Chapter 11Section 3

Testing the Difference Between Two Means: Dependent Samples

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Dependent Samples

Samples are considered to be Dependent when they are paired or matched in some way.

Examples:

Suppose a researcher wishes to see if a drug will affect a person’s reaction time.

The researcher will pretest the subject and then post-test the subject after administering the drug.

Or

Suppose you wish to test the effectiveness of a SAT prep course.

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Testing the Difference Between Two Means: Dependent Samples

When the values are dependent, do a t test on the differences. Denote the differences with the symbol D, the mean of the population differences with μD, and the sample standard deviation of the differences with sD.

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D

D

Dt

s n

with d.f. = n -1 and where

D =D∑n

and sD =n D2∑ −( D)2∑

n n−1( )

Hypotheses

When the samples are dependent, a special t-test for dependent means is used. This test employs the difference in values of the matched pairs. The hypotheses are as follows:

Two Tailed Left Tailed Right Tailed

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H0 :D =0H1 : D ≠0

H0 :D =0H1 : D < 0

H0 :D =0H1 : D > 0

Vitamin for StrengthA physical education director claims by taking a special vitamin, a weight lifter can increase his strength. Eight athletes are selected and given a test of strength, using the standard bench press. After 2 weeks of regular training, supplemented with the vitamin, they are tested again. Test the effectiveness of the vitamin regimen at α = 0.05. Each value in the data represents the maximum number of pounds the athlete can bench-press. Assume that the variable is approximately normally distributed.

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Vitamin for Strength

6

Step 1: State the hypotheses and identify the claim.

Here’s the tricky part: Since we are interested to see if there has been an increase, the weight they lifted before must be less than the weight they now lift, hence, the differences must be significantly less than before than they are now. Hence the difference in the means must be less than zero.

H0: μD = 0 and H1: μD <0 (claim)

Step 2: Find the critical value. The degrees of freedom are n – 1 = 8 – 1 = 7. The critical value for a left-tailed test with α = 0.05 is t = -1.895.

Vitamin for Strength

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Step 3: Compute the test value.

Before (X1) After (X2) D = X1 – X2 D2

210230182205262253219216

219236179204270250222216

-9-631

-83

-30

8136 9 164 9 9 0

Σ D = -19 Σ D2 = 20919

2.3758

D

Dn

22

1

D

n D Ds

n n 2

8 209 19

8 7

4.84

Vitamin for Strength

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Step 3: Compute the test value.

Step 4: Make the decision. Do not reject the null.

Step 5: Summarize the results.There is not enough evidence to support the claim that the vitamin increases the strength of weight lifters.

2.375, 4.84 DD s

D

D

Dt

s n

2.375 0

4.84 8

1.388

Cholesterol LevelsA dietitian wishes to see if a person’s cholesterol level will change if the diet is supplemented by a certain mineral. Six subjects were pretested, and then they took the mineral supplement for a 6-week period. The results are shown in the table. (Cholesterol level is measured in milligrams per deciliter.) Can it be concluded that the cholesterol level has been changed at α = 0.10? Assume the variable is approximately normally distributed.

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Cholesterol Levels

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Step 1: State the hypotheses and identify the claim. H0: μD = 0 and H1: μD 0 (claim)

Step 2: Find the critical value. The degrees of freedom are 5. At α = 0.10, the critical values are ±2.015.

≠

Cholesterol Levels

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Step 3: Compute the test value.

Before (X1) After (X2) D = X1 – X2 D2

210235208190172244

190170210188173228

2065-22-1

16

4004225

4 41

256

Σ D = 100 Σ D2= 4890100

16.76

DD

n

22

1

D

n D Ds

n n

26 4890 100

6 5

25.4

Cholesterol Levels

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Step 3: Compute the test value.

Step 4: Make the decision. Do not reject the null.

Step 5: Summarize the results.There is not enough evidence to support the claim that the mineral changes a person’s cholesterol level.

16.7, 25.4 DD s

D

D

Dt

s n

16.7 0

25.4 6

1.610

Confidence Interval for the Mean Difference

Formula for the t confidence interval for the mean difference

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2 2

d.f. 1

D DD

s sD t D t

n n

n

Confidence IntervalsFind the 90% confidence interval for the difference between the means for the data in Example 9–7.

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25.4 25.416.7 2.015 16.7 2.015

6 6 D

16.7 20.89 16.7 20.89 D

2 2 D DD

s sD t D t

n n

4.19 37.59 D

Since 0 is contained in the interval, the decision is to not reject the null hypothesis H0: μD = 0.