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

Chapter 8. Testing for Differences in Mean and Proportion: 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#:
\[\bar{D} = \cfrac{\sum{D}}{n}\phantom{00000} s_D = \sqrt{\cfrac{\sum(D - \bar{D})^2}{n-1}}\phantom{000}\text{or}\phantom{000}s_D = \sqrt{\cfrac{\sum{D^2}-\cfrac{(\sum{D})^2}{n}}{n-1}}\]
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 not 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 \leq \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 #11# 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	11
#X:\,\text{Before}#	#249#	#258#	#259#	#234#	#220#	#243#	#237#	#221#	#239#	#260#	#242#
#Y:\,\text{After}#	#249#	#253#	#252#	#238#	#216#	#242#	#234#	#215#	#240#	#255#	#237#

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=Y-X#.

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 = 0.05# significance level.

#p=0.021#

On the basis of this #p#-value, #H_0# should be rejected, because #\,p# #\lt# #\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=Y-X#:

City	1	2	3	4	5	6	7	8	9	10	11
#X:\,\text{Before}#	#249#	#258#	#259#	#234#	#220#	#243#	#237#	#221#	#239#	#260#	#242#
#Y:\,\text{After}#	#249#	#253#	#252#	#238#	#216#	#242#	#234#	#215#	#240#	#255#	#237#
#D:\,\text{Difference}#	#0#	#-5#	#-7#	#4#	#-4#	#-1#	#-3#	#-6#	#1#	#-5#	#-5#

Compute the mean of the difference scores #\bar{D}#:
\[\bar{D}=\cfrac{\displaystyle \sum D}{n}=\cfrac{(0) + (-5) + (-7) + (4) + (-4) + (-1) + (-3) + (-6) + (1) + (-5) + (-5)}{11}=-2.818182\]
Compute the standard deviation of the difference scores #s_{D}#:
\[
\displaystyle \sum D = (0) + (-5) + (-7) + (4) + (-4) + (-1) + (-3) + (-6) + (1) + (-5) + (-5) = -31
\\\phantom{0}\\
\displaystyle \sum D^2 = (0)^2 + (-5)^2 + (-7)^2 + (4)^2 + (-4)^2 + (-1)^2 + (-3)^2 + (-6)^2 + (1)^2 + (-5)^2 + (-5)^2 = 203
\\\phantom{0}\\
s_D = \displaystyle\sqrt{\cfrac{\sum D^2 - \cfrac{(\sum D)^2}{n}}{n-1}} = \displaystyle\sqrt{\cfrac{203 - \cfrac{(-31)^2}{11}}{11-1}} = 3.400535
\]
Compute the #t#-statistic:
\[t=\cfrac{\bar{D}}{s_D/\sqrt{n}}=\cfrac{-2.818182}{3.400535/\sqrt{11}} = -2.7486\]
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_{{10}}# 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:
\[
=2 \text{ * }(1 \text{ - } \text{T.DIST}(\text{ABS}(t),n \text{ - } 1,1))\\
\downarrow\\
=2 \text{ * }(1 \text{ - } \text{T.DIST}(\text{ABS}(\text{-}2.74864), 11 \text{ - } 1,1))
\]
This gives:
\[p = 0.021\]
Since #\,p# #\lt# #\alpha#, #H_0: \mu_D = 0# should be rejected.

R Calculation

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

City	1	2	3	4	5	6	7	8	9	10	11
#X:\,\text{Before}#	#249#	#258#	#259#	#234#	#220#	#243#	#237#	#221#	#239#	#260#	#242#
#Y:\,\text{After}#	#249#	#253#	#252#	#238#	#216#	#242#	#234#	#215#	#240#	#255#	#237#
#D:\,\text{Difference}#	#0#	#-5#	#-7#	#4#	#-4#	#-1#	#-3#	#-6#	#1#	#-5#	#-5#

has the #t_{n-1} = t_{{10}}# 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:
\[
2 \text{ * } \text{pt}(q = \text{abs}(t), df = n \text{ - } 1, lower.tail = \text{FALSE})\\
\downarrow\\
2\text{ * } \text{pt}(q = \text{abs}(\text{-}2.74864), df = 11 \text{ - } 1,lower.tail = \text{FALSE})
\]
This gives:
\[p = 0.021\]
Since #\,p# #\lt# #\alpha#, #H_0: \mu_D = 0# should be rejected.

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