ = −µ+µ−µ⁰

(

)

<0 if H₁

lower tail, lower limit

(minimum value is not reached)

0 0:

H µ≥µ H₁:µ < µ₀

The hypothesisedµ0=3.25 value is above the lower confidence limit (p>0.05), accepted:

failed to reject ...

not proved that at most...

0 0

H :µ µ≤ H :₁ µ µ> ₀

The hypothesisedµ₀^=3.25 value is below the upper confidence limit (p>0.05), accepted:

failed to reject ...

not proved that at most...

If the null hypothesis is accepted it does not „prove”

Comparing two variances ( F test)

2 2 2 1 1 :σ >σ H

2 2 2 1 0:

H σ =σ

The test statistic: 2^;

(

)

2 2 1

0 = n − ,n −

s F s

In case of one-sided alternative:

Fα

s s₁² / ₂² >

The null hypothesis is rejected if

5

In case of two-sided alternative: H₁:σ₁² ≠σ₂²

The null hypothesis is rejected if

2 2 1

2 2 1

F -a/

s

s < 2 2

2 2 1

Fa/

s s >

or

1 / ₂²

2

1 s ≥

s it is sufficient to check the upper limit

using

Two-sample t test

Two independent samples 1 2

2 2 2 1 2

1,n ; s ,s ; x,x n

Assuming the equality of variance for the two populations (to be checked through F-test):

( )

:

H₀ ^Ex₁−^x₂ =µ₁−µ₂ =

2 2 2

1 σ

σ =

7

( ) ( )

[

]

2 1

2 2 2 1

2 1 2 1

2 = s n − +s n −

- +n s n

( )

2 1

2 1 2 1

1 1

n s n

x x E x t=x

+

−

−

−

2 2

1+ −

=n n ν

( )

:

H₀ ^{E x}₁−^x₂ =µ₁−µ₂ =

The test statistic:

,

(

) (

)

= n n

ν

σ

σ

2 2

= 2

The assumption is checked through

F

2 1

2 1

0 1 1

n s n

x

= x t

+

−

Example 4

(Box-Hunter-Hunter: Statistics for Experimenters, J. Wiley, 1978, p. 97)

The wear of two kinds of raw material is compared as shoe soles on the foot of 10-10 boys (shoes1.xls).

Is the difference of means and variances significant at 0.05 level?

n mean sample variance

A 10 10.63 6.009

B 10 11.04 6.343

TABLE 4.3. Data on the amount of wear measured with two different materials A and B, boy’s shoes example*

boy material A material B B – A

difference d

1 13.2(L) 14.0(R) 0.8

2 8.2(L) 8.8(R) 0.6

3 10.9(R) 11.2(L) 0.3

4 14.3(L) 14.2(R) -0.1

5 10.7(R) 11.8(L) 1.1

6 6.6(L) 6.4(R) -0.2

7 9.5(L) 9.8(R) 0.3

8 10.8(L) 11.3(R) 0.5

9 8.8(R) 9.3(L) 0.5

10 13.3(L) 13.6(R) 0.3

average difference 0.41

Example 15

(Box-Hunter-Hunter: Statistics for Experimenters, J. Wiley, 1978, p. 97)

wear

material A material B boys

5 7 9 11 13 15

0 1 2 3 4 5 6 7 8 9 10 11

Paired t test ( ) ( )

:

H₀ E x_i =E y_i =

i i

i x y

d = −

( ) ( ) ( )

E = −

: H₀

n d

d ⁱ

∑

=

( )

1

2

2

n- d - d

s ⁱ

i d

∑

= s / n

= d t

d 0

one-sample

t

13

Paired t test ( )

0:E d =

H di =xi −yi

1. 1.

2. 2.

dependent samples

boys

B - A

-0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

0 1 2 3 4 5 6 7 8 9 10 11

149 .

2 =0

sd sd = 0.149 =0.386 0.122

10 386 .

0 =

n = s_d

4 . 122 3 . 0

41 . 0

0 = =

t

beta

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

two-sample paired

OC curve for the boys shoes example

Statistics > Basic Statistics/Tables t-test, dependent samples

Box Plot of multiple variables wear 5v*10c

Median; Box: 25%-75%; Whisker: Non-Outlier Range

Median 25%-75%

Non-Outlier Range

A B DIFF

-2 0 2 4 6 8 10 12 14 16

Testing goodness of fit

It is to be judged if the data may come from a certain distribution

Normality test

graphical tests statistical tests

Open:

body.sta

Statistics > Distribution fitting > continuous Distributions > Normal

Variable: WEIGHT, Distribution: Normal Kolmogorov-Smirnov d = 0,04966, p < 0,20, Lilliefors p < 0,01

Chi-Square test = 15,64733, df = 7 (adjusted) , p = 0,02854

140 145 150 155 160 165 170 175 180 185 190 195 200 205

Category (upper limits) 0

10 20 30 40 50 60 70 80 90 100 110

No. of observations

Graphs > 2D Graphs > Normal Probability Plots…

Normal Probability Plot of WEIGHT (body 4v*500c)

-4 -3 -2 -1 0 1 2 3 4

Expected Normal Value

Large samples:

Kolmogorov–Smirnov test

The data are grouped into classes, at least 5 classes are required.

χ² -test

The data are grouped into classes, at least 5 occurrences are required in a class.

Smaller samples:

Shapiro–Wilk test

Statistical tests for goodness of fit

Statistics > Distribution fitting > continuous Distributions > Normal

Options tab

Kolmogorov-Smirnov test: yes(continuous)

Variable: WEIGHT, Distribution: Normal

Kolmogorov-Smirnov d = 0,04966, p < 0,20, Lilliefors p < 0,01 Chi-Square test = 15,64733, df = 7 (adjusted) , p = 0,02854

30 40 50 60 70 80 90 100 110

No. of observations

Graphs > 2D Graphs > Normal Prbability Plots…

Shapiro–Wilk test

Normal Probability Plot of WEIGHT (body 4v*500c)

140 150 160 170 180 190 200 210

Obs erved V alue -4

-3 -2 -1 0 1 2 3 4

Expected Normal Value

WEIGHT: SW-W = 0,9927; p = 0,0160

Normal Probability Plot of maleWEIGHT (body unstack 4v *253c)

-1 0 1 2 3 4

pected Normal Value

Variable: maleWEIGHT, Distribution: Normal

Kolmogorov-Smirnov d = 0,07996, p < 0,10, Lilliefors p < 0,01 Chi-Square test = 4,56330, df = 4 (adjusted) , p = 0,33511

150 155 160 165 170 175 180 185 190 195 200 205

Category (upper limits) 0

10 20 30 40 50 60 70 80 90

No. of observations