Paired Samples t-test: Test Statistic and p-value

8. Testing for Differences in Means and Proportions: Paired Samples t-test

Paired Samples t-test: Test Statistic and p-value

Paired Samples t-test: Test Statistic

The test statistic of a paired samples $t$ -test is denoted $t$ .

To compute the $t$ -statistic, first compute the difference score $\boldsymbol{D}$ for each subject:
$D_i = X_i - Y_i \phantom{000}\text{or}\phantom{000} D_i = Y_i - X_i$

The resulting sample of $n$ difference scores will serve as the sample data for the hypothesis test.

Once the sample of difference scores has been constructed, calculate the sample mean $\boldsymbol{\bar{D}}$ and the sample standard deviation $\boldsymbol{s_D}$ for the sample of difference scores $^1$ :
$\bar{D} = \cfrac{\sum{D}}{n}\phantom{00000}s_D = \sqrt{\cfrac{\sum{D^2}-\cfrac{(\sum{D})^2}{n}}{n-1}}\phantom{000}\text{or}\phantom{000} s_D = \sqrt{\cfrac{\sum(D - \bar{D})^2}{n-1}}$
Once the statistics of the sample of difference scores have been calculated, the $t$ -statistic can be computed:
$t = \cfrac{\bar{D} - \mu_D}{s_{\bar{D}}} = \cfrac{\bar{D}}{s_D/\sqrt{n}}$

Under the null hypothesis of a paired samples $t$ -test, the $t$ -statistic follows a $t$ -distribution with $df = n - 1$ degrees of freedom.
$t \sim t_{n-1}$

Take note that, although it is true that $\bar{D} = \bar{X} - \bar{Y}$ , it is true that $s_D = s_X - s_Y$ .

Calculating the p-value of a Paired Samples t-test with Statistical Software

The calculation of the $p$ -value of a paired samples $t$ -test is dependent on the direction of the test and can be performed using either Excel or R.

To calculate the $p$ -value of a paired samples $t$ -test for $\mu_D$ in Excel, make use of one of the following commands:
$\begin{array}{llll} \phantom{0}\text{Direction}&\phantom{0000}H_0&\phantom{0000}H_a&\phantom{0000000000}\text{Excel Command}\\ \hline \text{Two-tailed}&H_0:\mu_D = 0&H_a:\mu_D \neq 0&=2 \text{ * }(1 \text{ - } \text{T.DIST}(\text{ABS}(t),n\text{ - }1,1))\\ \text{Left-tailed}&H_0:\mu_D \geq 0&H_a:\mu_D \lt 0&=\text{T.DIST}(t,n\text{ - }1,1)\\ \text{Right-tailed}&H_0:\mu_D \leq 0&H_a:\mu_D \gt 0&=1\text{ - }\text{T.DIST}(t,n\text{ - }1,1)\\ \end{array}$

To calculate the $p$ -value of a paired samples $t$ -test for $\mu_D$ in R, make use of one of the following commands:
$\begin{array}{llll} \phantom{0}\text{Direction}&\phantom{0000}H_0&\phantom{0000}H_a&\phantom{00000000000}\text{R Command}\\ \hline \text{Two-tailed}&H_0:\mu_D = 0&H_a:\mu_D \neq 0&2 \text{ * }\text{pt}(\text{abs}(t),n\text{ - }1,lower.tail=\text{FALSE})\\ \text{Left-tailed}&H_0:\mu_D \geq 0&H_a:\mu_D \lt 0&\text{pt}(t,n\text{ - }1, lower.tail=\text{TRUE})\\ \text{Right-tailed}&H_0:\mu_D \leq 0&H_a:\mu_D \gt 0&\text{pt}(t,n\text{ - }1, lower.tail=\text{FALSE})\\ \end{array}$

If $p \lt \alpha$ , reject $H_0$ and conclude $H_a$ . Otherwise, do not reject $H_0$ .

The government of Canada wants to know whether the legalization of marihuana has had any effect on the rate of drug-related offenses. To investigate this matter, a researcher selects a simple random sample of $10$ cities and compares the rates of drug-related offenses before $(X)$ and after $(Y)$ the legalization was implemented.

The values in the table below are the number of drug-related offenses per $100$ , $000$ residents:

City	1	2	3	4	5	6	7	8	9	10
$X:\,\text{Before}$	242	237	270	261	251	261	262	228	255	221
$Y:\,\text{After}$	237	221	267	251	254	256	265	218	257	226

You may assume that the population distributions of drug-related offenses both before and after the legalization are normal.

The researcher plans on using a paired samples $t$ -test to determine whether the legalization of marihuana has had a significant effect on the number of drug-related offenses.

Define $D=X-Y$ .

Calculate the $p$ -value of the test and make a decision regarding $H_0: \mu_D = 0$ . Round your answer to $3$ decimal places. Use the $\alpha = 0.09$ significance level.

$p=0.135$

On the basis of this $p$ -value, $H_0$ should not be rejected, because $\,p$ $\gt$ $\alpha$ .

There are a number of different ways we can calculate the $p$ -value of the test. Click on one of the panels to toggle a specific solution.

Excel Calculation

Compute the difference scores using $D=X-Y$ :

City	1	2	3	4	5	6	7	8	9	10
$X:\,\text{Before}$	242	237	270	261	251	261	262	228	255	221
$Y:\,\text{After}$	237	221	267	251	254	256	265	218	257	226
$D:\,\text{Difference}$	5	16	3	10	-3	5	-3	10	-2	-5

Compute the mean of the difference scores $\bar{D}$ :
$\bar{D}=\cfrac{\sum{D}}{n}=\cfrac{5+16+3+10-3+5-3+10-2-5}{10}=3.6$
Compute the standard deviation of the difference scores $s_{D}$ :
$\sum{D}=5+16+3+10-3+5-3+10-2-5=36$
$\begin{array}{rcl}\sum{D^2}&=&5^2+16^2+3^2+10^2+(-3)^2+5^2+(-3)^2+10^2+(-2)^2+(-5)^2\\&=&562\end{array}$
$s_{D}=\sqrt{\cfrac{\sum{D^2} - \cfrac{(\sum{D})^2}{n} }{n-1}}=\sqrt{\cfrac{562 - \cfrac{36^2}{10} }{10-1}}=6.9314$
Compute the $t$ -statistic:
$t = \cfrac{\bar{D}}{s_D/\sqrt{n}} = \cfrac{3.6}{6.9314/\sqrt{10}}=1.6424$
Assuming the population distributions of drug-related offenses are normal, we know that the test statistic
$t=\cfrac{\bar{D}}{s_D/\sqrt{n}}$

has the $t_{n-1} = t_{{9}}$ distribution, under the assumption that $H_0$ is true.

To calculate the $p$ -value of a $t$ -test, make use of the following Excel function:

T.DIST(x, deg_freedom, cumulative)

x: The value at which you wish to evaluate the distribution function.

deg_freedom: An integer indicating the number of degrees of freedom.

cumulative: A logical value that determines the form of the function.

TRUE - uses the cumulative distribution function, $\mathbb{P}(X \leq x)$

FALSE - uses the probability density function

Since we are dealing with a two-tailed $t$ -test, run the following command to calculate the $p$ -value:
$=2 \text{ * }(1 \text{ - } \text{T.DIST}(\text{ABS}(t),n \text{ - } 1,1))\\ \downarrow\\ =2 \text{ * }(1 \text{ - } \text{T.DIST}(\text{ABS}(1.6424), 10 \text{ - } 1,1))$
This gives:
$p = 0.135$
Since $\,p$ $\gt$ $\alpha$ , $H_0: \mu_D = 0$ should not be rejected.

R Calculation

Compute the difference scores using $D=X-Y$ :

City	1	2	3	4	5	6	7	8	9	10
$X:\,\text{Before}$	242	237	270	261	251	261	262	228	255	221
$Y:\,\text{After}$	237	221	267	251	254	256	265	218	257	226
$D:\,\text{Difference}$	5	16	3	10	-3	5	-3	10	-2	-5

has the $t_{n-1} = t_{{9}}$ distribution, under the assumption that $H_0$ is true.

To calculate the $p$ -value of a $t$ -test, make use of the following R function:

pt(q, df, lower.tail)

q: The value at which you wish to evaluate the distribution function.

df: An integer indicating the number of degrees of freedom.

lower.tail: If TRUE (default), probabilities are $\mathbb{P}(X \leq x)$ , otherwise, $\mathbb{P}(X \gt x)$ .

Since we are dealing with a two-tailed $t$ -test, run the following command to calculate the $p$ -value:
$2 \text{ * } \text{pt}(q = \text{abs}(t), df = n \text{ - } 1, lower.tail = \text{FALSE})\\ \downarrow\\ 2\text{ * } \text{pt}(q = \text{abs}(1.6424), df = 10 \text{ - } 1,lower.tail = \text{FALSE})$
This gives:
$p = 0.135$
Since $\,p$ $\gt$ $\alpha$ , $H_0: \mu_D = 0$ should not be rejected.

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