### 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.

\[t \sim t_{n-1}\]

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 #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}# | #270# | #246# | #233# | #257# | #245# | #225# | #267# | #261# | #230# | #221# |

#Y:\,\text{After}# | #271# | #246# | #231# | #260# | #239# | #230# | #260# | #256# | #234# | #217# |

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 = 0.05# significance level.

#p=0.443#

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.

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

City | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

#X:\,\text{Before}# | #270# | #246# | #233# | #257# | #245# | #225# | #267# | #261# | #230# | #221# |

#Y:\,\text{After}# | #271# | #246# | #231# | #260# | #239# | #230# | #260# | #256# | #234# | #217# |

#D:\,\text{Difference}# | #-1# | #0# | #2# | #-3# | #6# | #-5# | #7# | #5# | #-4# | #4# |

Compute the mean of the difference scores #\bar{D}#:

\[\bar{D}=\cfrac{\displaystyle \sum D}{n}=\cfrac{(-1) + (0) + (2) + (-3) + (6) + (-5) + (7) + (5) + (-4) + (4)}{10}=1.1\]

Compute the standard deviation of the difference scores #s_{D}#:

\[

\displaystyle \sum D = (-1) + (0) + (2) + (-3) + (6) + (-5) + (7) + (5) + (-4) + (4) = 11

\\\phantom{0}\\

\displaystyle \sum D^2 = (-1)^2 + (0)^2 + (2)^2 + (-3)^2 + (6)^2 + (-5)^2 + (7)^2 + (5)^2 + (-4)^2 + (4)^2 = 181

\\\phantom{0}\\

s_D = \displaystyle\sqrt{\cfrac{\sum D^2 - \cfrac{(\sum D)^2}{n}}{n-1}} = \displaystyle\sqrt{\cfrac{181 - \cfrac{(11)^2}{10}}{10-1}} = 4.332051

\]

Compute the #t#-statistic:

\[t=\cfrac{\bar{D}}{s_D/\sqrt{n}}=\cfrac{1.1}{4.332051/\sqrt{10}} = 0.803\]

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}(0.80297), 10 \text{ - } 1,1))

\]

This gives:

\[p = 0.443\]

Since #\,p# #\gt# #\alpha#, #H_0: \mu_D = 0# should not be rejected.

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

City | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

#X:\,\text{Before}# | #270# | #246# | #233# | #257# | #245# | #225# | #267# | #261# | #230# | #221# |

#Y:\,\text{After}# | #271# | #246# | #231# | #260# | #239# | #230# | #260# | #256# | #234# | #217# |

#D:\,\text{Difference}# | #-1# | #0# | #2# | #-3# | #6# | #-5# | #7# | #5# | #-4# | #4# |

Compute the mean of the difference scores #\bar{D}#:

\[\bar{D}=\cfrac{\displaystyle \sum D}{n}=\cfrac{(-1) + (0) + (2) + (-3) + (6) + (-5) + (7) + (5) + (-4) + (4)}{10}=1.1\]

Compute the standard deviation of the difference scores #s_{D}#:

\[

\displaystyle \sum D = (-1) + (0) + (2) + (-3) + (6) + (-5) + (7) + (5) + (-4) + (4) = 11

\\\phantom{0}\\

\displaystyle \sum D^2 = (-1)^2 + (0)^2 + (2)^2 + (-3)^2 + (6)^2 + (-5)^2 + (7)^2 + (5)^2 + (-4)^2 + (4)^2 = 181

\\\phantom{0}\\

s_D = \displaystyle\sqrt{\cfrac{\sum D^2 - \cfrac{(\sum D)^2}{n}}{n-1}} = \displaystyle\sqrt{\cfrac{181 - \cfrac{(11)^2}{10}}{10-1}} = 4.332051

\]

Compute the #t#-statistic:

\[t=\cfrac{\bar{D}}{s_D/\sqrt{n}}=\cfrac{1.1}{4.332051/\sqrt{10}} = 0.803\]

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 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}(0.80297), df = 10 \text{ - } 1,lower.tail = \text{FALSE})

\]

This gives:

\[p = 0.443\]

Since #\,p# #\gt# #\alpha#, #H_0: \mu_D = 0# should not be rejected.

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