The t Test for Two Dependent Samples

Data Set

The same group of people rated both brands of coffee. Half of the subjects tasted brand A first. Half of the subjects tasted brand B first. The order of tasting the two brands of coffee was completely counterbalanced. 

Is there a difference in the mean ratings between the two the two brands of coffee? 

The observed mean difference is 7 - 6.3 = .7.  

Hypotheses

The null hypothesis states that there is no difference in the mean ratings between the two brands of coffee. The difference between means equals zero.

The alternative hypothesis states that there is a significant difference in the mean ratings between the two brands of coffee. It is a two-sided test. 

Set a significance level

Use a .05 significant level.
 

Definitional Formula

Identify the Source of Variance

• True Variance Components

1. Variance Due to Column Means (the Experimental Effect)


(1) Compute the group means.




(2) Compute the variance due to group means.

The number of subjects in both experimental conditions is the same. Thus, we can compute the variance due to group means as shown below.

The variance due to different group means is .122.

2. Variance Due to Row Means (the Subject Effect)

Each row represents one subject. Observations from the same subject are more likely to be similar than observations from different subjects. 

First, sum the scores from two tests. Second, compute the mean (sum/2). Third, compute the variance of the row means. The variance due to different row means (subjects) is equal to .852.

3. Variance Due to the Total Group

Ignore the column and row information. Combine the scores from the two groups into one total group. The overall mean is 6.65 and the total variance is 1.227.

Residual Component

The residual component can be computed as

1.227 - .852 - .122 = .253
 

• Standard Variance Components
   

Divide each components by the total variance
 

Source of Variance	Standard Variance
Columns	.122 / 1.227 = .1
Rows	.852 / 1.227 = .694
Residual	.253 / 1.227 = .206
Total	1.227 / 1.227 = 1

• Compute the t ratio

There are ten pairs in the experiment. The degrees of freedom are 10-1 =9 

 

Note that a repeated measures design reduces the error term by removing the variance due to individual differences (1 - .1 - .694 = .206). 

t = 2.09

Computational Formula

SPSS

Input

Output

• Paired Samples Statistics and Paired Samples Correlations

The larger the positive correlation coefficient, the greater the benefit of pairing. For our example, the correlation between the two repeated measures was .545. However, the relationship was not statistically significant (p > .05).
  

• Paired Samples Test

The observed significance level is larger than .05. What do you conclude?
 

Using the Traditional Formula to Compute the Paired-Samples t Test

• Compute the unbiased standard deviation (s) for each variable.
  
  
  
   
• Compute the correlation between the two variables                                                                                

r = .545

The larger the positive correlation coefficient, the greater the benefit of pairing.
 

• Unbiased standard deviation of the difference scores
  
  
  Compute the unbiased standard deviation for the paired differences 

The unbiased standard deviation is 1.059. You may also compute the above value by using the following formula

 

S(A-B) = 1.059

Notice that the same person rated both brands of coffee, the two ratings are correlated. 
 

• Unbiased standard error of the mean difference
  
  The unbiased standard error of the mean difference would be Sm(1-2) and can be computed as

• List the mean difference, standard error of the mean difference, and the t value for the dependent samples. 

We know that the sample mean difference is .7. The standard error of the difference between two dependent means is .335. Thus, t = mean difference / standard error of the difference. 
 

 

t = .7 / .335 = 2.09

The observed significance level is larger than .05. What do you conclude?