The t Test - University of Arizona

The t Test

April 3-8, 2008

Again, we begin with independent normal observations X1. . . . , Xn with unknown mean ? and unknown variance 2. The likelihood function

L(?, 2|x)

=

1 (22)n/2

1 exp - 22

n

(xi - ?)2

i=1

Thus, ?^ = x?.

ln

L(?, 2|x)

=

n - (ln

2

2

+

ln

2)

-

1 22

n

(xi - ?)2

i=1

?

ln

L(?,

2|x)

=

-

1 2

n

(xi - ?)

i=1

Thus, For the hypothesis

2

ln L(?, 2|x)

=

n - 22

+

1 2(2)2

n

(xi - ?)2.

i=1

^2 = 1 n

n

(xi - x?)2.

i=1

H0 : ? = ?0 versus H1 : ? = ?0,

the likelihood ratio test

(x)

=

L(?0, ^02|x) L(?^, ^2|x)

where the value

^02

=

1 n

n

(xi - ?0)2

i=1

gives the maximum likelihood on the set ? = ?0.

L(?0, ^02|x)

=

1 (2^02)n/2

1 exp - 2^02

n

(xi

i=1

-

?0)2

=

1 (2^02)n/2

2 exp - ,

n

1

L(?^, ^2|x))

=

1 (2^2)n/2

1 exp - 2^2

n

(xi - x?)2

=

1

2

(2^2)n/2 , exp - n ,

i=1

and

^2 n/2

(x) = ^02

=

The critical region )x) 0 is equivalent to

ni=1(xi - ?0)2 -n/2 ni=1(xi - x?)2

c

ni=1(xi - ?0)2 ni=1(xi - x?)2

=

1+

n(x? - ?0)2

n i=1

(xi

-

x?)2

or T (x)2 (c - 1)(n - 1)

where

T (x) = x? -?0 s/ n

and we write s for the square root of the unbiased estimator of the variance

s2 = 1 n-1

n

(xi - x?)2.

i=1

1 Deriving the t distribution

For the hypothesis test, our next goal is to understand the distribution of T (X).

Step 1. n(X? - ?0)/ is a standard normal random variable.

For

this,

notice

that

X?

is

a

normal

random

variable

with

mean

?0

and

standard

deviation

/ n

Step 2. For each i, Xi - X? and X? are independent.

For normal random variables, uncorrelated random variables are independent. Thus, it suffices to show that the covariance is 0. To that end, note that

Cov(Xi - X? , X? ) = Cov(Xi, X? ) - Cov(X? , X? ).

For the first term, use the fact that Cov(X1, Xj) = 0 if i = j and Cov(X1, Xi) = Var(Xi) = 2. Then,

Cov(Xi, X? )

=

1 n

n

Cov(Xi, Xj)

=

1 2. n

j=1

From Step 1, we know that

Cov(X? , X? ) = Var(X? ) = 1 2. n

Now combine to see that Cov(Xi - X? , X? ) = 0.

Step 3. ni=1(Xi - X? )2/2 is a -square random variable with n - 1 degrees of freedom.

2

Let Zi = (Xi - ?)/ and Z? be the average of the Zi. Then Zi are independent standard normal random

variables.

1 2

n

(Xi - X? )2 =

n

(Zi - Z?)2 =

n

Zi2 - nZ?2

i=1

i=1

i=1

or

n

n

Zi2 = (Zi - Z?)2 + nZ?2.

i=1

i=1

Let's write this

Y = U + V.

By step 2, the sum is of independent random variables. So, if we use the properties of moment generating functions

MY (r) = MU (r) ? MV (r).

Now Y is a 2n random variable. So, MY (r) = (1 - 2r)-n/2. V is a 21 random variable. So, MY (r) = (1 - 2r)1/2. Consequently,

MU (r)

=

MY (r) MV (r)

=

(1 - 2r)-n/2 (1 - 2r)-1/2

=

(1 - 2r)-(n-1)/2

and U is a 2n-1 random variable. In summary, we can write

Z T=

U/(n - 1)

where Z is a standard random variable, U is a 2n-1 random variable, and Z and U are independent. Consequently, their densities are

fZ (z)

=

1 e-z2/2 2

and

fU (u)

=

un/2-3/2 2(n-1)/2((n -

e-u/2. 1)/2)

Step 4. Finding the density of T , fT (t). Z and U have joint density

fZ,U (z, u)

=

1 e-z2/2

un/2-3/2

e-u/2.

2

2(n-1)/2((n - 1)/2)

Define the one to one transformation t=

z and v = u.

u/(n - 1)

Then, the inverse transformation

z = t v and u = v.

n-1

3

the joint density

tv

fT ,V

(t,

v)

=

fZ,U

(

n

-

1

,

v)|J

(t,

v)|.

where |J(z, u)| is the absolute value of the Jacobian of the inverse transformation. In this case,

J(t, v) = det

z/t z/v u/t u/v

= det

v/ n - 1 t/(2 v(n - 1)

0

1

= v .

n-1

Then,

fT,V (t, v)

=

1

22(n-1)/2((n

vn/2-3/2 exp - 1)/2)

v -

2

t2 1+

n-1

v

n-1

Finally, to find the marginal density for T , we integrate with respect to v to obtain

1

fT (t)

=

22(n-1)/2((n

-

1)/2) n

-

1

vn/2-1 exp

0

v -

2

t2 1+

n-1

dv.

Change variables by setting w = v(1 + t2/(n - 1))/2.

fT (t)

=

1

22(n-1)/2((n - 1)/2) n - 1 0

2w 1 + t2/(n - 1)

n/2-1

e-w

2 1 + t2/(n - 1)

dw

=

1

t2 1+

-n/2

wn/2-1e-w dw

(n - 1)((n - 1)/2)

n-1

0

(n/2)

t2

-n/2

=

1+

.

(n - 1)((n - 1)/2)

n-1

Note that

t2 -n/2

t2

1+

exp -

n-1

2

and n . Thus, for large n, the t density is very close to the density of a standard normal.

2 Tests and confidence intervals using t distribution

We can now return to our original hypothesis. The critical region for an alpha level test is determined by the extreme values of the t statistic.

C = {x; |T (x)| t/2}

where

P {T

> t/2} =

. 2

Correspondingly, the -level confidence interval is obtained by choosing = 1 - and setting

s

x?

?

n

t/2.

4

For a one-sided hypothesis the critical region becomes

H0 : ? ?0 versus H1 : ? ?0, C = {x; T (x) t}

3 Two sample t test

We now have a three dimensional parameter space = {(?1, ?2, 2); ?1 R, ?2 R, 2 R}.. The two-sided test is H0 : ?1 = ?2 versus H1 : ?1 = ?2,

The data X1,j, . . . , Xnj,j are independent N (?j, 2) random variables, j = 1, 2. The likelihood function is

L(?1, ?2, 2|x1, x2)

=

1 (22)(n1+n2)/2

1 exp - 22

n1

n2

(xi,1 - ?1)2 + (xi,2 - ?2)2 .

i=1

i=1

Then, the likelihood ratio,

(x1, x2))

=

L(?^, ?^, ^122|x1, x2) L(?^1, ?^2, ^2|x1, x2)

Write the unbiased estimators for the mean and the variance

1 nj

x?j

=

nj

xi,j

i=1

and

s2j

=

1 nj -

1

nj

(xi,j

i=1

- x?j)2,

j = 1, 2.

Then, the overall maximum likelihood estimators are

?^1 = x?1

?^2 = x?2,

^2 = (n1 - 1)s21 + (n2 - 1)s22 . n1 + n2

The maximum likelihood estimators in the numerator takes place on the set ?1 = ?2.

?^ = n1x?1 + n2x?2 , n1 + n2

^122

=

n1

1 +

n2

n1

n2

(xi,1 - ?^)2 + (xi,2 - ?^)2

i=1

i=1

.

The yields the test statistic

T (x1, x2) =

x?1 - x?2

1 n1

+

1 n2

(n1 -1)s21 +(n2 -1)s22 n1 +n2 -2

from the fact that the likelihood ratio

(x1, x2) =

n1 + n2 - 2 n1 + n2 - 2 + T (x1, x2)2

2/(n1 +n2 )

.

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