chapter five: paired samples methods filethe paired samples (diﬀerence) t test the paired samples...

Chapter Five: Paired Samples Methods

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Introduction

Paired data arise with some frequency in a variety of research contexts.

Patients might have a particular type of laser surgery performed onone eye with a second form performed on the remaining eye. Visualacuity measures taken on the two eyes would constitute paired data.

Patients might have a baseline blood pressure measure taken, receivesome treatment with a subsequent blood pressure measure then beingtaken. The two blood pressure measures would constitute paired data.

Twins might be exposed to two different types of vaccine againstsome disease. An assessment as to whether the twins developed thedisease or not would constitute paired data.

5.1 Introduction 2/38

The Paired Samples (Difference) t Test

The paired samples t test is conducted on difference scores obtainedby subtracting one paired observation from the other.

The null hypothesis tested is

H0 : µd = µd0

where µd is the mean of a difference score population and µd0 is thehypothesized value of µd (usually but not always zero).

Methods Related To Mean Difference 3/38

The Paired Samples t Test (continued)

One-tailed alternatives are either

HA : µd > µd0

orHA : µd < µd0

The two-tailed alternative is

HA : µd 6= µd0


The Test Statistic

The paired samples t test is nothing more than a one mean t testconducted on difference (d) scores.

Therefore, the test statistic takes the form

t =d̄ − µd0

sd√n

where d̄ is the sample mean of the difference scores, s0 the samplestandard deviation of the difference scores and n the number of pairedobservations (or difference scores).


Example

Researchers are interested in comparing the effectiveness of two lasersurgery treatments for diabetic retinopathy.

Patients who manifest the disease in both eyes have one eye treatedby the first surgical method with the remaining eye then being treatedby the second method.

After a period of time, visual acuity is measured (via LogMar scores)for each eye.


Example (continued)

The question of interest is “Does one treatment method producebetter vision (as measured by acuity) than does the other?”

Use the data provided in the next slide (#8) to perform a two-tailedpaired samples t test with α = .05 in order to assist in determiningwhether one treatment was more effective than the other.

Based on this test, what do you conclude?


Example (continued)

Table: 5.3 LogMar Scores

Treatment One Treatment Two d d2

.0 .2 −.2 .04

.8 1.1 −.3 .09

.4 .9 −.5 .251.0 .5 .5 .25.5 .2 .3 .09.4 .7 −.3 .09.5 .5 .0 .00

σ 3.6 4.1 −.5 .81


Solution

Using the sums from Table 5.3,

sd =

√.81− (−.5)2

7

7− 1=

√.774

6=√

.129 = .359

and

d̄ =−.5

7= −.071.

By Equation 5.1 obtained t is

t =d̄ − µd0

sd√n

=−.071

.359√7

=−.071

.136= −.522.


Solution (continued)

Because the critical values for a two-tailed t test with six degrees offreedom conducted at the .05 level are +2.447 and −2.447, the nullhypothesis is not rejected.

We conclude that there was insufficient evidence to allow for aconclusion that the treatments differed in their effectiveness.


Introduction

It sometimes happens that researchers would like to establish thatthere is no practical difference between treatments or between pre-and post-treatment means rather than showing that there is adifference.

In such instances the paired samples t test can be used to establishequivalence.


Two-Tailed Null And Alternative Hypotheses

If we let EIU represent the upper end of EI and EIL the lower end, theequivalence null hypothesis for the (paired samples) two-tailedequivalence test is

H0E : µd ≤ EIL or µd ≥ EIU

The alternative isHAE : EIL < µd < EIU

Notice that the null hypothesis states that the population differencescore mean is not in EI while the alternative states that this mean isin EI .

The null hypothesis, then, is an assertion of non-equivalence while thealternative asserts equivalence.


Testing The Equivalence Null Hypothesis

Testing the null hypothesis for a two-tailed equivalence test requiresthat two one-tailed tests be conducted.

The null and alternative hypotheses for the two one-tailed tests forpaired samples means are as follows

Test One Test TwoH01 : µd = EIU H02 : µd = EILHA1 : µd < EIU HA2 : µd > EIL

Both of these tests must be significant in order to reject theequivalence null hypothesis.


Example

A manufacturer of blood pressure monitoring devices wishes todemonstrate that a new, less expensive, model produces equivalentresults to the older, more expensive, device. To this end, bloodpressures of 10 subjects are assessed by means of both devices.

Difference scores (d) are calculated for the paired readings with thefollowing results. ∑

d = 16∑

d2 = 94

Use these results to perform a two-tailed equivalence test with theequivalence interval of −4 to 4. Begin by stating the null andalternative equivalence hypotheses.


Solution

The null and alternative equivalence hypotheses are as follows.

H0E : µd ≤ −4 or µd ≥ 4

HAE : −4 < µd < 4

We calculate

sd =

√∑d2 − (

Pd)2

n

n − 1=

√94− (16)2

16

16− 1=

√78

15= 2.280

and

d̄ =

∑d

n=

16

16= 1.000.



Values of obtained t for Test One and Two are then

t1 =1.0− 4.0

2.28√16

= −5.263 and t2 =1.000− (−4.0)

2.28√16

= 8.772.

Critical t values based on 16− 1 = 15 degrees of freedom for the twoone-tailed tests are −1.753 and 1.753. It follows that both results aresignificant leading to rejection of the null hypothesis that the twodevices are not equivalent insofar as mean blood pressures areconcerned in favor of the alternative that maintains equivalence.


One-Tailed Null And Alternative Hypotheses

The null and alternative hypotheses for the one-tailed equivalence testare one of the following:

H0E : µd ≥ EIUOR

H0E : µd ≤ EIL

HAE : µd < EIU HAE : µd > EIL

The first one-tailed equivalence hypothesis given above is tested byTest One with the second being carried out by means of Test Two.


Example

Eighteen subjects who routinely take a standard dose of a cholesterollowering drug are switched to a lower does that is believed to produceless severe side effects. The question of interest is, “Will the reduceddose produce cholesterol levels that are not worse than those obtainedwith the higher dose?”

Cholesterol is measured while patients are still on the higher dose andagain after they have been on the lower dose for a specified period oftime. The researchers decide that if average cholesterol rises by lessthan six points the reduced dose can be declared to produce resultsthat are “no worse” than those produced by the higher dose.


Example (continued)

Difference scores are generated by subtracting cholesterol levelsobtained from the higher dosage from those obtained from the lowerdosage. The following results are obtained.∑

d = 40,∑

d2 = 730

Use a one-tailed equivalence test with α = .05 to make thisdetermination.


Solution

Because the researchers wish to determine if the rise in cholesterol isless than six points, a one-tailed equivalence test of the form

H0E : µd ≥ 6HAE : µd < 6



From the previously provided sums we calculate

sd =

√∑d2 − (

Pd)2

n

n − 1=

√730− (40)2

18

18− 1=

√641.111

17= 6.141

and

d̄ =

∑d

n=

40

18= 2.222.

Then,

t1 =d̄ − µd0

sd√n

=2.222− 6.0

6.141√18

= −2.610.



Critical t for a one-tailed test conducted at α = .05 with 17 degreesof freedom is −1.740. The null hypothesis is rejected leading to theconclusion that the reduced dose is not worse than the higher doseinsofar as control of cholesterol is concerned.


Introduction

The paired samples t test attempts to determine whether there is adifference between the means of two paired variables. A related andusually more informative question is, “How large is the differencebetween the means of the paired variables?” This difference can beestimated with a confidence interval.


Construction Of The Interval

Construction of L and U are as follows

L = d̄ − tsd√n

U = d̄ + tsd√n

where sd is the sample standard deviation of the difference scores andt is the appropriate value from the t table with n − 1 degrees offreedom.


Example

Use the data in slide number 8 to form a two-sided 95% confidenceinterval.

What is the meaning of this interval?


Solution

sd =

√.81− (−.5)2

7

7− 1=

√.774

6= .359

and

d̄ =−.5

7= −.071.



Noting from Appendix B that the appropriate t value (with sixdegrees of freedom) for a two-sided 95% confidence interval is 2.447,Equations 5.2 and 5.3 yield

L = d̄ − tsd√n

= −.071− 2.447.359√

7= −.403

and

U = d̄ + tsd√n

= −.071 + 2.447.359√

7= .261.



A statistical interpretation of this interval would maintain that we canbe 95% confident that µd is between −.403 and .261. From theresearcher’s point of view, it can be stated with 95% confidence thatthe average difference in visual acuity between patients receiving lasertreatment one and those receiving laser treatment two was between−.403 and .261 LogMar units. Based on this interval can you rule outa difference of zero?


Assumptions

Assumptions underlying the paired samples t test, equivalence testsbased on the paired samples t test, and the confidence interval formean difference are the same as those for the one mean t test andconsequently, the one mean Z test.

These are

normalityindependence

You should note that these assumptions apply to the difference scoresrather than to the two component distributions.


McNemar’s Test

McNemar’s test is used to assess outcomes for paired dichotomousdata.

The null hypothesis assessed isH0 : π = .5which is an assertion that the proportion of pairs in which onetreatment is favored is equal to the proportion of pairs in which theother treatment is favored.

Alternatives take one of the following forms.

HA : π < 0HA : π > 0HA : π 6= 0

Methods Related To Proportions 30/38

McNemar’s Test (continued)

There are both approximate and exact forms of the test.

The approximate form is based on the normal curve1 and uses thetest statistic

Z =p̂ − .5

.5√n

where p̂ is the proportion of pairs in which the designated treatmenthas the advantage and n is the number of pairs utilized in the analysis.

The exact test is based on the binomial distribution.

1An equivalent test utilizes the chi-square distribution.Methods Related To Proportions 31/38

Approximate Method: Example

Two forms of sunscreen are to be compared with respect to level ofprotection afforded against sun-induced skin damage.

One form of sunscreen is applied to a randomly selected arm of eachof 15 volunteer subjects with the second form then being applied tothe remaining arm.

After a specified period of sun exposure, the skin of each arm isexamined and graded as exhibiting a satisfactory level of protection Sor as not exhibiting a satisfactory level of protection S .


Approximate Method: Example (continued)

It was found that the number of pairs where both arms were graded Swas 2, both were S was 3, product one produced an S while producttwo produced an S was 9 and the number of pairs where product oneproduced an S while product two produced an S was 1.

Use the approximate method to perform a two-tailed McNemar’s testat α = .05 on the sunscreen data. What is your conclusionconcerning the relative effectiveness of the two sunscreen products?


Solution

Ignoring noninformative outcomes, we note that the proportion ofoutcomes favoring product one is 9

10 = .90 and n = 10. Equation 5.4then gives

Z =.90− .50

.50√10

= 2.530

Appendix A shows that the critical values for a two-tailed testconducted at α = .05 are +1.96 and −1.96 thereby leading torejection of the null hypothesis.



Viewed from a statistical point of view the null and alternativehypotheses are

H0 : π = .5 and HA : π 6= .5

When viewed from the context of this study, the null hypothesisasserts that there is no difference in the effectiveness of the twoproducts while the alternative maintains that such a difference doesexist. In this case, we rejected the claim of no difference in favor of astatement that the two products do differ in effectiveness. Further,we can conclude that product one is superior to product two in termsof protection provided.


Exact Method

The exact method employs the binomial distribution to test the nullhypothesis

H0 : π = .5


Example

Use the example outlined in slide 32 with the result reported in slide33 to perform an exact McNemar’s test with alternative

HA : π > .5


Solution

Noting that of the 10 pairs of arms, the number in which treatmentone afforded a satisfactory result while treatment two did not was 9.

By Equation 4.5 we calculateP (9) = .00977 and P (10) = .00098

The p-value for a test with alternative HA : π > .5 is then.00977 + .00098 = .01075. Because this value is less than α = .05,the null hypothesis is rejected.


chapter five: paired samples methods filethe paired samples (diﬀerence) t test the paired samples...

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