Paired t-test calculator dependent sample examples

Paired t-test 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. left-tailed, right-tailed or two-tailed.)

Paire t-test 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:
t-critical value(s):
p-value:

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 (left-tailed / right-tailed / two-tailed)

Step 4 - Click on "Calculate" button to get the result

Paired t-test 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 $d=x-y$.

$x$ $y$ $d$ $d-dbar$ $(d-dbar)^2$
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}{n-1}\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 $H_1 : \mu_d < 0$ ($\textit{left-tailed}$)

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 $\textit{left-tailed}$, 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{-1-0}{2/\sqrt{5}}\\ &= -1.118 \end{aligned}

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 $\textit{left-tailed}$ test, so p-value is the
area to the $\textit{left}$ of the test statistic ($t=-1.118$). That is p-value = $P(t\leq -1.118 ) = 0.1631$.

The p-value 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 t-test 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 $d=x-y$.

$x$ $y$ $d$ $d-dbar$ $(d-dbar)^2$
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}{n-1}\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 $H_1 : \mu_d > 0$ ($\textit{right-tailed}$)

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 $\textit{right-tailed}$, 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{1.8333-0}{1.7224/\sqrt{6}}\\ &= 2.6072 \end{aligned}

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 $\textit{right-tailed}$ test, so p-value is the
area to the $\textit{right}$ of the test statistic ($t=2.6072$). That is p-value = $P(t\geq 2.6072 ) = 0.0239$.

The p-value 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 t-test 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 $d=x-y$.

$x$ $y$ $d$ $d-dbar$ $(d-dbar)^2$
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}{n-1}\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 $H_1 : \mu_d \neq 0$ ($\textit{two-tailed}$)

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 $\textit{two-tailed}$, 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{-0.9-0}{1.3491/\sqrt{6}}\\ &= -1.6341 \end{aligned}

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 $\textit{two-tailed}$ test, so p-value is the area to the $\textit{extreme}$ of the test statistic ($t=-1.6341$). That is p-value = $2*P(t\geq 1.6341 ) = 0.1632$.

The p-value 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, p-value, and z-critical value.

Let me know in the comments if you have any questions on paired $t$-test calculator with examples and your thought on this article.