Comparing two means

Two-sample z distribution

Two independent samples t -distribution

Two sample t -test

Two-sample t -confidence interval

Robustness

Details of the two sample t procedures

We have two independent SRSs (simple random samples) coming maybe from two distinct populations with (μ
1
, σ
1
) and (μ
2
, σ
2
). We use x
1 x
2 to estimate the unknown μ
1 and μ
2
.
 
When both populations are normal, the sampling distribution of (
1
− is also normal, with standard deviation :
 n
1
1
2

 n
2
2
2 x x
2
)
Then the two-sample z statistic has the standard normal N( 0 , 1 ) z

( x
1
 x
2



1
2 n
1
(

1


2
2 n
2
 
2
) sampling distribution.

We have two independent SRSs (simple random samples) coming maybe from two distinct populations with
1
,
1
and (
2
,
2
unknown. Use the sample means and sample s.d.s to estimate these unknown parameters.
To compare the means, both populations should be normally distributed.
However, in practice, it is enough that the two distributions have similar shapes and that the sample data contain no strong outliers.

The two-sample t statistic follows approximately the t distribution with a standard error SE (spread) reflecting variation from both samples: SE
 s
1
2 n
1
 s
2
2 n
2
Conservatively, the degrees of freedom (df) is equal to the smallest of (n
1
− 1, n
2
− 1).
 df s
1
2 n
1
 s n
2
2
2

μ
1 -
μ
2 x
1
 x
2


The null hypothesis is that both population means μ
1 and μ
2 are equal, thus their difference is equal to zero.
H
0
: μ
1
= μ
2
<> μ
1
− μ
2
 0 with either a one-sided or a two-sided alternative hypothesis.
We find how many standard errors (SE) away from (
μ
1
− μ
2
) is ( x
1
− x
2
) by standardizing:
Because in a two-sample test H
0

μ − μ
2
) 
0, we simply use
With df = smallest(n
1
− 1, n
2
− 1)
 t t

( x
1
 x
2
)

(

1
 
2
)
SE
 x
1
 x
2 s
1
2 n
1
 s
2
2 n
2


Does smoking damage the lungs of children exposed to parental smoking?
Forced vital capacity (FVC) is the volume (in milliliters) of air that an individual can exhale in 6 seconds.
FVC was obtained for a sample of children not exposed to parental smoking and a group of children exposed to parental smoking.
Parental smoking FVC x s
Yes
No
75.5
88.2
9.3
15.1
 n
30
30
We want to know whether parental smoking decreases children’s lung capacity as measured by the FVC test.
Is the mean FVC lower in the population of children exposed to parental smoking?

H
0
: μ smoke
= μ no
H a
:μ smoke
< μ no
<=> (μ smoke
− μ no
) = 0
<=> (μ smoke
− μ no
) < 0 (one sided)
The difference in sample averages follows approximately the t distribution with 29 df:
We calculate the t statistic: t
 t
 x smoke
 x no
2 s smoke n smoke
 s
2 no n no


12 .
7
2 .
9

7 .
6
 
3 .
9
75 .
5

88 .
2
9 .
3
2
30

15 .
1
2
30
Parental smoking
Yes
No

In table 3, for df 29 we find:
FVC x
75.5
88.2
s
9.3
15.1
| t | > 3.659 => p < 0.0005 (one sided)
It’s a very significant difference, we reject H
0
.
n
30
30
Lung capacity is significantly impaired in children of smoking parents.


Because we have two independent samples we use the difference x
1
− x
2
) to estimate (μ
1
−μ
2
).
Practical use of t: t*


C

− t * and
We find t * in the line of Table 3 t *.
for df = smallest ( n
1
−1; n
2
−1) and the column for confidence level C.

The margin of error MOE is: 
C
SE
 s
1
2 n
1
 s
2
2 n
2
MOE
 t * s
1
2 n
1
 s
2
2 n
2
 t * SE − t * t *

Example: Can directed reading activities in the classroom help improve reading ability? A class of 21 third-graders participates in these activities for 8 weeks while a control classroom of 23 third-graders follows the same curriculum without the activities. After 8 weeks, all children take a reading test
(scores in table).

95% confidence interval for ( µ
1
− µ
2
), with df = 20 conservatively  t * = 2.086:
CI : ( x
1
 x
2
)

MOE ; MOE
 t * s
1
2 n
1
 s
2
2 n
2

2.086 * 4.31

8.99
With 95% confidence, ( µ
1
− µ
2
), falls within 9 .
96 ± 8.99 or 1.0 to 18.9.

“The two-sample t procedures are more robust than the one-sample t methods. When the sizes of the two samples are equal and the distributions of the two populations being compared have similar shapes, probability values from the t table are quite accurate for a broad range of distributions when the sample sizes are as small as n
1
= n
2
= 5”
 When planning a two-sample study, choose equal sample sizes if you can.
As a guideline, a combined sample size ( n
1
+ n
2
) of 40 or more will allow you to work even with the most skewed distributions. For very small samples though, make sure the data is very close to normal – no outliers, no skewness…

The true value of the degrees of freedom for a two-sample t distribution is quite lengthy to calculate. That’s why we use an approximate value, df = smallest( n
1
− 1, n
2
− 1), which errs on the conservative side (often smaller than the exact).
Computer software, though, gives the exact degrees of freedom —or the rounded value —for your sample data.
df
 n
1
1

1

 s
1 n
1
2

 s
1 n
1
2


2

 s
2
2


2 n
2 n
2
1

1

 s n
2
2
2


2


There are two versions of the two-sample t -test: one assuming equal variance (“pooled 2-sample test”) and one not assuming equal variance ( “unequal” variance , as we have studied) for the two populations. They have slightly different formulas and degrees of freedom.
The pooled (equal variance) twosample t -test was often used before computers because it has exactly the t distribution for degrees of freedom n
1
+ n
2
− 2.
Two normally distributed populations with unequal variances
However, the assumption of equal variance is hard to check, and thus the unequal variance test is safer.

When both populations have the same standard deviation, the pooled estimator of σ 2 is:
The sampling distribution for ( x
1
− x
2
) has exactly the t distribution with
(n
1
+ n
2
− 2) degrees of freedom .
A level C confidence interval for µ
1
− µ
2 is
(with area C between − t* and t *)
To test the hypothesis H
0
: µ
1
= µ
2 against a one-sided or a two-sided alternative, compute the pooled two-sample t statistic for the t(n
1
+ n
2
− 2 ) distribution.

For next time: Be sure to carefully read through sections 6.1 and 6.2
Then work on: #6.1, 6.4, 6.5, 6.10, 6.12