Table of Contents
h2>Introduction
An example of paired data is blood pressure of patients before yoga (i.e., readings before the intervention) and the blood pressure of patients after yoga (i.e., readings after the intervention). The most widely used technique, testing the impact of an intervention (or for analyzing paired data), when it is numeric, is the paired sample ttest. This is explained with the help of a medical example as follows.
Paired sample ttest
Using formulas
Suppose that a doctor has developed a new drug that is used to treat hypertension and anxiety. The doctor tested the level of hypertension and anxiety of 30 patients before giving them the drug and gave them a score from 0—10 with 0 indicating no hypertension and anxiety and 10 indicating the highest level of hypertension and anxiety (i.e., the patient is likely to suicide).
Then, in order to test the impact of the new drug that he has developed, he gave the drug to the same 30 patients and then tested the level of their hypertension and anxiety. The hypertension and anxiety scores before and after consuming the new drug are shown as follows:
# 
Patient 
Hypertension – Anxiety Score 

Before 
After 

1 
Alex 
8 
9 
2 
Bob 
7 
8 
3 
Cathay 
8 
9 
4 
Drake 
8 
7 
5 
Emily 
5 
4 
6 
Frank 
5 
3 
7 
George 
8 
4 
8 
Hood 
7 
3 
9 
Iris 
9 
8 
10 
Jack 
9 
7 
11 
Kate 
7 
4 
12 
Luis 
7 
6 
13 
Mathew 
6 
7 
14 
Nick 
4 
5 
15 
John 
2 
2 
16 
Paul 
8 
8 
17 
Ross 
8 
7 
18 
Rachel 
7 
6 
19 
Smith 
9 
8 
20 
Clark 
8 
4 
21 
David 
7 
1 
22 
Mark 
7 
2 
23 
Walker 
6 
4 
24 
Adam 
7 
5 
25 
Huss 
8 
5 
26 
Jones 
7 
4 
27 
Angelina 
5 
3 
28 
Brad 
8 
7 
29 
Tom 
8 
6 
30 
Dakota 
7 
5 
It is difficult to draw any conclusions about the impact of the new drug by just seeing the paired data above; thus, we need to conduct the paired sample ttest. The steps of the paired sample ttest are as follows:
Step 1: State the hypothesis of the paired test
Ho: Null Hypothesis: µd = 0: There is no statistically significant difference between the hyper tension and anxiety scores of patients before and after the intervention (i.e. the new drug).
HA: Alternative Hypothesis: µd ≠ 0: There is a statistically significant difference between the hyper tension and anxiety scores of patients before and after the intervention (i.e. the new drug).>
Step 2: Compute the test statistic
Mean of the difference = 1.663; Standard deviation of the difference = 1.829. These are computed as follows:
# 
Patient 
Hypertension – Anxiety Score 
Difference = After – Before 

1 
Alex 
1 
2 
Bob 
1 
3 
Cathay 
1 
4 
Drake 
1 
5 
Emily 
1 
6 
Frank 
2 
7 
George 
4 
8 
Hood 
4 
9 
Iris 
1 
10 
Jack 
2 
11 
Kate 
3 
12 
Luis 
1 
13 
Mathew 
1 
14 
Nick 
1 
15 
John 
0 
16 
Paul 
0 
17 
Ross 
1 
18 
Rachel 
1 
19 
Smith 
1 
20 
Clark 
4 
21 
David 
6 
22 
Mark 
5 
23 
Walker 
2 
24 
Adam 
2 
25 
Huss 
3 
26 
Jones 
3 
27 
Angelina 
2 
28 
Brad 
1 
29 
Tom 
2 
30 
Dakota 
2 
Mean of difference = d bar =
Sum of above differences / 30 
1.633 

Standard deviation of difference =
∑ (difference I – mean) 2 / 30 
1.829 
The tstatistic is thus equal to 1.663 / (1.829 / √ 30) = 4.98.
Step 3: Identify the critical values / rejection region
The n = sample size = 30 and we choose a 5% significance level; thus, the rejection region is less than 2.04 or more than +2.04.
Step 4: Make a conclusion
The t statistic in the example is 4.98. This is less than 2.04 and thus we can reject the Ho at the 5% significance level. Finally, we conclude that the new drug has been effective in terms of reducing the hypertension and anxiety of the patients.
Using SPSS
The above example can also be solved using the Statistical Package for Social Sciences (SPSS) as follows:
Go to ‘Analyze’ then ‘Compare Means’ then ‘Paired Sample ttest’. The SPSS output solution for the above example is as follows:
Ttest
Paired Sample Statistics
Mean 
N 
Std. Deviation 
Std. Error Mean 

Pair 1  Hypertension and anxiety after using the new drug 
5.3667 
30 
2.18905 
.39966 
Hypertension and anxiety score before the new drug 
7.0000 
30 
1.55364 
.28365 
Paired Sample Correlations
N 
Correlation 
Sig. 

Pair 1  Hypertension and anxiety after using the new drug and hypertension and anxiety score before the new drug 
30 
.568 
.001 
Paired Sample Test
Paired Differences 
t 
df 
Sig. (2tailed) 

Mean 
Std. Deviation 
Std. Error Mean 
95% Confidence Interval of the Difference 

Lower 
Upper 

Pair 1  Hyper tension and anxiety after using the new drug – hyper tension and anxiety score before the new drug 
1.6333 
1.82857 
.33385 
2.316 
0.950 
4.892 
29 
.000 
The pvalue of 0.000 is less than 0.01 and this indicates that we can reject the null hypothesis even at the 1% significance level.
Assumptions
The assumptions of the paired sample ttest are as follows:
As a parametric method (a technique which gauges obscure parameters), the paired ttest makes a few suppositions. In spite of the fact that ttests are very robust, it is great practice to assess the level of deviation from these suspicions, keeping in mind the end goal to evaluate the nature of the outcomes.
In a paired specimen ttest, the perceptions are characterized as the contrasts between two arrangements of qualities, and every supposition alludes to these distinctions, not the first information values. This special ttest has four primary presumptions:
 The response variable must not be categorical (interim/proportion).
 The values are not dependent on each other.
 The response variable ought to be around typically normally distributed.
 The response variable must not contain any extreme values.
McNemar’s test
Using SPSS
Suppose that a therapist has developed a new technique that is used to treat drug addicts. The therapist tested 30 patients before using the new technique on them and categorized them into ‘drug addicts’ and ‘nondrug addicts’. Then, in order to test the impact of the new technique that he has developed, he used the technique on the same 30 patients and then again categorized them into ‘drug addicts’ and ‘nondrug addicts’. The paired data for this example is shown as follows:
Before 
After 
NonDrugAddict 
Drug Addict 
Drug Addict 
NonDrugAddict 
Drug Addict 
NonDrugAddict 
Drug Addict 
NonDrugAddict 
Drug Addict 
NonDrugAddict 
Drug Addict 
Drug Addict 
Drug Addict 
Drug Addict 
Drug Addict 
NonDrugAddict 
Drug Addict 
NonDrugAddict 
NonDrugAddict 
Drug Addict 
NonDrugAddict 
NonDrugAddict 
NonDrugAddict 
NonDrugAddict 
Drug Addict 
NonDrugAddict 
Drug Addict 
NonDrugAddict 
Drug Addict 
Drug Addict 
Drug Addict 
Drug Addict 
Drug Addict 
NonDrugAddict 
Drug Addict 
NonDrugAddict 
NonDrugAddict 
Drug Addict 
Drug Addict 
NonDrugAddict 
Drug Addict 
NonDrugAddict 
Drug Addict 
NonDrugAddict 
Drug Addict 
NonDrugAddict 
Drug Addict 
NonDrugAddict 
Drug Addict 
Drug Addict 
Drug Addict 
NonDrugAddict 
NonDrugAddict 
NonDrugAddict 
NonDrugAddict 
Drug Addict 
Drug Addict 
NonDrugAddict 
Drug Addict 
NonDrugAddict 
Drug Addict 
NonDrugAddict 
It is difficult to draw any conclusions about the impact of the new technique by just seeing the paired data above; thus, we need to conduct the McNemar’s test. The test is conducted using SPSS and the output is shown as follows:
Descriptive Statistics
N 
Mean 
Std. Deviation 
Minimum 
Maximum 

Before 
30 
.77 
.425 
0 
1 
After 
30 
.29 
.461 
0 
1 
Wilcoxon Signed Ranks Test
N  Mean Rank  Sum of Ranks  
Before – After  Negative Ranks  19^{a}  12.00  228.00 
Positive Ranks  4^{b}  12.00  48.00  
Ties  8^{c}  
Total  31 
 After < Before
 After > Before
 After = Before
Test statistics
After – Before 

Z 
3.128^{b} 
Asymp. Sig. (2tailed) 
.002 
 Wilcoxon Signed Ranks Test
 Based on positive ranks
McNemar’s test
Crosstabs – Before and After
Before 
After 

Nondrug addict 
Drug Addict 

Nondrug addict  3  4 
Drug Addict  19  5 
Test statistics
Before – After 

N 
31 
Exact Sig. (2tailed) 
.003^{b} 
 McNemar’s Test
 Binomial distribution used
The pvalue of 0.003 is less than 0.05 and thus we reject the null hypothesis. Finally, we conclude that the new technique was effective in turning patients from addicts to nonaddicts.
Using SPSS
The steps of the McNemar’s test using formulas are as follows:
Step 1: State the hypothesis of the McNemar’s test
Ho: Null Hypothesis: µd = 0: There is no statistically significant impact of the new technique.
HA: Alternative Hypothesis: µd ≠ 0: There is a statistically significant impact of the new technique.
Step 2: Compute the test statistic
To run a McNemar test, your data must be placed in a 2×2 contingency table, with the cell frequencies that equals to the number of pairs as follows;
Test 2 positive  Test 2 negative  Row total  
Test 1 positive  a  b  a + b 
Test 1 negative  c  d  c + d 
Column total  a + c  b + d  n 
 Test statistic = (b – c) 2 / (b + c).
These boxes for this example are as follows:
Before 
After 

Drug addict 
NonDrug Addict 

Drug addict 
5 
19 
NonDrug Addict 
4 
3 
Thus, test statistic = (4 – 19) 2 / (4 + 19) = 9.78
Step 3: Identify the critical values / rejection region
The n = sample size = 30 and we choose a 5% significance level; thus, the rejection region is less than 2.04 or more than +2.04.
Step 4: Make a conclusion
The test statistic in the example is 9.78. This is more than +2.04 and thus we can reject the Ho at the 5% significance level. Finally, we conclude that the new treatment has been effective in terms of treating addictions.
This test is a nonparametric test for paired ostensible information. It’s utilized when you are occupied with finding an adjustment in extent for the linked information. For instance, you could utilize this test to break down review casecontrol experiments, where every treatment is matched with a control. It could likewise be utilized to break down a trial where two medicines are given to coordinated sets. The method is simple, quick and easy to perform. It also enables an appropriate confirmatory data analysis for situations dealing with paired dichotomous responses to surveys or experiments.
This test is sometimes referred to as McNemar’s ChiSquare test in light of the fact that the test measurement has a chisquare model. The McNemar’s test is utilized to decide whether there are contrasts on a dichotomous response variable between two related gatherings. It can be thought to be like the matched specimens ttest; however, for a dichotomous as opposed to a nonstop ward variable.
In any case, not at all like the combined specimens ttest, it can be conceptualized to test two unique properties of a rehashed measure dichotomous variable, as is clarified beneath. The McNemar’s test is utilized to break down pretestposttest plan and also being generally utilized in examining coordinated matches and casecontrol surveys. On the offchance that you have more than two repeated estimates, you could utilize the Cochran’s Qtest.
A limitation of the Mcnemar test includes that it was made for to be used with large samples. It also assumes that the discordant pair i.e. (b+c) is equal to or larger than 10; hence, the use of an exact binomial test is recommended if the discordant pair is less than 10.
Assumptions
 There must be one continuous variable with two categories and one independent variable with two connected groups.
 The two groups in your response variable must be mutually exclusive. This means that participants cannot appear in more than one group.
 Your sample must be a random sample.
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