Hypothesis Tests for Two Means: Paired-Sample T-Test

Conduct a paired-sample t-test for the mean difference between two populations when the population standard deviation of the differences is unknown. Compute the test statistic using the sample standard deviation, compare to critical values from the t-distribution with n-1 degrees of freedom, and reach a conclusion for one-tailed and two-tailed alternatives.

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Tutorial

Introduction

In a paired-sample t-test, we have nn paired observations (X1,Y1),,(Xn,Yn)(X_1, Y_1), \ldots, (X_n, Y_n) and we want to test a hypothesis about the population mean of the differences μd,\mu_d, but the population standard deviation σd\sigma_d is unknown.

The setup mirrors the paired-sample z-test, with two changes: we use the sample standard deviation sds_d in place of σd,\sigma_d, and we compare to a tt-distribution with n1n-1 degrees of freedom instead of the standard normal.

We form the differences Di=XiYiD_i = X_i - Y_i and compute

dˉ=1ni=1nDi,sd=1n1i=1n(Didˉ)2.\bar{d} = \dfrac{1}{n}\sum\limits_{i=1}^n D_i, \qquad s_d = \sqrt{\dfrac{1}{n-1}\sum\limits_{i=1}^n (D_i - \bar{d})^2}.

The test statistic for H0:μd=μd,0H_0: \mu_d = \mu_{d,0} is

t=dˉμd,0sd/n,t = \dfrac{\bar{d} - \mu_{d,0}}{s_d / \sqrt{n}},

which follows a tt-distribution with n1n-1 degrees of freedom under H0.H_0.

For instance, if n=4n=4 paired observations yield dˉ=5\bar{d} = 5 and sd=4,s_d = 4, then to test H0:μd=0H_0: \mu_d = 0 we have

t=504/4=52=2.5,df=3.t = \dfrac{5 - 0}{4/\sqrt{4}} = \dfrac{5}{2} = 2.5, \qquad \text{df} = 3.
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