The paired sample t-test is used to compare two sample populations that are correlated. This test is most commonly employed for “before and after” or pretest-postest design. Below are the assumptions of paired sample t-test.

• Only the matched pairs are used to perform the analysis
• The data is normally distributed
• The variances of the two samples are homogeneous
• The observations are independent of each other

Below are the steps involved in conducting a paired sample t-test

1. Set up the hypotheses
   • H0: The mean of the paired samples are the same or (μ1 = μ2)
   • H1: The means of the paired samples are not equal or
     (μ1 ≠ μ2, μ1 > μ2, μ1 < μ2)
2. Determine the level of statistical significance (.1, .05, or .01) and if it is two-tailed or one-tailed
   • Two-tailed means there are two choices. One mean can be greater or lesser than the other.
   • One-tailed means there is one choice. One of the means is greater or it lesser but not vice versa.
3. You also must take into account the degree of freedom which is sample size – 1 this information is useful when looking at the t-test chart to calculate the t critical value
4. Calculate the paired t-test. The formulas are below. Take note that there are three separate formulas labeled A, B, and CFORMULA A   t computed = Mean difference
                                                                    standard deviation of the mean /                                                                               square root of the  sample sizeFORMULA B    mean difference = sum of the difference
                                                                                 sample size

FORMULA C    standard deviation of the difference =
√ΣD² – (ΣD)²
                 n     
n – 1
Sorry there is no simple way to explain formula C

4. Make Statistical decision

5. State conclusion

Below is an example

A teacher develops an incentive plan for his students. Students who were quiet got additional stickers in their notebook. The teacher picked 10 students at random to see if the number of stickers they earned was more after the incentive program was adopted. Here are the results

Student               Before             After
1                               20                        35
2                               30                        41
3                                25                       38
4                                31                       42
5                                19                       18
6                                18                       16
7                                23                       34
8                                32                       19
9                                24                       24
10                             19                       33

Step 1 State the hypotheses
H0: μ1 < 0 or the number of stickers after the incentive plan are not more than before
H2: μ1 > 0 or the number of stickers is greater after the incentive plan

Step 2: The level of statistical significance is .05. This is also a one-tailed test

Step 3: Calculate the critical region which is
degrees of freedom = sample size – 1 = tcritical
df = 10 – 1= 9 and the tcritical is 1.83 according to the table

Step 4: Compute t computed for paired samples

Student    Before     After      Difference       Difference²
1                       20                35           15                             225
2                       30                41           11                             121
3                       25                38           13                             169
4                       31                42            11                             121
5                       19                18            -1                                    1
6                       18                16            -2                                    4
7                        23                34           11                              121
8                        32                19         -13                              169
9                        24                24             0                                     0
10                     19                33           14                               196
Sum of difference     59 Sum of the difference² 1127

Find the mean of the difference
59 / 10 = 5.9

Find the standard deviation of the differences (the entire equation below is squared)

1127 – (5.9)²
                10      
10-1

the standard deviation is 11.17

Find the t computed

            59              = 16.69
11.17 / √10

Step 5: Decision

Since the t computed 16.69 is greater than the t critical of 1.83 we reject the null hypothesis

Step 6 Conclusion

Since we reject the null we can conclude that there is evidence that the incentive program has increased the number of stickers the students earn.