Hypothesis Testing: Paired t-Test  (aka Matched Pairs Test)


Paired t-tests are used when...

  • You take one random sample to study two paired variables.
    • These paired variables are DEPENDENT.
    • Usually, you are studying some measurable quality about the same subject before and after a treatment.
    • For example, let's say you want to study the difference in productivity before an after a nap. You would measure each subject's work output before and after napping and compare the two values. These values are dependent upon each other because they come from the same subject, just at different times. 
  • You want to prove that mean difference (\(\mu_{diff}\)) is not zero

A few symbols need to be defined before we dive in: 

  • \(\mu_{diff}\) refers to the mean difference between the two data sets.
  • \(\bar{x}_{diff}\) refers to the mean difference derived from your sample, which you will use to disprove the null. 
  • \(s_{diff}\) refers to the sample standard deviation.
  • \(n\) refers to the sample size.



You are interested in whether drinking green tea can assist weight loss. You take a random sample of 35 individuals and record each individual weight. You then instruct the subjects to drink a cup of green tea every day for a month. At the end of the month, you record their weights again. Using the data below, conduct a hypothesis test at the significance level of 0.05 to determine whether green tea helps people lose weight. 

Subject Weight Before (lbs) Weight After (lbs)
1 181 176
2 210 212
3 154 153
... 35 166 161

** Abbreviated table. Assume \(\bar{x}_{diff} = -2.72\) and \(s_{diff} = 5.28\).


Step 1: Name Test: Paired t-test

Step 2Define Test: 

The null hypothesis assumes that there is no difference (\(H_0: \mu_{diff} = 0\))

With this null hypothesis, the options for the alternative hypothesis are as follows: 

Left-Sided Test Two-Sided Test Right-Sided Test

\(H_{o}: \mu_{diff} = \mu_{o}\)
\(H_{A}: \mu_{diff} < \mu_{o}\)

\(H_{o}: \mu_{diff} = \mu_{o}\)
\(H_{A}: \mu_{diff} \neq \mu_{o}\)

\(H_{o}: \mu_{diff} = \mu_{o}\)
\(H_{A}: \mu_{diff} > \mu_{o}\)

In this case, we think that there should be weight-loss. Thus, the mean difference (weight after - weight before) should be negative. 

\(H_{o}: \mu_{diff} = \mu_{o}\)

\(H_{A}: \mu_{diff} < \mu_{o}\)


Step 3: Assume \(H_0\) is true and define its normal distribution. Then check the conditions.

1.  Data collected from a random sample.

2.  N ≥ 10n

3.  The sampling distribution of \(\bar{x}_{diff}\) is approximately normal.

Note: This can be checked using the Central Limit Theorem and/or a Normal Probability Plot. 


Step 4: Using the normal distribution, calculate the test statistics and p-value.

Although the full formula is \(t = \frac{\bar{x}_{diff}\, -\, \mu_{o}}{s_{diff} \, /\sqrt{n}}\)it can be simplified. Recall that the null is \(H_{o}: \mu_{diff} = \mu_{o}\). Thus, \(\bar{x}_{diff} - \mu_{diff} = \bar{x}_{diff} - 0 = \bar{x}_{diff}\) , leaving us with the formula below:
Test Statistic (paired t-test):   \(t = \frac{\bar{x}_{diff}}{s_{diff} \, /\sqrt{n}}\)

The first step is to plug the table values into your calcuator. If you are not given the difference, enter each subject's scores into List 1 and List 2 (making sure to keep them matched!), and then plug L1 - L2 into List 3 in order to compute the difference. Then, take the list with the difference values (L3) and run them through 1-Var Stats. The mean your calculator computes from the difference values will be \(\bar{x}_{diff}\), and the standard deviation it computes will be your \(s_{diff}\) value.

In this case,  \(\bar{x}_{diff} = -2.72\) and \(s_{diff} = 5.28\).

Test Statistic: 

\(t = \frac{-2.72}{5.28 \, /\sqrt{35}}\)  → \(t = -3.048\)


The p-value will be found by using the t cdf function on your calculator: 
  • lower limit: -999
  • upper limit: \(t\)
  • degrees of freedom: \(n-1\)
  • All together, it looks like this: tcdf (-999, \(t\), \(n-1\))

*Note: If it was a right-sided test and the test statistic was positive (t > 0), then your lower limit would be the test statistic (t) and your upper limit would be 999. 

In this case, we do tcdf (-999, -3.048, 34) to get a p-value of 0.0022. 


Step 5: Analyze your results and determine if they are statistically significant. 

We calculated a p-value of 0.0022. This p-value is less than the significance level of 0.05. Therefore, we reject the null hypothesis. The data supports the claim that there is a drop in one's weight after drinking green tea.