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Student's t test

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Student’s t test

 This test was invented by a

statistician working for the

brewer Guinness. He was called

WS Gosset (1867-1937), but

preferred to keep anonymous so

wrote under the name “Student”.

The t-distribution









William Gosset

lived from 1876 to 1937



Gosset invented the t -test to handle small samples for quality

control in brewing. He wrote under the name "Student".

t-Statistic

x

t

s/ n



 When the sampled population is

normally distributed, the t statistic is

Student t distributed with n-1 degrees

of freedom.

T-test

1. Test for single mean

Whether the sample mean is equal to the predefined

population mean ?





2. Test for difference in means

Whether the CD4 level of patients taking treatment A is

equal to CD4 level of patients taking treatment B ?





3. Test for paired observation

Whether the treatment conferred any significant benefit ?

T- test for single mean

The following are the weight (mg) of each of 20

rats drawn at random from a large stock. Is it

likely that the mean weight for the whole stock

could be 24 mg, a value observed in some previous

work?.



9 18 21 26

14 18 22 27

15 19 22 29

15 19 24 30

16 20 24 32

Steps for test for single mean

1. Questioned to be answered

Is the Mean weight of the sample of 20 rats is 24 mg?

N=20, x =21.0 mg, sd=5.91 ,  =24.0 mg



2. Null Hypothesis

The mean weight of rats is 24 mg. That is, The

sample mean is equal to population mean.

x

3. Test statistics t --- t (n-1) df

s/ n

4. Comparison with theoretical value

if tab t (n-1) cal t (n-1) accept Ho,

5. Inference

t –test for single mean

 Test statistics

n=20, x =21.0 mg, sd=5.91 ,

 =24.0 mg

l 21.0  24l

t   2.30

5.91 20

t = t .05, 19 = 2.093 Accept H0 if t = 2.093

Inference :



There is no evidence that the sample is taken

from the population with mean weight of 24 gm

Determining the p-Value





Area = .025

Area = .025





Area =.005 Area = .005







Z









1.96

-2.575









2.575

-1.96









0

f(t)



.9

.025 5 .025

-1.96 0 1.96 t

red area = rejection region for 2-sided test

T-test for difference in means

Given below are the 24 hrs total energy

expenditure (MJ/day) in groups of lean and

obese women. Examine whether the obese

women‟s mean energy expenditure is

significantly higher ?.



Lean Obese

6.1 7.0 7.5 8.8 9.2 9.2

7.5 5.5 7.6 9.7 9.7 10.0

7.9 8.1 8.1 11.5 11.8 12.8

8.1 8.4 10.2

10.9

Two sample t-test



Difference

between means

+

Sample size + t-test t

Variability

of data

T-test for difference in means

Null Hypothesis

Obese women’s mean energy expenditure is

equal to the lean women’s energy expenditure.

Test statistics :

x1  x 2

t t(n1+n2-2)

1 1

 

n1 n2

(n1  1)s1  (n 2  1)s2

2 2



n1  n 2  2





x 1, x 2 - means of sample 1 and sample 2



 1,  2 – sd of sample 1 and sample 2

n1 , n2 – number of study subjects in sample 1 and

sample 2

T-test for difference in means

Data Summary

lean Obese

l 8.1  10.3l

N 13 9 t  3.82

x 8.10 10.30 1.32 1.25



2 2









S 1.38 1.25 9 13



tab t 9+13-2 =20 df = t 0.05,20 =2.086



Inference : The cal t (3.82) is higher than tab t at

0.05, 20. ie 2.086 . This implies that there is a

evidence that the mean energy expenditure in obese

group is significantly (p 0

So, need to know ∑D and ∑D2:

Before After

Student Program Program D D2

1 520 555 35 1225



2 490 510 20 400



3 600 585 -15 225



4 620 645 25 625



5 580 630 50 2500



6 560 550 -10 100



7 610 645 35 1225



8 480 520 40 1600



∑D = 180 ∑D2 = 7900

Recall that for single samples:

X   score - mean

tobt  

sX standard error



For related samples:



D  D

tobt 

sD

where:

 D  2





sD 

sD

and D  N

2





N sD 

N 1

Mean of D:



D

 D  180  22.5

N 8



Standard deviation of D:



 D  2

180 2



D  N

2

7900 

8

sD    23.45

N 1 8 1



Standard error:



sD 23.45

sD    8.2908

N 8

D  D

tobt 

sD

Under H0, µD = 0, so:



D 22.5

tobt    2.714

sD 8.2908

From Table B.2: for α = 0.05, one-tailed, with df = 7,



tcrit = 1.895



2.714 > 1.895 → reject H0



The program is effective.

t-Value

t is a measure of:

How difficult is it to believe the null hypothesis?

High t

Difficult to believe the null hypothesis -

accept that there is a real difference.



Low t

Easy to believe the null hypothesis -

have not proved any difference.

In Conclusion !

 Student „s t-test will be used:

--- When Sample size is small

and for the following situations:

(1) to compare the single sample mean

with the population mean

(2) to compare the sample means of

two indpendent samples

(3) to compare the sample means of

paired samples



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