T test

One sample
t test,

independent
& paired

by Nur Zakiah Mohd Saat

One sample, independent sample t test & paired t test

One sample t test
Example

A study was conducted to examines the stress scores , stress reactions towards mental heals
based on General Health Questionaire (GHQ28) among adults age 15-20 in Selangor
Question
Is the total mean score of GHQ28 differed than 84
84 is the standard value(stress)

Example
Hypothesis testing
(ghq=84, stress)
Two sided
Ho: µ=84
Ha: µ≠84

One –sided
Ho: µ=84
Ha: µ>84 OR
Ha: µ<84

Solution
One sample t-test
Ho:=0
H1: <0

t  x  0

s/ n

If -t<-t(df=n-1,) Reject Ho
If t>t(df=n-1,  ) Do not reject Ho/Accept Ho

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Formulae mean and standard deviation

n

 xi

x  i1
n

n  xi  x 2



s i 1

n 1

Hypothesis testing Rejection area
Finding the critical value

Two sided
Ho: µ=84
Ha: µ≠84

Rejection area

T critical T critical

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One –sided
Ho: µ=84
Ha: µ>84 OR
Ha: µ<84

What is the value of tcritical and draw the rejection area

t-table Rejection area(two-sided)
Rejection area(one sided right)

What is the value of tcritical and draw the rejection area

Rejection area(one-sided left)
T table

Solution
Two SIDED
Df=4, =0.05/2=0.025
tcritical=2.776

Rejection area(two-sided)
reject Ho if t>2.776 or -t<-2.776
T table
ONE SIDED
Df=4, =0.05
tcritical=2.132

Rejection area(two-sided)
Reject Ho if t >2.132

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Example

Suppose the mean birthweight is 120 oz , based on a sample size 25. The mean birthweight is
found to be 115 oz with standard deviation 25 oz. Assess the result of the study
Hypotheses
Ho: =120

Test value t  115 120  1
25 / 25

Critical value t(24,0.05 )=1.711
Decision : Since the test value is more than -1.711, the decision is do not reject Ho

Two sided alternatives
Ho: = 
H1: 0

Examples
Suppose we assume that the mean of cholesterol levels in women aged 21-40 is 180 mg/dL .
Blood test is performed on 15 female from Semenyih aged 21-40, and is found to be 175
mg/dL with standard deviation 20 mg/dL. What can be concluded on the basis of this
evidence.

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Hypotheses
Ho:=180
H1: ≠180
Compute test value

t  175180  0.968
20/ 15

Critical value
Df=14, 0.05=2.145
Reject the Ho if t>2.145 or if t<-2.145
Decision: Do not reject Ho, thus there is not enough evidence to support the claim that the
cholesterol level of women aged 21-40 in Malaysia is differ than 180 mg/dL

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Independent t test

Example
A study was conducted to examines the stress scores , stress reactions towards mental heals
based on General Health Questionaire (GHQ28) among adults age 15-20 in Selangor
Question
Is the total mean score of GHQ28 differed between male and female

Hypothesis testing
Two sided
Ho: µ1=µ2
Ha: µ1≠µ2

One –sided
Ho: µ1=µ2
Ha: µ1>µ2 OR
Ha: µ1<µ2
For Testing the Difference between Two Means
Variances are assumed to be unequal
Df1= n1+n2-2

t  (X1  X 2)
s12  s22
n1 n2

Variances are assumed to be equal
Df=n1+n2-2

t  x1  x2
s12 (n1 1)  s22 (n2 1) 1  1
n1  n2  2 n1 n2

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Confidence interval for equal variance

 x1  x2  tcritical s12(n1 1)  s22(n2 1)
n1  n2  2

Confidence interval for UNequal variance

 x1  x2  tcritical s12  s22
n1 n2

Example
A researcher wishes to determine whether the salaries of professional nurses employed by
private hospitals are higher than those of nurses employed by government-owned hospitals.

Private
Mean=26800
sd=600
N=10
Government
Mean=25400
sd=450
N=8

Test of equality of variance

H0 :  2   2
1 2

H1 :  2   2
1 2

F  s12  6002  1.777
s22 4502
7

F table
Df1=9
Df2=7
=0.05
Fcritical=F(9,7,0.05)=3.67
F calc<F critical
Do not reject Ho
Variance group 1 equal to variance group 2

Independent t test

t  x1  x2
s12 (n1 1)  s22 (n2 1) 1  1
n1  n2  2 n1 n2

Hypotheses

Ho : 1  2
H1 : 1  2

t=5.47
Critical value
Df=16, ( 1.746, =0.05 )
Decision: The salaries paid to nurses employed by private hospitals are higher than those paid
to nurses employed by government hospital.

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PAIRED T -TEST

Paired t test
Example
A study was conducted to examines the stress scores , stress reactions towards mental heals
based on General Health Questionaire (GHQ28) among adults age 15-20 in Selangor
Question
Is the total mean score of GHQ28 differed during week 1 and week 5

Hypothesis testing
Two sided
Ho: µd=0
Ha: µd≠0

One –sided
Ho: µd=0
Ha: µd>0 OR
Ha: µd<0

The paired t Test
A dietitian wishes to see if a person’s cholesterol will change if the diet supplemented by a
certain mineral

Subject 1 2 34 5 6

Before 210 235 208 190 172 244

After 190 170 210 188 173 228

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hypothesis

Ho:d=0
Ha: d≠0
Degree of freedom=np-1=5
t critical =0.05 , tc=2.571
Critical region
Reject Ho if -t<-2.571 or t>2.571

Dependent Samples

t D
s/ n

D D

n

 D2  ( D)2

s n

n 1

Subject 1 2 3 4 5 6

Before 210 235 208 190 172 244

After 190 170 210 188 173 228

D 20 65 -2 2 -1 16

D2 400 4225 4 4 1 256

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Mean of the difference =16.7

 D2  4890

Standard deviation= 25.4
t=1.610
Critical value= 2.571
Decision: There is not enough evidence to support the claim that the mineral changes a
person’s cholesterol level

Example 4.1 Using SPSS

First, we must setup the variables in SPSS.
DependentVariable1 = preseas
(for Pre-season scores)
DependentVariable2 = postseas
(for Post-season scores)

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p value=0.009<0.01.
Reject Ho. There was significance
mean difference between pre and
post season at =0.01

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