Session 20 Notes, Section 9-5

Testing the Difference Between Two Means when there are small dependent samples

Special  t tests allow researchers to compare population parameters, such as means, when we have small dependent samples.

At the end of this session you will be able to:

• Test the difference between two small dependent samples
  • Example one tailed
  • Example two tailed
• Finding the confidence interval for small dependent samples
  • Example
• Your Turn:  Summary

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Test the difference between two small dependent samples

• Independent and Dependent Samples
  • Dependent Samples are samples that are paired or matched in some way.
  • Independent and Dependent Samples
• Examples of Dependent Samples
  • Samples in which the same subjects are used in a pre-post situation are dependent.
  • Another type of dependent samples are samples matched on the basis of variables extraneous to the study.
• Two Notes of Caution
  • When subjects are matched according to one variable, the matching process does not eliminate the influence of other variables.
  • When the same subjects are used for a pre-post study, sometimes the knowledge that they are participating in a study can influence the results.
• Special t test for Dependent Means
  • Hypotheses: Possible hypotheses are:
    •  
    • µ_D is the expected mean of the differences of the matched pairs.
  • General Procedure—Finding the Test Value
    • The procedure is outlined on page 466.
  • t Test Formulas
    • The formula for the t test for dependent samples:
      • .
  • Procedures for testing the difference between Means for Dependent Samples:
    1. State the hypotheses and identify the claim
    2. Find the Critical Value(s)
    3. Compute the test value
       1. Make a table of the score and the differences: .
       2. Find the differences and place the results in column A
          • D = X₁ - X₂
       3. Find the mean of the differences
          • D = Σ D / n
       4. Square the differences and place the results in column B.  Complete the table
          • D₂ = (X₁ - X₂)²
       5. Find the standard deviation of the differences
          • .
       6. Find the test value
          • 
    4. Make the decision
    5. Summarize the results

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Example: one tail

A 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 two 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 these data represents the maximum number of pounds the athlete can bench press. Assume that the variable is approximately normally distributed. Determine whether or not there is enough evidence to support the claim.

Athlete			1		2		3		4		5		6		7		8
Before (X1)			204		229		190		207		266		256		217		216
After (X2)			217		240		187		205		277		254		228		216

1. State the hypotheses and identify the claim.
   • In order for the vitamin to be effective, the before weights must be significantly less than the after weights; hence, the mean of the differences must be less than zero.
   • H₀: µ_D ≥0 and H₁:µ_D < 0 (claim)
2. Find the critical value.
   • The degrees of freedom are n - 1. In this case, d.f. = 8 - 1 = 7.
   • The critical value for a left-tailed test with α= 0.05 is -1.895.
3. Compute the test value.
   • Make a table.
   • Find the differences and place the results in column A.
     • 204 - 217 = -13
     • 229 - 240 = -11
     • 190 - 187 = +3
     • 207 - 205 = +2
     • 266 - 277 = -11
     • 256 - 254 = +2
     • 217 - 228 = -11
     • 216 - 216 = 0
     •           ΣD = -39
   • Find the mean of the differences.
   •  Square the differences and place the results in column B.
     • (-13)² =169
     • (-11)² = 121
     • (+3)² = 9
     • (+2)² = 4
     • (-11)² = 121
     • (+2)² = 4
     • (-11)² = 121
     • 0² = 0
     • ΣD² = 549
   • The completed table is:
   • Find the standard deviation of the differences.
   • Find the test value.
4. Make the decision.
   • The decision is to reject the null hypothesis at α= 0.05, since -1.926 < -1.895, as shown below.
     • 
5. Summarize the results.
   • There is enough evidence to support the claim that the vitamin increases the strength of weight lifters.

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Example: two tail

A dietician 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. (Cholesterol level is measured in milligrams per deciliter.) Can it be concluded that the cholesterol level has been changed at α= 0.10? Assume that the variable is approximately normally distributed.

Subject			1		2		3		4		5		6
Before (X1)			213		230		201		188		173		245
After (X2)			179		153		204		174		174		204

1. State the hypotheses and identify the claim.
   • If the diet is effective, the before cholesterol levels should be different from the after levels.
     • H₀: µ_D = 0 and H₁:µ_D not = 0 (claim)
2. Find the critical value.
   • The degrees of freedom are 6-1=5.
   • The critical value for a two tailed test with α= 0.10 is 2.015.
3. Compute the test value.
   • Make a table.
   • Find the differences and place the results in column A.
     • 213 - 179 = 34
     • 230 - 153 = 77
     • 201 - 204 = -3
     • 188 - 174 = 14
     • 173 - 174 = -1
     • 245 - 204 = 41
     •            ΣD = 162
   • Find the mean of the differences.
   • Square the differences and place the results in column B.
     • (34)² = 1156
     • (77)² = 5929
     • (-3)² = 9
     • (14)² = 196
     • (-1)2 = 1
     • (41)² = 1681
     • ΣD² = 8972
   • The completed table is shown next.
   • Find the standard deviation of the differences.
   • Find the test value.
4. Make the decision.
   • The decision is to reject the null hypothesis, since the test value is in the critical region.
5. Summarize the results.
   • There is enough evidence to support the claim that the mineral changes a person's cholesterol level.

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Finding the confidence interval for small dependent samples

The formula for calculating the confidence interval for the mean difference:

Example.

A dietician 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. (Cholesterol level is measured in milligrams per deciliter.) Find the 90% confidence interval for the data. Assume that the variable is approximately normally distributed.

Subject			1		2		3		4		5		6
Before (X1)			214		235		203		192		170		240
After (X2)			172		160		204		188		173		193

1. State the hypotheses and identify the claim.
   • If the diet is effective, the before cholesterol levels should be different from the after levels.
     • H₀: µ_D = 0 and H₁:µ_D not = 0 (claim)
2. Calculate the confidence interval
   • Locate the formula for confidence interval for small dependent samples
   • Find the mean of the differences.
   • Square the differences and find the sum.
     • Σ D² = 9624
   • Find the standard deviation of the differences.
   • Find the degrees of Freedom
     • 6-1= 5 degrees of freedom.
   • Look up the t_α/2 score for α = 0.10 for a two tailed t. (Note: 90% confidence interval  gives an alpha of 1-.90 = 0.10)
     •   t_α/2 = 2.015.
   • Substitute the values calculated in the above steps in the formula for confidence interval
     • 27.333 - 26.379 < µ_D < 27.333 + 26.379
     • 0.954 < µ_D< 53.712
3. Make Decision
   • Since 0 is not contained in the interval, the decision is to reject the null hypothesis H₀: µ_D = 0
4. Summarize findings
   • There is enough evidence to support the claim that the mineral changes a person's cholesterol level.

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Summary

• For dependent samples the dependent samples t test is used

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