An Introduction To Statistical Inference And Data Analysis

210 CHAPTER 10. 2-SAMPLE LOCATION PROBLEMS

Since j1:81j < 2:00, we accept H0 at signi¯cance level ® = :05. (The

:

signi¯cance probability is P = :067.)

10.1.3The Normal Behrens-Fisher Problem

² Now we must estimate both variances, ¾12 and ¾22; hence, we let

²Unfortunately, the distribution of TW is unknown. However, Welch (1937, 1947) argued that TW »t(º), with

² Since ¾12 and ¾22 are unknown, we estimate º by

The approximation TW »t(^º) works well in practice.

² Interval Estimation. A (1 ¡ ®)-level con¯dence interval for ¢ is

²Example. Suppose that n1 = 57, x¹ = :0167, and s1 = :0042; suppose that n2 = 12, y¹ = :0144, and s2 = :0024. Then

²In general:

{If n1 = n2, then t = tW .

{If the population variances are (approximately) equal, then t and tW will tend to be (approximately) equal.

{If the larger sample is drawn from the population with the larger variance, then t will tend to be less than tW . All other things equal, this means that Student's test will tend to produce significance probabilities that are too large.

{If the larger sample is drawn from the population with the smaller variance, then t will tend to be greater than tW . All other things equal, this means that Student's test will tend to produce significance probabilities that are too small.

{If the population variances are (approximately) equal, then º^ will be (approximately) n1 + n2 ¡ 2.

{It will always be the case that º^ · n1 + n2 ¡ 2. All other things equal, this means that Student's test will tend to produce significance probabilities that are too large.

²Conclusions:

{If the population variances are equal, then Welch's approximate t- test is approximately equivalent to Student's 2-sample t-test.

{If the population variances are unequal, then Student's 2-sample t-test may produce misleading signi¯cance probabilities.

{\If you get just one thing out of this course, I'd like it to be that you should never use Student's 2-sample t-test." (Erich L. Lehmann)

10.2The 2-Sample Location Problem for a General Shift Family

10.3The Symmetric Behrens-Fisher Problem

10.4Exercises

is an unbiased estimator of ¾2.

² If H0 is true, then

¹1 = ¢¢¢ = ¹k = ¹

and

¹¢ = Xk Nni ¹ = ¹;

i=1

it follows that the quantity

Xk

° = ni(¹i ¡ ¹¢)2

i=1

measures departures from H0. An estimator of this quantity is the between-groups or treatment sum of squares

which is the variation of the sample means about the sample grand mean.

² Fact: Under H0,

SSB=¾2 » Â2(k ¡ 1);

where Â2(º) denotes the chi-squared distribution with º degrees of freedom.

²If we knew ¾2, then we could test H0 by referring SSB=¾2 to a chisquared distribution. We don't know ¾2, but we can estimate it. Our test statistic will turn out to be SSB=S2 times a constant.

²Let

=1 j=1

denote the within-groups or error sum of squares. This is the sum of the variations of the individual observations about the corresponding sample means.

² Fact: Under H0, SSB and SSW are independent random variables and

where F (º1; º2) denotes the F distribution with º1 and º2 degrees of freedom.

² The anova F -test of H0 is to reject if and only if

P = PH0 (F ¸ f) · ®;

i.e. if and only if

f ¸ q = qf(1-®,df1=k-1,df2=N-k);

where f denotes the observed value of F and q is the ® quantile of the appropriate F distribution.

² Let

=1 j=1

the total sum of squares. This is the variation of the observations about the sample grand mean.

²Fact: SST =¾2 » Â2(N ¡ 1).

²Fact: SSB + SSW = SST . This is just the Pythagorean Theorem; it is one reason that squared error is so pleasant.

²The above information is usually collected in the form of an anova table:

with R2 = :9424.

11.1.2Planned Comparisons

²Rejecting H0 : ¹1 = ¢¢¢ = ¹k leaves numerous alternatives. Typically, the investigator would like to say more than simply \H0 is false." Often, one can determine speci¯c comparisons of interest in advance of the experiment.

²Example: Heyl (1930) attempted to determine the gravitational constant using k = 3 di®erent materials|gold, platinum, and glass. It seems natural to ask not just if the three materials lead to identical determinations of the gravitational constant, by testing H0 : ¹1 = ¹2 = ¹3, but also to ask:

1.If glass di®ers from the two heavy metals, by testing

H0 : ¹1 + ¹2 ¡ 2¹3 = 0 vs: H1 : ¹1 + ¹2 ¡ 2¹3 6= 0; or, equivalently,

H0 : µ1 = 0 vs: H1 : µ1 6= 0;

where µ1 = ¹1 + ¹2 ¡ 2¹3.

2. If the two heavy metals di®er from each other, by testing

H0 : ¹1 = ¹2 vs: H1 : ¹1 6= ¹2; or, equivalently,

H0 : ¹1 ¡ ¹2 = 0 vs: H1 : ¹1 ¡ ¹2 6= 0; or, equivalently,

H0 : µ2 = 0 vs: H1 : µ2 6= 0; where µ2 = ¹1 ¡ ¹2.

² De¯nition 11.1 A contrast is a linear combination of the k population means,

Xk

µ = ci¹i;

i=1

for which Pki=1 ci = 0.

²For example, in the contrasts suggested above,

1.µ1 = 1 ¢ ¹1 + 1 ¢ ¹2 + (¡2) ¢ ¹3 and 1 + 1 ¡ 2 = 0; and

2.µ2 = 1 ¢ ¹1 + (¡1) ¢ ¹2 + 0 ¢ ¹3 and 1 ¡ 1 + 0 = 0.

We usually identify di®erent contrasts by their coe±cients, e.g. c = (1; 1; ¡2).

Orthogonal Contrasts

²We want to test H0 : µ = 0 vs. H1 : µ 6= 0. An unbiased estimator of µ is

^Xk ¹

µ= ciXi¢;