Paired ttest examples
Paired ttest calculator with examples
In this tutorial we will discuss some numerical examples on paired $t$test (dependent sample $t$test).
paired $t$test calculator
The $t$test calculator for testing two population means makes it easy to calculate the test statistic, $t$ critical value and the $p$value given the sample information, level of significance and the type of alternative hypothesis (i.e. lefttailed, righttailed or twotailed.)
Paire ttest Calculator  

Sample 1  Sample 2  
Enter Data (Separated by comma ,)  
Level of Significance ($\alpha$)  
Tail  Left tailed Right tailed Two tailed 

Results  
Number of pairs of Obs. (n):  
Mean of Diff.: $\overline{d}$  
Std. Dev. of D: $s_d$  
Standard Error of $\overline{d}$  
Test Statistics t:  
Degrees of Freedom:  
tcritical value(s):  
pvalue:  
How to use paired $t$test calculator?
Step 1 – Enter the sample data for first sample separated by commas and second sample separated by commas
Step 2 – Enter the level of significance $\alpha$
Step 3 – Select the alternative hypothesis (lefttailed / righttailed / twotailed)
Step 4 – Click on "Calculate" button to get the result
Paired ttest Example 1
A new prep class was designed to improve AP statistics test scores. Five students were selected at random. The numbers of correct answers on two practice exams were recorded; one before the class and one after. The data recorded in the table below. We want to test if the numbers of correct answers, on average, are higher after the class.
Student  1  2  3  4  5 

Before Class  12  15  9  12  12 
After Class  14  18  11  10  12 
Is there evidence to suggest that the mean number of correct answers after the class exceeds the mean number of correct answers before the class? Use $\alpha = 0.1$.
Solution
Let $x$ denote the number of correct answer before the class and $y$ denote the number of correct answer after the class.
Because the two samples are dependent, we use paired $t$test for testing the hypothesis problem.
The sample size $n = 5$. Let $d=xy$.
$x$  $y$  $d$  $ddbar$  $(ddbar)^2$  

12  14  2  1  1  
15  18  3  2  4  
9  11  2  1  1  
12  10  2  3  9  
12  12  0  1  1  
Total  5  16 
The sample mean of the difference is
$$ \begin{aligned} \overline{d}&= \frac{1}{n}\sum_{i=1}^n d_i\\ &=\frac{5}{5}\\ &=1 \end{aligned} $$
and the sample standard deviation of the difference is
$$ \begin{aligned} s_d&= \sqrt{\frac{1}{n1}\sum_{i=1}^n (d_i\overline{d})^2}\\ &=\sqrt{\frac{16}{4}}\\ &=2 \end{aligned} $$
Step 1 Hypothesis
The hypothesis testing problem is
$H_0 : \mu_d = 0$ against $H_1 : \mu_d < 0$ ($\textit{lefttailed}$)
Step 2 Test Statistic
The test statistic for testing $H_0$ against $H_1$ is
$$ \begin{aligned} t=\frac{\overline{d} \mu_d}{s_d/\sqrt{n}} \end{aligned} $$
Step 3 Level of Significance
The significance level is $\alpha = 0.1$.
Step 4 Critical Value(s)
As the alternative hypothesis is $\textit{lefttailed}$, the critical value of $t$ for $4$ degrees of freedom and $\alpha = 0.1$ level of significance $\text{is}$ $\text{1.533}$.
The rejection region (i.e. critical region) is $\text{t < 1.533}$.
Step 5 Computation
The test statistic for testing above hypothesis testing problem under the null hypothesis is
$$ \begin{aligned} t&=\frac{\overline{d} \mu_d}{s_d/\sqrt{n}}\\ &= \frac{10}{2/\sqrt{5}}\\ &= 1.118 \end{aligned} $$
Step 6 Decision (Traditional approach)
The test statistic is $t =1.118$ which falls $\textit{outside}$ the critical region, we $\textit{fail to reject}$ the null hypothesis.
OR
Step 6 Decision ($p$value approach)
The test is $\textit{lefttailed}$ test, so pvalue is the
area to the $\textit{left}$ of the test statistic ($t=1.118$). That is pvalue = $P(t\leq 1.118 ) = 0.1631$.
The pvalue is $0.1631$ which is $\textit{greater than}$ the significance level of $\alpha = 0.1$, we $\textit{fail to reject}$ the null hypothesis at $0.1$ level of significance.
Interpretation
There is no sufficient evidence to claim that the mean number of correct answers after the class exceeds the mean number of correct answers before the class.
Paired ttest Example 2
A manufacturer claims that they have designed a new keyboard that will increase the speed of a typist. A following study was done : Six students were selected at random to type a term paper on a regular keyboard and on a newly designed keyboard. Which keyboard the student types on the first was chosen at random. The following table shows the number of minutes each student took to type the term paper on each keyboard. At $\alpha =0.05$, did the newly designed keyboard reduce the number of minutes to type the paper?
Student  A  B  C  D  E  F 

Regular  54  63  47  61  52  59 
Newly designed  52  61  48  58  51  55 
Solution
Let $x$ denote the number of minutes to type term paper using regular keyboard and $y$ denote the number of minutes to type term paper using newly designed keyboard.
Because the two samples are dependent, we use paired $t$test for testing the hypothesis problem.
The sample size $n = 6$. Let $d=xy$.
$x$  $y$  $d$  $ddbar$  $(ddbar)^2$  

54  52  2  0.167  0.028  
63  61  2  0.167  0.028  
47  48  1  2.833  8.028  
61  58  3  1.167  1.361  
52  51  1  0.833  0.694  
59  55  4  2.167  4.694  
Total  11  14.8333 
The sample mean of the difference is
$$ \begin{aligned} \overline{d}&= \frac{1}{n}\sum_{i=1}^n d_i\\ &=\frac{11}{6}\\ &=1.8333 \end{aligned} $$
and the sample standard deviation of the difference is
$$ \begin{aligned} s_d&= \sqrt{\frac{1}{n1}\sum_{i=1}^n (d_i\overline{d})^2}\\ &=\sqrt{\frac{14.8333}{5}}\\ &=1.7224 \end{aligned} $$
Step 1 Hypothesis
The hypothesis testing problem is
$H_0 : \mu_d = 0$ against $H_1 : \mu_d > 0$ ($\textit{righttailed}$)
Step 2 Test Statistic
The test statistic for testing $H_0$ against $H_1$ is
$$ \begin{aligned} t=\frac{\overline{d} \mu_d}{s_d/\sqrt{n}} \end{aligned} $$
Step 3 Level of Significance
The significance level is $\alpha = 0.05$.
Step 4 Critical Value(s)
As the alternative hypothesis is $\textit{righttailed}$, the critical value of $t$ for $5$ degrees of freedom and $\alpha = 0.05$ level of significance $\text{is}$ $\text{2.015}$.
The rejection region (i.e. critical region) is $\text{t > 2.015}$.
Step 5 Computation
The test statistic for testing above hypothesis testing problem under the null hypothesis is
$$ \begin{aligned} t&=\frac{\overline{d} \mu_d}{s_d/\sqrt{n}}\\ &= \frac{1.83330}{1.7224/\sqrt{6}}\\ &= 2.6072 \end{aligned} $$
Step 6 Decision (Traditional approach)
The test statistic is $t =2.6072$ which falls $\textit{inside}$ the critical region, we $\textit{reject}$ the null hypothesis.
OR
Step 6 Decision ($p$value approach)
The test is $\textit{righttailed}$ test, so pvalue is the
area to the $\textit{right}$ of the test statistic ($t=2.6072$). That is pvalue = $P(t\geq 2.6072 ) = 0.0239$.
The pvalue is $0.0239$ which is $\textit{less than}$ the significance level of $\alpha = 0.05$, we $\textit{reject}$ the null hypothesis at $0.05$ level of significance.
Interpretation
There is sufficient evidence to support the claim that the newly designed keyboard reduce the number of minutes to type the term paper.
Paired ttest Example 3
Listed below are body temperatures (in $^oF$) of subjects measured at 8:00 AM and at 12:00 AM by a physician.
Is body temperature basically the same at both times? Use $\alpha = 0.05$.
At 8:00 AM : 97.0 96.2 97.6 96.4 97.8 99.2
At 12:00 AM : 98.0 98.6 98.8 98.0 98.6 97.6
Solution
Let $x$ denote the body temperature (in $^oF$) of subjects at 8:00 AM and $y$ denote the body temperature (in $^oF$) of subjects at 12:00 AM.
Because the two samples are dependent, we use paired $t$test for testing the hypothesis problem.
The sample size $n = 6$. Let $d=xy$.
$x$  $y$  $d$  $ddbar$  $(ddbar)^2$  

97  98  1  0.1  0.01  
96.2  98.6  2.4  1.5  2.25  
97.6  98.8  1.2  0.3  0.09  
96.4  98  1.6  0.7  0.49  
97.8  98.6  0.8  0.1  0.01  
99.2  97.6  1.6  2.5  6.25  
Total  5.4  9.1 
The sample mean of the difference is
$$ \begin{aligned} \overline{d}&= \frac{1}{n}\sum_{i=1}^n d_i\\ &=\frac{5.4}{6}\\ &=0.9 \end{aligned} $$
and the sample standard deviation of the difference is
$$ \begin{aligned} s_d&= \sqrt{\frac{1}{n1}\sum_{i=1}^n (d_i\overline{d})^2}\\ &=\sqrt{\frac{9.1}{5}}\\ &=1.3491 \end{aligned} $$
Step 1 Hypothesis
The hypothesis testing problem is
$H_0 : \mu_d = 0$ against $H_1 : \mu_d \neq 0$ ($\textit{twotailed}$)
Step 2 Test Statistic
The test statistic for testing $H_0$ against $H_1$ is
$$ \begin{aligned} t=\frac{\overline{d} \mu_d}{s_d/\sqrt{n}} \end{aligned} $$
Step 3 Level of Significance
The significance level is $\alpha = 0.05$.
Step 4 Critical Value(s)
As the alternative hypothesis is $\textit{twotailed}$, the critical value of $t$ for $5$ degrees of freedom and $\alpha = 0.05$ level of significance $\text{are}$ $\text{2.571 and 2.571}$.
The rejection region (i.e. critical region) is $\text{t < 2.571 or t > 2.571}$.
Step 5 Computation
The test statistic for testing above hypothesis testing problem under the null hypothesis is
$$ \begin{aligned} t&=\frac{\overline{d} \mu_d}{s_d/\sqrt{n}}\\ &= \frac{0.90}{1.3491/\sqrt{6}}\\ &= 1.6341 \end{aligned} $$
Step 6 Decision (Traditional approach)
The test statistic is $t =1.6341$ which falls $\textit{outside}$ the critical region, we $\textit{fail to reject}$ the null hypothesis.
OR
Step 6 Decison ($p$value approach)
The test is $\textit{twotailed}$ test, so pvalue is the area to the $\textit{extreme}$ of the test statistic ($t=1.6341$). That is pvalue = $2*P(t\geq 1.6341 ) = 0.1632$.
The pvalue is $0.1632$ which is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis at $0.05$ level of significance.
Interpretation
We conclude that the body temperature of the subjects are same at both the times.
Endnote
In this tutorial, you learned the about how to solve numerical examples on paired $t$test. You also learned about the step by step procedure to apply paired $t$test and how to use paired $t$test calculator to get the value of test statistic, pvalue, and zcritical value.
To learn more about other hypothesis testing problems, hypothesis testing calculators and step by step procedure, please refer to the following tutorials:
Let me know in the comments if you have any questions on paired $t$test calculator with examples and your thought on this article.