Download Lecture No.2 Inference For Two Population Mean The pooled t test

Lecture No.2
Inference For Two Population Mean
The pooled t test for two population means
(critical-value approach)
Assumptions
1.independent samples
2.normal populations or large samples
3.equal population standard deviations
Step 1: the null hypothesis is
H O : 1  2
and the alternative
hypothesis is
H a : 1  2 (two  tailed ) H O : 1   2 (left tailed )
H O : 1  2 ( right tailed )
Step 2 : decide on the significance level, 
Step 3: compute the value of the test statistic
T
Where
x1  x2
S p / (1 / n1 1 / n2 )
Sp 
(n1  1) s12  (n2  1) s22
n1  n2  2
Step 4: the critical value (s) are
 t / 2 (two  tailed ) or
 t (left  tailed )
or
t (right  tailed ) with degrees of freedom (df= n1 + n2 -2)
Step 5 : if the value of the t test statistics falls in the rejection
region, reject HO ; otherwise, fail to reject H0
Step 6 : interpret the results of the hypothesis test.
The pooled t test for two population means
(p-value approach)
Assumptions
1.independent samples
2.normal populations or large samples
3.equal population standard deviations
Step 1: the null hypothesis is
H O : 1  2
and the alternative
hypothesis is
H a : 1  2 (two  tailed ) H O : 1   2 (left tailed )
H O : 1  2 ( right tailed )
Step 2 : decide on the significance level, 
Step 3: compute the value of the test statistic
T
Where
Sp 
x1  x2
S p / (1 / n1 1 / n2 )
(n1  1) s12  (n2  1) s22
n1  n2  2
Step 4: the value of t-statistics has df= n1 + n2 -2. Use a table to
estimate the p-value or obtain it exactly by using technology.
Step 5 : if the P- value less than or equal  , ( p  value   ), reject
HO ; otherwise, fail to reject H0
Step 6 : interpret the results of the hypothesis test.
Example : we perform a hypotheses test to decide
whether there is a difference between the mean salaries
of faculty in public and private institutions. Independent
random samples of 20 faculty members in public
institutions and 35 faculty
members in private
institutions yield in the data in table below. At the 5%
significance level, do the data provide sufficient
evidence to conclude that means salaries for faculty in
public and private institutions differ?
Annual salary ($1000s)for 30 faculty members in public institutions and 35 faculty
members in private institutions
Sample 1 (public institutions)
Sample 2 (private institutions)
34.2
56.8 58.2 29.2 60.2
92.9
62.9
45.2 66.3 47.2 71.0
90.0
41.4 76.8 15.8 88.2
52.0
53.8
76.0 31.1 59.3 97.3
100.4 35.0 84.2 33.8 44.6
63.1
101.0 56.1 71.1 97.5 92.6
24.6
54.2 79.4 40.2 64.4 118.5
68.6
77.6 73.5 27.2 56.0
37.7
51.5
61.6 67.6 81.2 62.3
56.0 81.8 41.2 71.0 102.2
46.4
78.3 52.4 24.8
107.4 24.4 42.2 51.2 74.0
63.6
Solution:
Summary statistics for the samples
public institutions private institutions
x1  57.48
s1  23.95
n1  30
x2  66.39
s2  22.26
n2  35
Step 1: statethe null hypothesis and the alternative hypothesis
H O : 1  2 ( mean salaries are the same)
H a : 1  2
( mean salaries are the different)
Step 2 : decide on the significance level, 
  0.05
Step 3: compute the value of the test statistic
T
x1  x2
S p / (1 / n1 1 / n2 )
Where
Sp 
T
Sp 
(n1  1) s12  (n2  1) s22
n1  n2  2
(30  1)( 23.95) 2  (35  1) (22.26) 2
 23.05
30  35  2
57.48  66.39
 1.554
23.05 / (1 / 30 1 / 35)
Critical-value approach
Step 4: the critical value (s) are
 t / 2 (two  tailed ) with degrees of freedom (df= n1
+ n2 -2)
From a table the critical values wit (df = 30+35-2=63) are
 t / 2   t0.05 / 2  1.998
Step 5 : if the value of the t test statistics falls in the rejection
region, reject HO ; otherwise, fail to reject H0
From step 3 the value of the test statistics is t =-1.554, which
does not fall in the rejection region, thus we do not reject HO .
Step 6 : interpret the results of the hypothesis test.
at 5% significance level, the data do not provide
sufficient evidence to conclude that a difference exist
between the mean salaries of faculty in public and
private institutions .
p-value approach
Step 4: from a table(with df = 63) the p-value ( in two tailed)
greater than 0.1 and less than 0.20 ( 0.1 < p < 0.2) , and by
using technology, we obtain the p-value = 2 p ( t>= 1.554) =
0.125 (with df = 63)
Step 5: p value < 0.05) so we reject HO
at 5% significance level, the data do not provide
sufficient evidence to conclude that a difference exist
between the mean salaries of faculty in public and
private institutions .
use SPSS program
use the SPSS program to perform the hypothesis in
previous example
STEP 1: Enter The Data As Shown Below
Step 3 : the result shown below
Tests of Normality
a
SALARY
TYPE
PRIV
PUBL
Kolmogorov-Smirnov
Statistic
df
Sig.
.080
35
.200*
.105
30
.200*
Shapiro-Wilk
Statistic
df
.980
35
.975
30
*. This is a lower bound of the true significance.
a. Lilliefors Significance Correction
Group Statistics
SALARY
TYPE
PUBL
PRIV
N
30
35
Mean
57.480
66.394
Std. Deviation
23.9528
22.2611
Std. Error
Mean
4.3732
3.7628
Sig.
.755
.680
Independent Sample s Te st
SALARY Levene's Test for
F
Equality of Variances Sig.
t-test for Equality of t
Means
df
Sig. (2-tailed)
Mean Difference
Std. Error Difference
95% Confidence Interval
of the Difference
Lower
Upper
Equal variances
assumed
.458
.501
-1.554
63
.125
Equal variances
not assumed
-8.914
-8.914
5.7363
5.7692
-20.3774
2.5488
-20.4549
2.6264
-1.545
59.853
.128